tag:blogger.com,1999:blog-89171248410135637092024-02-20T06:05:23.025+05:30Solutions to I E Irodov - ElectrodynamicsElectromagnetism has had a profound effect on human civilization. Power generated from large electric generators is distributed around the world bringing light into our lives. Maxwell united electricity to magnetism and uncovered the true nature of light. He also arguably motivated Einstein into developing relativity. Hopefully solving Irodov's problems will deepen your appreciation of the beauty and power of Maxwell's equations.Unknownnoreply@blogger.comBlogger66125tag:blogger.com,1999:blog-8917124841013563709.post-72786098902672871202010-11-28T09:22:00.031+05:302010-11-28T11:44:31.726+05:30Irodov Problem 3.66<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiNJN05-BET5DyvJuNxAs0ImfP-T6aP4KPg-YBnHv3NYspJx4r82weCn7hP4scbaI9EIibsgvM2xhWvgudbeLgZKkltg_5sZzY4dqvVl6oiHlMThScfSgNdpGKMfcWBtVqS4wjhkqV9OQE/s1600/3_66_fig1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 400px; height: 266px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiNJN05-BET5DyvJuNxAs0ImfP-T6aP4KPg-YBnHv3NYspJx4r82weCn7hP4scbaI9EIibsgvM2xhWvgudbeLgZKkltg_5sZzY4dqvVl6oiHlMThScfSgNdpGKMfcWBtVqS4wjhkqV9OQE/s400/3_66_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5544443874300401378" border="0" /></a><span style="font-weight: bold;">(a)</span> Say, plates <span style="font-weight: bold;">2</span> an <span style="font-weight: bold;">3</span> are connected across the terminals of a battery as shown. The battery sucks out electrons from plate <span style="font-weight: bold;">2</span> making it positively charged and deposits them on plate <span style="font-weight: bold;">3</span> making it negatively charged. Since the total charge has to be conserved, the total charges on plates <span style="font-weight: bold;">2</span> and <span style="font-weight: bold;">3</span> must be equal and opposite. Further, all charge within the plates will be uniformly distributed since electrons repel each other and would like to be as far from each other as possible. So let the charges on plates <span style="font-weight: bold;">2</span> and <span style="font-weight: bold;">3</span> be <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_T-jBrKq_qGgduyGEu-LqpT40OX9kxodKXCB_nBLUdsDaGbG-qBiS9ul52p4DhhBO33_CTHRMjj5pErWS6AcxjyDDOWjx71VsYs-sSJe_4tmLoskuB9qf8cWItFIvjEj8udFwWfdsFnI/s1600/3_66_1.jpg"><img style="cursor: pointer; width: 13px; height: 18px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_T-jBrKq_qGgduyGEu-LqpT40OX9kxodKXCB_nBLUdsDaGbG-qBiS9ul52p4DhhBO33_CTHRMjj5pErWS6AcxjyDDOWjx71VsYs-sSJe_4tmLoskuB9qf8cWItFIvjEj8udFwWfdsFnI/s400/3_66_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5544448161556284962" border="0" /></a> and <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjq3f8FKOPIaQu4ondj_RMDvLhUVVvoBQZbY4PbsY1dPMge1dRUi7KG1JgTrnkWTkcjbqNMZQB1wm85lvEeEu-0pPupf-GRYznnopFyYqTfKSA8LdDRoOBc_9EdvtLYtJRJA0yxpY5fSYQ/s1600/3_66_2.jpg"><img style="cursor: pointer; width: 19px; height: 15px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjq3f8FKOPIaQu4ondj_RMDvLhUVVvoBQZbY4PbsY1dPMge1dRUi7KG1JgTrnkWTkcjbqNMZQB1wm85lvEeEu-0pPupf-GRYznnopFyYqTfKSA8LdDRoOBc_9EdvtLYtJRJA0yxpY5fSYQ/s400/3_66_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5544448261079156146" border="0" /></a> respectively.<br /><br />Since plates <span style="font-weight: bold;">1</span> and <span style="font-weight: bold;">4</span> are connected through a conducting wire, electrons are also free to flow between the plates. Consequently the total charge across plates <span style="font-weight: bold;">1</span> and <span style="font-weight: bold;">4</span> must also be conserved. Further, plates <span style="font-weight: bold;">1</span> and <span style="font-weight: bold;">4</span> must be at the same potential since they are connected by a wire. Hence, we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1t_uEt1oji0IKOf7oj_jL0y3hvvX_NeY812gUsFaIjqWt6uyUEXVCxTmlYooStQVNZJD9U9KHov8rAbo5vLBEqXjusensS9Ghykwl_si46x33Zz5VTk-L6OvYbZ5ZCAvZwJZQImDCotw/s1600/3_66_3.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 74px; height: 25px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1t_uEt1oji0IKOf7oj_jL0y3hvvX_NeY812gUsFaIjqWt6uyUEXVCxTmlYooStQVNZJD9U9KHov8rAbo5vLBEqXjusensS9Ghykwl_si46x33Zz5VTk-L6OvYbZ5ZCAvZwJZQImDCotw/s400/3_66_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5544449028241068114" border="0" /></a><br /><br />Since the potential difference between plates and <span style="font-weight: bold;">2</span> ans <span style="font-weight: bold;">3</span> is <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjurLilfowl-ocUZMUSJK-X6Zw7qUgFCRhMkkKOfHMwKWLeocNdz8PvNVjBzd17F4NcGJsI280dGBoudP8gUKJar0Jbi9xtpV0lX0d4BO93Y-OtM1-vd-fIgfOOh6pJ12p6SaY5c9CDbvk/s1600/3_66_4.jpg"><img style="cursor: pointer; width: 22px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjurLilfowl-ocUZMUSJK-X6Zw7qUgFCRhMkkKOfHMwKWLeocNdz8PvNVjBzd17F4NcGJsI280dGBoudP8gUKJar0Jbi9xtpV0lX0d4BO93Y-OtM1-vd-fIgfOOh6pJ12p6SaY5c9CDbvk/s400/3_66_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5544449441898551490" border="0" /></a>, thus,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiVR1IYt4ZuW2c381FyieXR-7fb2302r0FAAcFBSQeVHxoVJuigNB5xoBkmwstoPi0DWbs9kG-Ru_fGR0sVajr1MjjvEYjH_sHTaEjphUYlXteJ5XhML-hKn8W97Fjb5jMAgLmj3As7XHg/s1600/3_66_5.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 108px; height: 24px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiVR1IYt4ZuW2c381FyieXR-7fb2302r0FAAcFBSQeVHxoVJuigNB5xoBkmwstoPi0DWbs9kG-Ru_fGR0sVajr1MjjvEYjH_sHTaEjphUYlXteJ5XhML-hKn8W97Fjb5jMAgLmj3As7XHg/s400/3_66_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5544449801448605090" border="0" /></a><br /><br />Let us arbitrarily fix <span style="text-decoration: underline;"><br /><br /></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpLDSZbmtzfpXBYw3YTuYzYUncGHZys9jxcNC3ajvf0PgaseLDkrbyk645z0vGkmHQAHz93juTKDLSM-Dp6hDex9wsLhmEOzp4_69n23-liPA020VO19Ie-T0eNOG09iP3erAIGTtAMY8/s1600/3_66_6.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 64px; height: 23px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpLDSZbmtzfpXBYw3YTuYzYUncGHZys9jxcNC3ajvf0PgaseLDkrbyk645z0vGkmHQAHz93juTKDLSM-Dp6hDex9wsLhmEOzp4_69n23-liPA020VO19Ie-T0eNOG09iP3erAIGTtAMY8/s400/3_66_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5544450726330344002" border="0" /></a><br /><span style="text-decoration: underline;"><br /></span>From <span style="font-weight: bold;">(2)</span> and <span style="font-weight: bold;">(3)</span> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjZmwMNwm0RfJmlLzTl8uOpgrYu798bVMNtimxcwe6UbN0TSehxwZGH-tUknZdFA4plX8i_oZQvqYlnKEjjxWHC0aw0oppWH0VSZS8DDuHoj7-TMmlnL9S9HdG3a6tSKjVsdenvj1ci8MU/s1600/3_66_6.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 76px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjZmwMNwm0RfJmlLzTl8uOpgrYu798bVMNtimxcwe6UbN0TSehxwZGH-tUknZdFA4plX8i_oZQvqYlnKEjjxWHC0aw0oppWH0VSZS8DDuHoj7-TMmlnL9S9HdG3a6tSKjVsdenvj1ci8MU/s400/3_66_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5544451043706286130" border="0" /></a><br /><br />Since, the potentials on plates <span style="font-weight: bold;">1</span> and <span style="font-weight: bold;">4</span> are the same, this must imply that <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjL1FJiy9d-qGPQ62hn4aNUskjEk95B2pgfinEiUoDJ4eCnqBuoEZzkeFFnmbSagvmyGesNcfEhTiKkcBgGLhXqp60keoL-Q-tXg9JJ9DEgPn_XpjI-RaNXcKnAak54n9ki0wC9wlGlGm0/s1600/3_66_7.jpg"><img style="cursor: pointer; width: 48px; height: 25px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjL1FJiy9d-qGPQ62hn4aNUskjEk95B2pgfinEiUoDJ4eCnqBuoEZzkeFFnmbSagvmyGesNcfEhTiKkcBgGLhXqp60keoL-Q-tXg9JJ9DEgPn_XpjI-RaNXcKnAak54n9ki0wC9wlGlGm0/s400/3_66_7.jpg" alt="" id="BLOGGER_PHOTO_ID_5544451960933021282" border="0" /></a> and <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJi07rTNyrXpxcqktHw4JlZ7a87XG3148JH82YMnWBfoeeitn3lggmC7rJU8oKrhXmICI1DkJUBmmdewO1HhqsWepmHRyRLVADebHu8-cMdFkj_R4iA0UMvdJEPXeP43CxXBfm7LeV5vk/s1600/3_66_8.jpg"><img style="cursor: pointer; width: 48px; height: 25px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJi07rTNyrXpxcqktHw4JlZ7a87XG3148JH82YMnWBfoeeitn3lggmC7rJU8oKrhXmICI1DkJUBmmdewO1HhqsWepmHRyRLVADebHu8-cMdFkj_R4iA0UMvdJEPXeP43CxXBfm7LeV5vk/s400/3_66_8.jpg" alt="" id="BLOGGER_PHOTO_ID_5544452306845084562" border="0" /></a>. This means that the direction of electric <span style="font-weight: bold;">E12</span> will be from plate <span style="font-weight: bold;">2</span> towards plate <span style="font-weight: bold;">1</span> and that of <span style="font-weight: bold;">E34</span> will be from plate <span style="font-weight: bold;">4</span> towards plate <span style="font-weight: bold;">3 </span>as shown in the figure.<br /><br />Now let us put a Gaussian cylinder of area <span style="font-weight: bold;">A</span> as shown in <span style="font-weight: bold;">Figure 2</span>.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjLIYuYB8SGvPRDXX6t3n0c6v_Vvm5APwi_DqKcOHn8dirUsNXnjcSibQia_aA_sO2FHKiE3Pvi6S4aVAFnJdHRdNa_OQmsWD4_rU4ebqONsSzYiVjWDMCA7KoHqha08rkRl5-J0ZbIigg/s1600/3_66_fig2.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 400px; height: 226px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjLIYuYB8SGvPRDXX6t3n0c6v_Vvm5APwi_DqKcOHn8dirUsNXnjcSibQia_aA_sO2FHKiE3Pvi6S4aVAFnJdHRdNa_OQmsWD4_rU4ebqONsSzYiVjWDMCA7KoHqha08rkRl5-J0ZbIigg/s400/3_66_fig2.jpg" alt="" id="BLOGGER_PHOTO_ID_5544455883519978946" border="0" /></a><br /><br />The total electric flux through the cylinder will be <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiKKdcIZnAXqhTxJwEWY7kkfu7ATHvISYgKl2q2ocCNV6-B_6KecEWM0gHyPF2iZCoUKKtF2MGCQ83mLPZoBmYJzSub8e3JS6ScTLEXDU4_LITzFxaMntJ9bOS0TcaETYDzdCYKdLRFP0k/s1600/3_66_9.jpg"><img style="cursor: pointer; width: 73px; height: 22px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiKKdcIZnAXqhTxJwEWY7kkfu7ATHvISYgKl2q2ocCNV6-B_6KecEWM0gHyPF2iZCoUKKtF2MGCQ83mLPZoBmYJzSub8e3JS6ScTLEXDU4_LITzFxaMntJ9bOS0TcaETYDzdCYKdLRFP0k/s400/3_66_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5544456606536006802" border="0" /></a>. The total charge contained in the Gaussian cylinder is <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipqQpFXK2A-1YM6JRkwTP51s8i0-laBv7uAy_bV1A5HqRTW-QSx6sVvp9Lzq3cmoj6AyGUTxSZnwtpPEb894tWz0yqsWtLIfvF9lL_jLeL5WuMXomu0oCLR0KvEP2ZpM_eMfJHB1zsoCE/s1600/3_66_10.jpg"><img style="cursor: pointer; width: 71px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipqQpFXK2A-1YM6JRkwTP51s8i0-laBv7uAy_bV1A5HqRTW-QSx6sVvp9Lzq3cmoj6AyGUTxSZnwtpPEb894tWz0yqsWtLIfvF9lL_jLeL5WuMXomu0oCLR0KvEP2ZpM_eMfJHB1zsoCE/s400/3_66_10.jpg" alt="" id="BLOGGER_PHOTO_ID_5544456996403879074" border="0" /></a>. From Gauss law we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgTRoceNP5hCHrNB-BsTVYo9qoacD31qOVg5oTNa0mjwksFUfU8yvLc-8g-XxzaJ2DsI80OJ3w3QmJXyZIHtIOfA9Z57YPtGaMq768VzO_Fyska-VAsEBFA21LruAu1cHAxD3H5WjbsJR4/s1600/3_66_11.jpg"><img style="cursor: pointer; width: 121px; height: 55px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgTRoceNP5hCHrNB-BsTVYo9qoacD31qOVg5oTNa0mjwksFUfU8yvLc-8g-XxzaJ2DsI80OJ3w3QmJXyZIHtIOfA9Z57YPtGaMq768VzO_Fyska-VAsEBFA21LruAu1cHAxD3H5WjbsJR4/s400/3_66_11.jpg" alt="" id="BLOGGER_PHOTO_ID_5544457589459508274" border="0" /></a><br /><br />The potential at plate <span style="font-weight: bold;">1</span> will be <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgm6QKq4LdT2hgy39pBiwvPeuFyu7Ozo6ciVAdZ6mp0kxG0isFLnKEC22YwaUXQZ2JSag9i9WDotYUCUrL8w0TY3Sz7AGSSlcJB7p5BW1h9GvpXVWhDV6J_akSdCAxmQmYs0mJtwMPuhuI/s1600/3_66_12.jpg"><img style="cursor: pointer; width: 129px; height: 23px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgm6QKq4LdT2hgy39pBiwvPeuFyu7Ozo6ciVAdZ6mp0kxG0isFLnKEC22YwaUXQZ2JSag9i9WDotYUCUrL8w0TY3Sz7AGSSlcJB7p5BW1h9GvpXVWhDV6J_akSdCAxmQmYs0mJtwMPuhuI/s400/3_66_12.jpg" alt="" id="BLOGGER_PHOTO_ID_5544458711740737282" border="0" /></a> and that of plate <span style="font-weight: bold;">2</span> will be <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhbP8_xFnuYw4w2O9yzikB6OR06jOQWgHU4Onbvago2ZKiXMHy1xnu4tycCkPrCh7SALINPtPECb9xGSIBcvDLXk87ueeLbLKGhA_AKyQrthwuOg7pYjgrpsuyG4WLuQPu2kWOMSOBWOJM/s1600/3_66_13.jpg"><img style="cursor: pointer; width: 143px; height: 23px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhbP8_xFnuYw4w2O9yzikB6OR06jOQWgHU4Onbvago2ZKiXMHy1xnu4tycCkPrCh7SALINPtPECb9xGSIBcvDLXk87ueeLbLKGhA_AKyQrthwuOg7pYjgrpsuyG4WLuQPu2kWOMSOBWOJM/s400/3_66_13.jpg" alt="" id="BLOGGER_PHOTO_ID_5544459186372996642" border="0" /></a>. Since both these potentials must be the same <span style="font-weight: bold;">(1)</span>, we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhryD_Mjt6yKKL_0kIv5aNUc2-Md2X_TcKYOttJpxEPzbkAXbFU2lfeF9sieDuctKUmOdlXKZZ0Xvk-w8_IPJW3DzbMojcTMRyomk6AjU8kYSeUmOPJ1Up6xnV08x8dEcXMmLZcHWFoBf4/s1600/3_66_14.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 126px; height: 73px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhryD_Mjt6yKKL_0kIv5aNUc2-Md2X_TcKYOttJpxEPzbkAXbFU2lfeF9sieDuctKUmOdlXKZZ0Xvk-w8_IPJW3DzbMojcTMRyomk6AjU8kYSeUmOPJ1Up6xnV08x8dEcXMmLZcHWFoBf4/s400/3_66_14.jpg" alt="" id="BLOGGER_PHOTO_ID_5544460027412914498" border="0" /></a><br /><br /><br /><br /><br />Similarly, since the potential between plates <span style="font-weight: bold;">2</span> and <span style="font-weight: bold;">3</span> is <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjurLilfowl-ocUZMUSJK-X6Zw7qUgFCRhMkkKOfHMwKWLeocNdz8PvNVjBzd17F4NcGJsI280dGBoudP8gUKJar0Jbi9xtpV0lX0d4BO93Y-OtM1-vd-fIgfOOh6pJ12p6SaY5c9CDbvk/s1600/3_66_4.jpg"><img style="cursor: pointer; width: 22px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjurLilfowl-ocUZMUSJK-X6Zw7qUgFCRhMkkKOfHMwKWLeocNdz8PvNVjBzd17F4NcGJsI280dGBoudP8gUKJar0Jbi9xtpV0lX0d4BO93Y-OtM1-vd-fIgfOOh6pJ12p6SaY5c9CDbvk/s400/3_66_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5544449441898551490" border="0" /></a> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8sxTglPYMXqLE-2zOnQ36pPnpiKbDiOvBUd0cqNeTKkc8jm_-UYWai3t4XNgd-E_dbUoS_epdP5ZErTmNI0z7Fg8vDGvtTwfh-vzm_1GQvdTTqfJ7GnrM-ub_Yqj0UUcP-MxDzeIP1oc/s1600/3_66_15.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 92px; height: 65px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8sxTglPYMXqLE-2zOnQ36pPnpiKbDiOvBUd0cqNeTKkc8jm_-UYWai3t4XNgd-E_dbUoS_epdP5ZErTmNI0z7Fg8vDGvtTwfh-vzm_1GQvdTTqfJ7GnrM-ub_Yqj0UUcP-MxDzeIP1oc/s400/3_66_15.jpg" alt="" id="BLOGGER_PHOTO_ID_5544461491046402226" border="0" /></a><br /><br /><br /><br /><span style="font-weight: bold;">(b)</span> The charge densities of plates <span style="font-weight: bold;">1</span> and <span style="font-weight: bold;">4</span> will be equal and opposite as the total charge across them is conserved. Let these charges by <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgFuPnmYMD3uJTwb9kb-zNJTsGvfMWFzTVmAc6GMVnlFA4RFoR5trJiYybWh_hv5T1bMKlTWPoM8LZJt5582vhqVz5EWbcdg7vBJzpzXGEnZHGjfCLbhGoGkc__WZBzcWZfxFy-1EnW950/s1600/3_66_16.jpg"><img style="cursor: pointer; width: 15px; height: 19px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgFuPnmYMD3uJTwb9kb-zNJTsGvfMWFzTVmAc6GMVnlFA4RFoR5trJiYybWh_hv5T1bMKlTWPoM8LZJt5582vhqVz5EWbcdg7vBJzpzXGEnZHGjfCLbhGoGkc__WZBzcWZfxFy-1EnW950/s400/3_66_16.jpg" alt="" id="BLOGGER_PHOTO_ID_5544468508544763346" border="0" /></a> and <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiHOL0Zo-99Y9tEgJ7XjB4czMHGSfju6R__3fJaKQeyFSvb10ZNzaOyCcP6zw8Cc9SVFy6WMKamrJmffdN51PLnb4OI7GSPmOWsQy9Q_Zifo79g87yx3UtrFUZVzPolQmgNcmnkij9CxzM/s1600/3_66_17.jpg"><img style="cursor: pointer; width: 19px; height: 18px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiHOL0Zo-99Y9tEgJ7XjB4czMHGSfju6R__3fJaKQeyFSvb10ZNzaOyCcP6zw8Cc9SVFy6WMKamrJmffdN51PLnb4OI7GSPmOWsQy9Q_Zifo79g87yx3UtrFUZVzPolQmgNcmnkij9CxzM/s400/3_66_17.jpg" alt="" id="BLOGGER_PHOTO_ID_5544468712459758114" border="0" /></a> respectively. Let us first find the electric field due to a charged plate with surface charge density <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_T-jBrKq_qGgduyGEu-LqpT40OX9kxodKXCB_nBLUdsDaGbG-qBiS9ul52p4DhhBO33_CTHRMjj5pErWS6AcxjyDDOWjx71VsYs-sSJe_4tmLoskuB9qf8cWItFIvjEj8udFwWfdsFnI/s1600/3_66_1.jpg"><img style="cursor: pointer; width: 13px; height: 18px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_T-jBrKq_qGgduyGEu-LqpT40OX9kxodKXCB_nBLUdsDaGbG-qBiS9ul52p4DhhBO33_CTHRMjj5pErWS6AcxjyDDOWjx71VsYs-sSJe_4tmLoskuB9qf8cWItFIvjEj8udFwWfdsFnI/s400/3_66_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5544448161556284962" border="0" /></a>.<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj7gkhS6bcva7C22m8QaHwQsGRGe0QRXFRYi0xMfEd9bWqsdAZTV4Q4RSw_FpbNIoG8fMD-oCp6dRaY6Snm6ljnj9eOeFW3ZAI_o4TpidA3Ktt1mV_s9bofgP5DOPKyNiJ57Dgrb2ePGYM/s1600/3_66_18.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 182px; height: 294px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj7gkhS6bcva7C22m8QaHwQsGRGe0QRXFRYi0xMfEd9bWqsdAZTV4Q4RSw_FpbNIoG8fMD-oCp6dRaY6Snm6ljnj9eOeFW3ZAI_o4TpidA3Ktt1mV_s9bofgP5DOPKyNiJ57Dgrb2ePGYM/s400/3_66_18.jpg" alt="" id="BLOGGER_PHOTO_ID_5544470051083824530" border="0" /></a><br /><br />Let us consider a Gaussian cylinder across the charged conductor with area <span style="font-weight: bold;">A</span>. Let the electric field be <span style="font-weight: bold;">E</span> due to the charged conductor. The total electric flux through the cylinder will be <span style="font-weight: bold;">2EA</span>. The total charge contained within this cylinder is <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgk1AZpTBDoye7dv1SC2eM1QyBoSuBPfiHj0D2zV09Z6NDmAqgIR6TRMpditXDY8Nl_0CMaFr8q-I-3h8yDWs8mHdGnusvXPJCGHJM9cFhItK9CdKT8Dudfgpm5jpE3tRMMhkm9_Joi0HM/s1600/3_66_19.jpg"><img style="cursor: pointer; width: 21px; height: 18px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgk1AZpTBDoye7dv1SC2eM1QyBoSuBPfiHj0D2zV09Z6NDmAqgIR6TRMpditXDY8Nl_0CMaFr8q-I-3h8yDWs8mHdGnusvXPJCGHJM9cFhItK9CdKT8Dudfgpm5jpE3tRMMhkm9_Joi0HM/s400/3_66_19.jpg" alt="" id="BLOGGER_PHOTO_ID_5544471121305340226" border="0" /></a>. Hence, from Gauss Law we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdyMJcOcKzSW1BdhUvHam10ajRMXPmOmBsgprr8p9EuztsfEslBhQbjxbNccvmdfbeki2DEwlT_Ho2zvpkN6dRVRXTrDG-gQJWQwXwO1PNtcr33AdJ1joLo7jXzC19gCKzadbKCCue1zw/s1600/3_66_20.jpg"><img style="cursor: pointer; width: 105px; height: 56px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdyMJcOcKzSW1BdhUvHam10ajRMXPmOmBsgprr8p9EuztsfEslBhQbjxbNccvmdfbeki2DEwlT_Ho2zvpkN6dRVRXTrDG-gQJWQwXwO1PNtcr33AdJ1joLo7jXzC19gCKzadbKCCue1zw/s400/3_66_20.jpg" alt="" id="BLOGGER_PHOTO_ID_5544471581384816882" border="0" /></a><br /><br />For our problem then, the net electric filed is the superposition of the electric fields from all the plates. This is illustrated in <span style="font-weight: bold;">Figure 4</span>.<br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPLXG_XOqXzFfURVq3X1aNrtzPgj6r2FzPtxWu2TkUZm5i68WcK-funN2j0XQrSyCGCXmyXfvIMDAFwIinTg_EeFtRThD5Z41r13MbJcEdkCDrLZHcIwtZ-8W-1cPc3R77HdAkZVtprIE/s1600/3_66_fig4.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 400px; height: 219px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPLXG_XOqXzFfURVq3X1aNrtzPgj6r2FzPtxWu2TkUZm5i68WcK-funN2j0XQrSyCGCXmyXfvIMDAFwIinTg_EeFtRThD5Z41r13MbJcEdkCDrLZHcIwtZ-8W-1cPc3R77HdAkZVtprIE/s400/3_66_fig4.jpg" alt="" id="BLOGGER_PHOTO_ID_5544477149431913394" border="0" /></a><br />As seen from <span style="font-weight: bold;">Figure 4</span>,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj4y2Djfq2hCl0a6xDip9IZ69uqSMHgQ0Evyo28ZavdKKZbq7X-Qvju5O8VlgwEF1bWIdVrdNAshsrKI7Wj_InEJEeAZKo8Mlm6NCwadTGsjiDnaWoBDnQHZm0MqlLniDYFruNWOEdmfsI/s1600/3_66_21.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 174px; height: 94px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj4y2Djfq2hCl0a6xDip9IZ69uqSMHgQ0Evyo28ZavdKKZbq7X-Qvju5O8VlgwEF1bWIdVrdNAshsrKI7Wj_InEJEeAZKo8Mlm6NCwadTGsjiDnaWoBDnQHZm0MqlLniDYFruNWOEdmfsI/s400/3_66_21.jpg" alt="" id="BLOGGER_PHOTO_ID_5544478707579618946" border="0" /></a><br /><br /><br /><br /><br /><br />Also we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBBAO94uP7uVbcd4gJm3lapRoYdtNWyUISOOvNZksfwYoKPCkwDQte8DNh_CV-nz8CO3_IUGXSUgBe3Jdib_sl5TVWQcz4y5DPDXmdRumoRQ_iFchxRrRbgn1iWboMHsgoy1-NAdtaCEs/s1600/3_66_22.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 179px; height: 99px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBBAO94uP7uVbcd4gJm3lapRoYdtNWyUISOOvNZksfwYoKPCkwDQte8DNh_CV-nz8CO3_IUGXSUgBe3Jdib_sl5TVWQcz4y5DPDXmdRumoRQ_iFchxRrRbgn1iWboMHsgoy1-NAdtaCEs/s400/3_66_22.jpg" alt="" id="BLOGGER_PHOTO_ID_5544479967696717234" border="0" /></a>Unknownnoreply@blogger.com6tag:blogger.com,1999:blog-8917124841013563709.post-40741320196282708722010-11-27T18:01:00.012+05:302010-11-27T19:26:45.960+05:30Irodov Problem 3.65<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVkMYhyphenhyphenvs2BGWjKq_ey_mwLgDcu1Tb7ggqHhyCE6UepWRTEI24rJGUFYmQxMnJLssJJh31v9-nk4eK1_k_VilPukK_KznJeIIK5DhtBuplAiTtCiY_2sz536w0gJmHTPpvDVbawqyyCLE/s1600/3_65_fig1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 323px; height: 343px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVkMYhyphenhyphenvs2BGWjKq_ey_mwLgDcu1Tb7ggqHhyCE6UepWRTEI24rJGUFYmQxMnJLssJJh31v9-nk4eK1_k_VilPukK_KznJeIIK5DhtBuplAiTtCiY_2sz536w0gJmHTPpvDVbawqyyCLE/s400/3_65_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5544227929722156354" border="0" /></a>Since electrons are free to move about within a conductor, and two electrons repel from each other, they will tend to spread themselves uniformly across the surface of the conductor. Given the spherical symmetry of the problem, the spherical shells will have a uniform charge density. Thus, the surface charge densities of the spheres will be<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi-FDcDOYDMxWvzojbClfqlQyCAcI2DpAPSveDV4EQYSatMxVITcajIaVMN-klWPmO9YZn5tRe-JaseLgBxFRCwlqgfHRbIMJq6OcnmmmI3LaKjxJXXiU8sSeMKZUQcw9Kv8eRtcxgbA_Q/s1600/3_65_1.jpg"><img style="cursor: pointer; width: 106px; height: 90px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi-FDcDOYDMxWvzojbClfqlQyCAcI2DpAPSveDV4EQYSatMxVITcajIaVMN-klWPmO9YZn5tRe-JaseLgBxFRCwlqgfHRbIMJq6OcnmmmI3LaKjxJXXiU8sSeMKZUQcw9Kv8eRtcxgbA_Q/s400/3_65_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5544207970778419122" border="0" /></a><br /><br />The electric potential at the center of the sphere <span style="font-weight: bold;">O</span>, will be due to each of the two spheres, equal to,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZS7JYJa8cDm07sDVG2Ohb1DfM3KmoA_OqmeN0eoeJ7FNfdqFNuA46naNQSdZk-0Rnso6HEUcCv-gz7lsQByXfFmfDTGSlbJcRGvsruA2j0TPOXwEschRvvErslxuk9uUofGcxqHTZntw/s1600/3_65_2.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 195px; height: 104px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZS7JYJa8cDm07sDVG2Ohb1DfM3KmoA_OqmeN0eoeJ7FNfdqFNuA46naNQSdZk-0Rnso6HEUcCv-gz7lsQByXfFmfDTGSlbJcRGvsruA2j0TPOXwEschRvvErslxuk9uUofGcxqHTZntw/s400/3_65_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5544210496340397138" border="0" /></a><br /><br /><br /><br /><br /><br /><br />If this potential <span style="font-weight: bold;">Vo</span> is zero then from <span style="font-weight: bold;">(2)</span> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_5cCUAnWQia0XGYMDV1OSmgN5n0wwJxEPN3A2VcBuScSJnmdur9QriE61wqEycLqNEnsW-Phpkjk-eRsm88DY3-XzDCYs_QAvAJBfQ2Jov8QNt-eU3a3BVXadGl91PCmT1k4n1Qtw5JM/s1600/3_65_3.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 89px; height: 39px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_5cCUAnWQia0XGYMDV1OSmgN5n0wwJxEPN3A2VcBuScSJnmdur9QriE61wqEycLqNEnsW-Phpkjk-eRsm88DY3-XzDCYs_QAvAJBfQ2Jov8QNt-eU3a3BVXadGl91PCmT1k4n1Qtw5JM/s400/3_65_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5544210999546962354" border="0" /></a><br /><br /><br />Consider a spherical Gaussian surface within the innermost sphere (<span style="font-weight: bold;">G1 </span>as shown in the figure). Since there is no charge within this sphere, there must be no electric flux through <span style="font-weight: bold;">G1</span>, in other words the net electric filed inside the sphere will be zero at all points.<br /><br />Consider another spherical Gaussian surface (<span style="font-weight: bold;">G2</span> as shown in the figure) with radius <span style="font-weight: bold;">r</span> such that <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhzitBSgRB2rXoo6ZzAK6trGtaE-63otRTOsOxONLZTIkauZQP6Ra1jc3vLaJ5ZK8WEA56Ysr76FBaZNRplJVp9Bf-6fVB8NDdAkEe97RPiz1YhuDf_q53BiQo8Oikmxo8xxK-NrugvCN4/s1600/3_65_9.jpg"><img style="cursor: pointer; width: 60px; height: 18px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhzitBSgRB2rXoo6ZzAK6trGtaE-63otRTOsOxONLZTIkauZQP6Ra1jc3vLaJ5ZK8WEA56Ysr76FBaZNRplJVp9Bf-6fVB8NDdAkEe97RPiz1YhuDf_q53BiQo8Oikmxo8xxK-NrugvCN4/s400/3_65_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5544226396719964882" border="0" /></a><span style="font-weight: bold;">. </span><span>Since the net charge inside this sphere is <span style="font-weight: bold;">q1</span>, the electric field in this region can be determined using Gauss law and is given by,</span><span style="font-weight: bold;"><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEit2yeIZXhZ6snmXudkk8hZ2wf1vKjZed8xwM0DyOHFfsP1YkqR7eyXjqh6X5RrP5DgFXB4vxrVIy9BcS_PTe0uO0lDd4p4ylSEWj5zsGd7RwfzX6JB6czdvJPKfXOPbAVtHCZWopD0jA4/s1600/3_65_4.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 208px; height: 38px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEit2yeIZXhZ6snmXudkk8hZ2wf1vKjZed8xwM0DyOHFfsP1YkqR7eyXjqh6X5RrP5DgFXB4vxrVIy9BcS_PTe0uO0lDd4p4ylSEWj5zsGd7RwfzX6JB6czdvJPKfXOPbAVtHCZWopD0jA4/s400/3_65_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5544213921305343618" border="0" /></a><br /><br /><br /></span><span>Now consider another spherical Gaussian surface (<span style="font-weight: bold;">G3</span> as shown in the figure) with radius <span style="font-weight: bold;">r</span> such that</span><span style="font-weight: bold;"> </span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgbP1JKKxfNH92zKOAJOAf2Q2gO7KDqv10QFzGX0JW5ZP4HbQOa-Bux07fEJ7BseNlESp3xMIQJaCQV-CsJ8pHMkMdrBlK-zw7Vwm5H0JHJvPt1SebcyPm7uEr-9BZtmNkDepeGA-l5gss/s1600/3_65_10.jpg"><img style="cursor: pointer; width: 32px; height: 18px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgbP1JKKxfNH92zKOAJOAf2Q2gO7KDqv10QFzGX0JW5ZP4HbQOa-Bux07fEJ7BseNlESp3xMIQJaCQV-CsJ8pHMkMdrBlK-zw7Vwm5H0JHJvPt1SebcyPm7uEr-9BZtmNkDepeGA-l5gss/s400/3_65_10.jpg" alt="" id="BLOGGER_PHOTO_ID_5544226748597787314" border="0" /></a><span>.</span><span style="font-weight: bold;"> </span><span>The net charge within this Gaussian surface is</span><span style="font-weight: bold;"> <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh7J8JLlY5QU33gQcMm_1gMhtki6WBUjr2OHPWTgcNkN8bNyKzebKDCj82NsWIq-d3Bip7bVYKq6irYmtE614B0cc0WOPFdTFJpwt4hc8-lWMf64Mr2aJ6g1q2PBW51WCs0Vcn-Xrg6zlI/s1600/3_65_5.jpg"><img style="cursor: pointer; width: 76px; height: 18px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh7J8JLlY5QU33gQcMm_1gMhtki6WBUjr2OHPWTgcNkN8bNyKzebKDCj82NsWIq-d3Bip7bVYKq6irYmtE614B0cc0WOPFdTFJpwt4hc8-lWMf64Mr2aJ6g1q2PBW51WCs0Vcn-Xrg6zlI/s400/3_65_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5544215265965579746" border="0" /></a></span><span>. Hence, from Gauss law, the electric field in this region is given by,</span><span style="font-weight: bold;"><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg46wPqk2WgJ_kB1V5qZMu9Wu3200FYE5w0oUhRuRIkZ0vjeTGANVG9tfxY6IZbqFeqh51Wp9MpAEL7ilgh3pr8iKBjh5sSycf1mnW-5z0hbaxsiVA7sAEw6wsnWCcEhE5gNfoTkC5K9VE/s1600/3_65_6.jpg"><img style="cursor: pointer; width: 202px; height: 43px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg46wPqk2WgJ_kB1V5qZMu9Wu3200FYE5w0oUhRuRIkZ0vjeTGANVG9tfxY6IZbqFeqh51Wp9MpAEL7ilgh3pr8iKBjh5sSycf1mnW-5z0hbaxsiVA7sAEw6wsnWCcEhE5gNfoTkC5K9VE/s400/3_65_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5544216315843397282" border="0" /></a><br /><br /></span><span>The electric potential <span style="font-weight: bold;">V(r)</span> as a function of distance is given by,<br /></span><span style="font-weight: bold;"><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_vwkz2LwR96DrJsyajib6H1ODXW8qCC-dNbhyphenhyphentTCc3r67r7puiEhkkYrZH4IVYhSB2y4-PXTMU3kDnShKpv52JjEXXlgf0wriuQ9wUmAGm2FLwqi9a3BwQ5r7cQGpHPKd-I6e72UvlqI/s1600/3_65_7.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 142px; height: 39px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_vwkz2LwR96DrJsyajib6H1ODXW8qCC-dNbhyphenhyphentTCc3r67r7puiEhkkYrZH4IVYhSB2y4-PXTMU3kDnShKpv52JjEXXlgf0wriuQ9wUmAGm2FLwqi9a3BwQ5r7cQGpHPKd-I6e72UvlqI/s400/3_65_7.jpg" alt="" id="BLOGGER_PHOTO_ID_5544217428928909266" border="0" /></a><br /><br /><br /></span><span>Using <span style="font-weight: bold;">(4a)</span> and <span style="font-weight: bold;">(4b)</span> we have,</span><span style="font-weight: bold;"><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjejTsBm8DQlmA7B6T57Y_sQMMtPH5acpKy-qVvKSEl6lRPk107dWI9QSK1FiFowpbwuMJ9TSUdWEu0-Tw5l4xufKbK429mbNMe8C-LN-QxRzXmtVqfSkqDzkGbzpkF2VtEcsu5Qs_49iQ/s1600/3_65_8.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 385px; height: 279px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjejTsBm8DQlmA7B6T57Y_sQMMtPH5acpKy-qVvKSEl6lRPk107dWI9QSK1FiFowpbwuMJ9TSUdWEu0-Tw5l4xufKbK429mbNMe8C-LN-QxRzXmtVqfSkqDzkGbzpkF2VtEcsu5Qs_49iQ/s400/3_65_8.jpg" alt="" id="BLOGGER_PHOTO_ID_5544225823495120290" border="0" /></a><br /></span>Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-8917124841013563709.post-65302384570303880612010-11-22T08:29:00.017+05:302010-11-23T07:56:40.856+05:30Irodov Problem 3.64<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2Zm_Y0XuynK3sC-s6rX-aKByutx9EjYywz1Lnz0pqa7BunUFygKHEgZl6jQcHHcjy9tTRSyUQJH6ERX4dwaPOE1Dtqhq0WeHE1i7glM9jV4sKoXozcOPKvqsA50w1x37DMsmyfFdpuTY/s1600/3_64_fig1.jpg"><img style="margin: 0px 10px 10px 0px; width: 400px; float: left; height: 227px;" id="BLOGGER_PHOTO_ID_5542203521770284386" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2Zm_Y0XuynK3sC-s6rX-aKByutx9EjYywz1Lnz0pqa7BunUFygKHEgZl6jQcHHcjy9tTRSyUQJH6ERX4dwaPOE1Dtqhq0WeHE1i7glM9jV4sKoXozcOPKvqsA50w1x37DMsmyfFdpuTY/s400/3_64_fig1.jpg" border="0" /></a> First let us start by understanding some important concepts. As described in Problem <strong>3.63</strong>, the entire conductor must have the same potential and hence, there cannot be any electric field inside the conductor. In other words there is no electric field within the conducting shell. <div><div><div><div><div><div><div><div><div><div><div><br />Consider a Gaussian spherical surface of radius <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZ16NE-CyQOzzVNFvSItBtAlw-TilRfA8yy_eI8irr5BMou9m3Gr2xzuBMYKmix3RYvjzMrnx55rQPkTRfcCVNz0MwlTdJUpv6HBTVxYOXc-LclkIdUUDQhnu_traBdIbUPSlivwAJF-U/s1600/3_64_1.jpg"><img style="width: 45px; height: 18px;" id="BLOGGER_PHOTO_ID_5542204948020348930" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZ16NE-CyQOzzVNFvSItBtAlw-TilRfA8yy_eI8irr5BMou9m3Gr2xzuBMYKmix3RYvjzMrnx55rQPkTRfcCVNz0MwlTdJUpv6HBTVxYOXc-LclkIdUUDQhnu_traBdIbUPSlivwAJF-U/s400/3_64_1.jpg" border="0" /></a>such that .<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhAfPGsj7YNkZbQnu2v_eIm3LDXKQVWIDKz98mHVlv22vYYX-R6h6aDUL7cQxgeSVpc7hLVDj5_NHKKgro-sZQ42Dws2uNVZIynz-UoxNuCrcb9OqW-Sfr1snq1fnzNQZxGfshZATiUgD4/s1600/3_64_2.jpg"><img style="width: 43px; height: 18px;" id="BLOGGER_PHOTO_ID_5542205598849437234" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhAfPGsj7YNkZbQnu2v_eIm3LDXKQVWIDKz98mHVlv22vYYX-R6h6aDUL7cQxgeSVpc7hLVDj5_NHKKgro-sZQ42Dws2uNVZIynz-UoxNuCrcb9OqW-Sfr1snq1fnzNQZxGfshZATiUgD4/s400/3_64_2.jpg" border="0" /></a>, in other words this Gaussian surface is just above the inner circle within the conductor. Since there cannot is any electric field in side the conductor, there is no electric flux through the Gaussian surface. From Gauss Law in turn, this imples that the net charge contained within this Gaussian sphere must be <strong>0</strong>. This can only happen if there is a net charge of <strong>-q</strong> on inner the surface of the sphere. Let <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCilRC4F3QGkzOAUXXqH83vlHkSBJEw6zcHadk9HGI6FrQQ80anKnD_-Nrq9NLvYJjkFm3fcKZIHJv-1X1Pz5yCkAlCSksm1qI2quc66FRkv5mfzl-RwQhSNOdDwWYCNmCu9TjI32QgRw/s1600/3_64_4.jpg"><img style="width: 56px; height: 22px;" id="BLOGGER_PHOTO_ID_5542209999933795522" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCilRC4F3QGkzOAUXXqH83vlHkSBJEw6zcHadk9HGI6FrQQ80anKnD_-Nrq9NLvYJjkFm3fcKZIHJv-1X1Pz5yCkAlCSksm1qI2quc66FRkv5mfzl-RwQhSNOdDwWYCNmCu9TjI32QgRw/s400/3_64_4.jpg" border="0" /></a> be the charge density at a location <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXON_3E77cbCkpe3tIZS6RQR7jxdcDZxDqluZQF2qNxLS29rg2VCuoCwowJkm22_aWw7zPdWMM-0d4oW1JB8S2SwdTJMN8DgVLi_ssP3y0x4m4OMqRk1AvekZLjGMC6LJfAAwBQsQdbUE/s1600/3_64_5.jpg"><img style="width: 48px; height: 22px;" id="BLOGGER_PHOTO_ID_5542210379707038850" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXON_3E77cbCkpe3tIZS6RQR7jxdcDZxDqluZQF2qNxLS29rg2VCuoCwowJkm22_aWw7zPdWMM-0d4oW1JB8S2SwdTJMN8DgVLi_ssP3y0x4m4OMqRk1AvekZLjGMC6LJfAAwBQsQdbUE/s400/3_64_5.jpg" border="0" /></a> in spherical coordinates on the inner surface of the conductor. Then we have,<br /><div><br /></div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPv3z4t9ywVu7IFIHW011BeRWg81oLdZkdXdqqra_48-LerLg9VbbssQU17zw2dzHasTpevGqtdkw4EMBzNAMDA6FTU8wNeiToyElXv-nfu8WS74eaYG305CYEsHBIdmhyMav0o7jM9-0/s1600/3_64_6.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 345px; height: 50px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPv3z4t9ywVu7IFIHW011BeRWg81oLdZkdXdqqra_48-LerLg9VbbssQU17zw2dzHasTpevGqtdkw4EMBzNAMDA6FTU8wNeiToyElXv-nfu8WS74eaYG305CYEsHBIdmhyMav0o7jM9-0/s400/3_64_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5542564994411253474" border="0" /></a><br /><br /><br /><br />Since the electric field anywhere within the conductor is <strong>0</strong>, for all Gaussian surfaces in the conducting shell with radius <strong>R</strong> such that <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhYedoA1ftDx2Awz1fQwEO73sRkhRQaTnJyvDHxEC8JTfAHCtvvK8-nqOWqCiylesDgipyhQnkMdl8yioNZpcZKb_FN8swJ2alQkfZbRr2AMU9S92XGKOPIwot1tIbNm881-N55nUMcUM/s1600/3_64_3.jpg"><img style="width: 60px; height: 20px;" id="BLOGGER_PHOTO_ID_5542213083311721986" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhYedoA1ftDx2Awz1fQwEO73sRkhRQaTnJyvDHxEC8JTfAHCtvvK8-nqOWqCiylesDgipyhQnkMdl8yioNZpcZKb_FN8swJ2alQkfZbRr2AMU9S92XGKOPIwot1tIbNm881-N55nUMcUM/s400/3_64_3.jpg" border="0" /></a> the net flux through them will be zero, in turn implying that no free charges now exist anywhere within the shell. Since the total charge in the conductor must be consrved, this in turn implies that all the positive charges in the shell are concentrated on the outer surface of the conducting shell as shown in the figure. Further, the sum total of all these positive charges will be equal to <strong>q</strong>. Let <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiu71aaOY9IOS37aSAb669oWd4jL_wlGVzWn1KzHzf80Gxs8H8iHlHu-g-SBFBqMzLaY45RK2-C4Nx0zvjheVCm5m_iHKVTIMH_84qS3jKHS6c2X9jtggn0REVsIqAnEEc4oLYzwgZ0ejc/s1600/3_64_7.jpg"><img style="width: 58px; height: 22px;" id="BLOGGER_PHOTO_ID_5542214149636194386" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiu71aaOY9IOS37aSAb669oWd4jL_wlGVzWn1KzHzf80Gxs8H8iHlHu-g-SBFBqMzLaY45RK2-C4Nx0zvjheVCm5m_iHKVTIMH_84qS3jKHS6c2X9jtggn0REVsIqAnEEc4oLYzwgZ0ejc/s400/3_64_7.jpg" border="0" /></a>be the surface charge density on the outer surface of the conductor at a location with sperical coordinates <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjkQZllj5KUXUR9Z9Tq0dHjHhUHK4hfk9daWym6TPY-tWEKUxW0CmmzvGxna_FgEJH9AXVJubNMadlR5__R3G2ieDMqN4gmtK693WNf53A66EmltYRgv4ekoky1WzIsRdro9F3ghNigol0/s1600/3_64_8.jpg"><img style="width: 48px; height: 23px;" id="BLOGGER_PHOTO_ID_5542214577521921634" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjkQZllj5KUXUR9Z9Tq0dHjHhUHK4hfk9daWym6TPY-tWEKUxW0CmmzvGxna_FgEJH9AXVJubNMadlR5__R3G2ieDMqN4gmtK693WNf53A66EmltYRgv4ekoky1WzIsRdro9F3ghNigol0/s400/3_64_8.jpg" border="0" /></a>. Then, we have,<br /></div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhqFwqatj40ddz1Jz_acLBQxOEFxNdibXMF-o8SLStp7Eej8UAnGkXlvxSycmi_6d3K1yUkZmzA4FKrzijw7L5uHHJlPImbya77skMl-Cd-eAGad_vuUSNOWrBw6UQFZ4Z7qQ3yU4SpSKs/s1600/3_64_9.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 352px; height: 50px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhqFwqatj40ddz1Jz_acLBQxOEFxNdibXMF-o8SLStp7Eej8UAnGkXlvxSycmi_6d3K1yUkZmzA4FKrzijw7L5uHHJlPImbya77skMl-Cd-eAGad_vuUSNOWrBw6UQFZ4Z7qQ3yU4SpSKs/s400/3_64_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5542565063770499330" border="0" /></a><br /><div></div><br /><br /><br /><div></div>The potential at the center of the sphere is due to three sources, <span style="font-weight: bold;">i)</span> the charge <span style="font-weight: bold;">q</span>, <span style="font-weight: bold;">ii)</span> the surface charge density at the inner surface of the conducting sphere and <span style="font-weight: bold;">iii)</span> the surface charge density at the outer surface of the conducting sphere and is given by.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEijMuO0NvJkmefJ2lK2h18Drhh_jn13kFs6F1W9ulvHeanr0aQ-eDPdfpifEUen6wGvS9T5YD6dkfdZV3DRqfawxYUgnPJxV46yKHMMaaAGmZAMTXlAUHBXQusuTNpcTtypJLZo2k3Hp_M/s1600/3_64_10.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 320px; height: 115px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEijMuO0NvJkmefJ2lK2h18Drhh_jn13kFs6F1W9ulvHeanr0aQ-eDPdfpifEUen6wGvS9T5YD6dkfdZV3DRqfawxYUgnPJxV46yKHMMaaAGmZAMTXlAUHBXQusuTNpcTtypJLZo2k3Hp_M/s400/3_64_10.jpg" alt="" id="BLOGGER_PHOTO_ID_5542565117988272466" border="0" /></a><br /><br /><div></div><br /><br /><br /><br /><br /><div></div></div></div></div></div></div></div></div></div></div></div>Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-8917124841013563709.post-55496132309031716562010-11-20T10:05:00.010+05:302010-11-23T08:02:01.787+05:30Irodov Problem 3.63<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixUZiCD6hlxk2YEUAzx73TRe5p_x3j7KKZ_2oI5Wwb7OpGRPItAvbxzF6O6k8Dok_t1VjycaQtKhdxMNRTu-eos2nMwwvL7PS2iuIQtKZUluqBpoGIPx3ld51yCAZCvVEjuuBQ_zlgXOI/s1600/3_63_fig1.jpg"><img style="width: 331px; height: 282px;" id="BLOGGER_PHOTO_ID_5541819938912778162" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixUZiCD6hlxk2YEUAzx73TRe5p_x3j7KKZ_2oI5Wwb7OpGRPItAvbxzF6O6k8Dok_t1VjycaQtKhdxMNRTu-eos2nMwwvL7PS2iuIQtKZUluqBpoGIPx3ld51yCAZCvVEjuuBQ_zlgXOI/s400/3_63_fig1.jpg" border="0" /></a> Couple of concepts need explanation here first before the problem can be solved. So I shall first provide some explanations and then proceed to solve the problem.<br /><div><div><div><br /><div> </div><div><strong>All charges will reside only on the surface of the solid spherical conductor</strong></div><div>The basic thing to understand in this problem is that electrons flow freely within conductors. As described in problem <strong>3.54</strong>, the electric potential at all points in a conducting body should be equal. Otherwise since in a conductor charges are free flowing, the charges would freely flow and redistribute themselves until potential is equalized. Since there is no potential difference between any two points within the conducting body, there cannot be any electric field inside the solid conducting sphere. Since there is not electric filed inside the solid sphere, there cannot be any charge distribution inside the sphere (from Gauss law if there where any charge distribution inside the sphere, there would be an electric field). <em>This leads to an inescapable conclusion that charges in the conducting sphere will only reside on the surface of the sphere. In other there will only be a surface charge density in the solid sphere.<br /><br /></em></div><div> </div><div><strong>The total charge in the sphere is zero</strong></div><div>Since the sphere had no charge of its own to being with, and charges have only redistributed themselves, the net charge of the sphere must be zero. In other words if the surface charge density of the sphere at a location <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJPsV8EtHlqQkaGg-bQne-c4t3V7EtZru_JMTCcuM1MEtyDhHOH86D_Lz0wp-MBqfZXCDBj3uqZNEglwq4HRd4JnalZkv7mIyYWrX1GgUxAPYGrEIKUvQfrUR-TxPHd2LQYj-wDLll7Co/s1600/3_63_2.jpg"><img style="width: 40px; height: 21px;" id="BLOGGER_PHOTO_ID_5541816321113741266" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJPsV8EtHlqQkaGg-bQne-c4t3V7EtZru_JMTCcuM1MEtyDhHOH86D_Lz0wp-MBqfZXCDBj3uqZNEglwq4HRd4JnalZkv7mIyYWrX1GgUxAPYGrEIKUvQfrUR-TxPHd2LQYj-wDLll7Co/s400/3_63_2.jpg" border="0" /></a>, <strong>R</strong> being the radius of the sphere, is <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjaegjdYDj3P2YYSHbc7q-lLndxeSSi4eqg_AI3i9mOXJQHoamwIzQFnNNDelShG6-6A56kWM1_uIga6ntIRp7XiNmmHLl_Y7875d4NzrG-TuKzCDa4WKPed0-x1TXVktTtT0hFytz61Rs/s1600/3_63_1.jpg"><img style="width: 37px; height: 22px;" id="BLOGGER_PHOTO_ID_5541816552987754098" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjaegjdYDj3P2YYSHbc7q-lLndxeSSi4eqg_AI3i9mOXJQHoamwIzQFnNNDelShG6-6A56kWM1_uIga6ntIRp7XiNmmHLl_Y7875d4NzrG-TuKzCDa4WKPed0-x1TXVktTtT0hFytz61Rs/s400/3_63_1.jpg" border="0" /></a>(<strong>R</strong> is ignored since all charge resides at the surface only), then,</div><br /><div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgcPClR27Q1CruFzKstpMSjchm5ozRWYzD4X4xxpfEam18WXqXFaGw6u-fleen6m9yRNTpZMp2DMJK_zh9bX5k2hQ6NB-jNu_WSQOQiFErtcCsv7j2UIqZjT3TyW735ROaq88qVmtwmC48/s1600/3_63_3.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 293px; height: 48px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgcPClR27Q1CruFzKstpMSjchm5ozRWYzD4X4xxpfEam18WXqXFaGw6u-fleen6m9yRNTpZMp2DMJK_zh9bX5k2hQ6NB-jNu_WSQOQiFErtcCsv7j2UIqZjT3TyW735ROaq88qVmtwmC48/s400/3_63_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5542567084982804546" border="0" /></a><br /><br /></div><br /><br /><div><strong>Potential of the sphere<br /></strong>Since the entire conducting sphere is at the same potential, determing the potential at any one point within the sphere is sufficient to determine the potental at any point in the sphere. It turns out that it the easiest to determine the potential at the center of the sphere. The total potential at the center of the sphere is given by, the potential due to <strong>i)</strong> the charge <strong>q</strong> outside the sphere and <strong>ii)</strong> due to the surface charge density induced in the surface of the conducting sphere <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgL9pbMk_9XZuvB9b9SueqUdVUbOuSBskUv9sAYLUYjrY_bPNMKkwYYk9P-SZrWWnhZjvNm86v06tcb069_Fel3_HekP_ZalBBXlG0-0-uPqQPN0-xhiKtYShLXlyFYVL6EMaBfuHHxHQM/s1600/3_63_1.jpg"><img style="width: 37px; height: 22px;" id="BLOGGER_PHOTO_ID_5541823135729335170" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgL9pbMk_9XZuvB9b9SueqUdVUbOuSBskUv9sAYLUYjrY_bPNMKkwYYk9P-SZrWWnhZjvNm86v06tcb069_Fel3_HekP_ZalBBXlG0-0-uPqQPN0-xhiKtYShLXlyFYVL6EMaBfuHHxHQM/s400/3_63_1.jpg" border="0" /></a>. </div><div>This is given by,</div><div> </div><br /><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgIPbHRhN1WqIKg_GzHUBzpcViFhovrx4e6_DMYCdyteXsaUsXrHMp0jIsf8NHO0eFBKdZFQC7EqapDXFQUse71qxWd61v-0GpC79-QnBfew5lX674CmTsG4BNs7OR7jKOKQwKF_EmpTR8/s1600/3_63_4.jpg"><img style="margin: 0px 10px 10px 0px; width: 248px; float: left; height: 241px;" id="BLOGGER_PHOTO_ID_5541825742806043042" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgIPbHRhN1WqIKg_GzHUBzpcViFhovrx4e6_DMYCdyteXsaUsXrHMp0jIsf8NHO0eFBKdZFQC7EqapDXFQUse71qxWd61v-0GpC79-QnBfew5lX674CmTsG4BNs7OR7jKOKQwKF_EmpTR8/s400/3_63_4.jpg" border="0" /></a></div></div></div></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-35682119949245996352010-11-17T16:16:00.012+05:302010-11-20T07:33:16.701+05:30Irodov Problem 3.62<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg3sjR4IRDUTpKFh3TC33CEzA7QDCUY_8jXttyuvycsFmt80-gTSRFkR32LfMOaqsCfg5qFhhHoT7UbGSmjcfghnnbJpJH2lik6dNfFg2Qsy4huuGXi50zNNV-CrOXVl3yH5rpFdJyUZXk/s1600/3_62_fig1.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 400px; FLOAT: left; HEIGHT: 252px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5540468957194488546" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg3sjR4IRDUTpKFh3TC33CEzA7QDCUY_8jXttyuvycsFmt80-gTSRFkR32LfMOaqsCfg5qFhhHoT7UbGSmjcfghnnbJpJH2lik6dNfFg2Qsy4huuGXi50zNNV-CrOXVl3yH5rpFdJyUZXk/s400/3_62_fig1.jpg" /></a> Using the method of images we can replace the conductor by an image of the positively charged ring i.e. a negatively charged ring at a distance <strong>2l</strong> from the original ring as shown in the figure.<br /><br />From problem <strong>3.9</strong>, we know the electric field due to a positively charged ring at a distance <strong>l</strong> along the axis as,<br /><br /><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjnXs7ogs1mz5iPbDrKwmZOJSPkcU9euzaDl5IMNpt74Yrav8BEUa9kx05kHc4RfVdIZEmYs7w1eJOVgLvTDhaEogxioEPqMbBHgfbDabaV_iD_8wFL3ZlinaoaRJjeBjrOaOqG6V7iYcU/s1600/3_62_1.jpg"><img style="WIDTH: 143px; HEIGHT: 44px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5540470728704030562" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjnXs7ogs1mz5iPbDrKwmZOJSPkcU9euzaDl5IMNpt74Yrav8BEUa9kx05kHc4RfVdIZEmYs7w1eJOVgLvTDhaEogxioEPqMbBHgfbDabaV_iD_8wFL3ZlinaoaRJjeBjrOaOqG6V7iYcU/s400/3_62_1.jpg" /></a><br /><br /><br />The net electric field due to both the rings (image and real) on the conductor at the center of the two rings is given by,<br /><br /><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj95bkDxPdUhjZj80S1VoROMeWpZtV5Fvk3vfoYUO0-dqJGLHRFxcCjRyj8hNzzM_DhR6eyQelH9K5Sl6qNLmQtpXYPXK2Wo7Jz9NMiOW8xxAfE2NeuHNFsQz-135sm5A5kEhwXpwH5DVQ/s1600/3_62_2.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 155px; FLOAT: left; HEIGHT: 46px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5540471848425232530" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj95bkDxPdUhjZj80S1VoROMeWpZtV5Fvk3vfoYUO0-dqJGLHRFxcCjRyj8hNzzM_DhR6eyQelH9K5Sl6qNLmQtpXYPXK2Wo7Jz9NMiOW8xxAfE2NeuHNFsQz-135sm5A5kEhwXpwH5DVQ/s400/3_62_2.jpg" /></a><br /><br /><br /><br /><br />As seen from problem <strong>3.59</strong>, if the surface charge density at the conductor is <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiadwrG4g6e3_uWjeYOIH2rLzlM24wW6FgwhrjfitPKGbtj8JUzLeqcQfBUNduwY4FHFIr2_d90r9T_XtDyhEcnxh9Uo_0KiQvXsLUxepJNWYFrS5g_6iTdk39_GUbSr0MF0aSeZ5hxNsE/s1600/3_62_3.jpg"><img style="WIDTH: 11px; HEIGHT: 17px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5540472589286900610" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiadwrG4g6e3_uWjeYOIH2rLzlM24wW6FgwhrjfitPKGbtj8JUzLeqcQfBUNduwY4FHFIr2_d90r9T_XtDyhEcnxh9Uo_0KiQvXsLUxepJNWYFrS5g_6iTdk39_GUbSr0MF0aSeZ5hxNsE/s400/3_62_3.jpg" /></a> at this location then we have,<br /><br /><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3ToX0qjkjjCRzYOhmvMn2wksmEq2icx4nC0OLrxqiFnplYmR8CBCK8NK8rGIpcelIcs3gFLvrWZzBeWTf0wx2PtDVI7DPd2OJvSqrSpgRYARY0RRbgvQv1oxVHQVG-1KMc7eyNNmCFL0/s1600/3_62_4.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 122px; FLOAT: left; HEIGHT: 72px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5540473598018588290" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3ToX0qjkjjCRzYOhmvMn2wksmEq2icx4nC0OLrxqiFnplYmR8CBCK8NK8rGIpcelIcs3gFLvrWZzBeWTf0wx2PtDVI7DPd2OJvSqrSpgRYARY0RRbgvQv1oxVHQVG-1KMc7eyNNmCFL0/s400/3_62_4.jpg" /></a><br /><br /><br /><br /><br /><br /><br /><strong>b)</strong><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhYI6_E0aduJExEXvrCE1C9zzvuRO4255VrvTVWItM-u4FWb-bXEnKPqxAso3exyldBbdHbM4sitvW5Nm0qT8O9ZKG0q1bVIwKPRTTU0R0PUt4owGXFO2DpTJc8EQM2QoHlIttqjM0ghSE/s1600/3_62_fig2.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 276px; FLOAT: left; HEIGHT: 400px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5540474263707960866" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhYI6_E0aduJExEXvrCE1C9zzvuRO4255VrvTVWItM-u4FWb-bXEnKPqxAso3exyldBbdHbM4sitvW5Nm0qT8O9ZKG0q1bVIwKPRTTU0R0PUt4owGXFO2DpTJc8EQM2QoHlIttqjM0ghSE/s400/3_62_fig2.jpg" /></a><br />The electric field strength at the center of the ring will be entirel due to negatively charged image ring and can be computed using <strong>eqn</strong> <strong>(1)</strong> as,<br /><br /><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixSZYWCQItzdXzTXpk9sGeJnFZ7BKNxocXqjdIZ7QTAtBJD1R78_3X1_9KllvxKOlvpwFBmrZZrbVW4WoVU_RsJk5bjdty-gRNbiqHiQH-GUaJPXJkR7zpvm5kEWMGG5zCnLTjEWT-7XY/s1600/3_62_5.jpg"><img style="WIDTH: 152px; HEIGHT: 112px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5540476739279703682" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixSZYWCQItzdXzTXpk9sGeJnFZ7BKNxocXqjdIZ7QTAtBJD1R78_3X1_9KllvxKOlvpwFBmrZZrbVW4WoVU_RsJk5bjdty-gRNbiqHiQH-GUaJPXJkR7zpvm5kEWMGG5zCnLTjEWT-7XY/s400/3_62_5.jpg" /></a><br /><br />The electric potential due the positively charged ring at its center is given by,<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjEWSWIZtcR4vNwGgR1iwt3H8Z-QFw6fXdrvJV3MULGH7cXjIhoRW-iQCi5xPAjcwpdtxi_x9kgu9SUIJAclXQSWL65qHJRMD-R49ePpLucIY0IwyR7IqpgRANzo94nqMNE-aCnG-Z97DM/s1600/3_62_6.jpg"><img style="WIDTH: 44px; HEIGHT: 34px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5540478048657820258" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjEWSWIZtcR4vNwGgR1iwt3H8Z-QFw6fXdrvJV3MULGH7cXjIhoRW-iQCi5xPAjcwpdtxi_x9kgu9SUIJAclXQSWL65qHJRMD-R49ePpLucIY0IwyR7IqpgRANzo94nqMNE-aCnG-Z97DM/s400/3_62_6.jpg" /></a>. The electric potential due to the negatively charged ring at the center of the positively charged ring is given by,<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjmPhZMYypoWyY6I27B70JnYlrZPHfnOjKmYdCIdpRiXt0_eJtkN4auwAYwma1GvArETC64yOtTKIV_hsJ4753soOSXwah2aXNpH_GwwRjy8tn6FC0o7kxux69QLnv4lcCvJ-W9PPVtJTg/s1600/3_62_7.jpg"><img style="WIDTH: 94px; HEIGHT: 46px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5540479464664035778" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjmPhZMYypoWyY6I27B70JnYlrZPHfnOjKmYdCIdpRiXt0_eJtkN4auwAYwma1GvArETC64yOtTKIV_hsJ4753soOSXwah2aXNpH_GwwRjy8tn6FC0o7kxux69QLnv4lcCvJ-W9PPVtJTg/s400/3_62_7.jpg" /></a>. The net potential is thus given by,<br /><br /><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi68d6lc5rl619HHl0VmFKnTXEXfBsIaSY76kHipC8xyAWVZY9AF1ODwzR85NPyOayMsJiS37zwstxjY0CyyjMG6X4Xuw1c3vaVoaU3E5BbUGygUM_8SxdMDo1nDBT_DwPOAu_b7zXtk0c/s1600/3_62_8.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 194px; FLOAT: left; HEIGHT: 55px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5540480904236359394" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi68d6lc5rl619HHl0VmFKnTXEXfBsIaSY76kHipC8xyAWVZY9AF1ODwzR85NPyOayMsJiS37zwstxjY0CyyjMG6X4Xuw1c3vaVoaU3E5BbUGygUM_8SxdMDo1nDBT_DwPOAu_b7zXtk0c/s400/3_62_8.jpg" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-38176382183410113072010-11-14T08:21:00.008+05:302010-11-14T09:10:55.477+05:30Irodov Problem 3.61<div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEglyvNS0cCMbO9cMiHHyv1TVRhJ43wLOa7uiYfHy5HvTxjeBzr1Ydu9mmV5CkRqxgMaa9JunB5RPBGTt_ztX0Zzdj3J27geJci_YyxFHQ9BvYKFrSjQfwh7KWGjIVvVqmZq0x8QZepwOSA/s1600/3_61_fig1.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 400px; FLOAT: left; HEIGHT: 347px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5539240685052855650" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEglyvNS0cCMbO9cMiHHyv1TVRhJ43wLOa7uiYfHy5HvTxjeBzr1Ydu9mmV5CkRqxgMaa9JunB5RPBGTt_ztX0Zzdj3J27geJci_YyxFHQ9BvYKFrSjQfwh7KWGjIVvVqmZq0x8QZepwOSA/s400/3_61_fig1.jpg" /></a> Following the method of images we replace the infinite plane conductor by a negatively charged rod placed at the mirror image location of the conductor as shown in the figure. <div><div><br /><div>Now let us determine the electric field due to the infintely long charged rods at a location that is at a distance<strong> r</strong> on the plane of the <img class="gl_italic" border="0" alt="Italic" src="http://www.blogger.com/img/blank.gif" />conductor from its axis as shown in the figure.<br /></div><div>Consider an infinitesimally small section of the rod of length <strong>dx</strong>, at a height <strong>x</strong> from the plane. The electric field due to this infinitesimally small section of the rod will be,<br /></div><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjcXgLdIv2gZboB9sN57f2zmVQh0tWRyFr9tlpDs-ozW4tUPJE8bAfCZMosjDAVYtyCcbb7W_6p8WsMEm72-Tv9nFcDaZMNNJcXgjTVGHw2WAq_s7punb8mia51zGDggawAoiJQ2AavNeA/s1600/3_61_1.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 400px; FLOAT: left; HEIGHT: 106px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5539239914414950354" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjcXgLdIv2gZboB9sN57f2zmVQh0tWRyFr9tlpDs-ozW4tUPJE8bAfCZMosjDAVYtyCcbb7W_6p8WsMEm72-Tv9nFcDaZMNNJcXgjTVGHw2WAq_s7punb8mia51zGDggawAoiJQ2AavNeA/s400/3_61_1.jpg" /></a></div><br /><div><br /></div><br /><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjcXgLdIv2gZboB9sN57f2zmVQh0tWRyFr9tlpDs-ozW4tUPJE8bAfCZMosjDAVYtyCcbb7W_6p8WsMEm72-Tv9nFcDaZMNNJcXgjTVGHw2WAq_s7punb8mia51zGDggawAoiJQ2AavNeA/s1600/3_61_1.jpg"></a></div><br /><div><br /><br /><br /></div><div>Thus,</div><br /><div><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiuhZ-Xv8JlpHTP45paJXhbP-y-4R7UKBATlJseW1w7NzuhDUuYfVqEtwPujCmr2tGgUtx7UZICoyM31foTJD0aNIS_6u3TDdpqFtfbA-Gt83A1To5xIAxjKXww3SUZhzsnf2cZrnO9T1g/s1600/3_61_2.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 286px; FLOAT: left; HEIGHT: 69px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5539243152915773506" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiuhZ-Xv8JlpHTP45paJXhbP-y-4R7UKBATlJseW1w7NzuhDUuYfVqEtwPujCmr2tGgUtx7UZICoyM31foTJD0aNIS_6u3TDdpqFtfbA-Gt83A1To5xIAxjKXww3SUZhzsnf2cZrnO9T1g/s400/3_61_2.jpg" /></a><br /></div><br /><div></div></div><br /><div></div><div> </div><div>As described in Problem <strong>3.59</strong>, the charges induced in the conductor at this location can be determined as,</div><br /><div></div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjz4DLFdZji8pbEWsj4UIzuc0pSKh6PWbnfQHgkIzYwJ0kFxm77nSEUgDZ4LxH0jhYAawr9Fjp5P-McOlwvON-oRFfL2OfXByGOY4HatQ51uEpJpvMstfkNzqlUjp_boZo1oqZdpVqfmls/s1600/3_61_3.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 170px; FLOAT: left; HEIGHT: 220px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5539245054574942562" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjz4DLFdZji8pbEWsj4UIzuc0pSKh6PWbnfQHgkIzYwJ0kFxm77nSEUgDZ4LxH0jhYAawr9Fjp5P-McOlwvON-oRFfL2OfXByGOY4HatQ51uEpJpvMstfkNzqlUjp_boZo1oqZdpVqfmls/s400/3_61_3.jpg" /></a><br /><div></div><br /><div><br /><br /></div><br /><div></div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_RQgGCKmUqjU7RgGMhasTQ35m8TpTgSXDmxPb5a5AM2YEztNtpKz39qCF9Wvihc9Kr3X1zVTEraqe_cEbUA4se-tV_hAchXWgLbJFxSJAYbaqf5gXzKpVadBUUsYwsI2Cc8jmHPGKHRQ/s1600/3_61_2.jpg"></a><br /><div><br /><br /><br /></div><br /><div></div></div></div>Unknownnoreply@blogger.com7tag:blogger.com,1999:blog-8917124841013563709.post-46992464920892482142010-11-10T07:16:00.010+05:302010-11-10T08:03:53.606+05:30Irodov Problem 3.60<div><div><div><div><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2mhr_Ulj7CKVKNwafqUTI_Js7KhPSauhgsVqBpvVzmK41Kc-oqZcS4QtnwbD0vZdzKXn5BsLeyQ4h14TauWNJ_zqUbRV3EYoEFYvy7ftfGe0K3UHra_zY0DsMt7qhTXZZWqLqQmcTbfc/s1600/3_60_fig1.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 400px; FLOAT: left; HEIGHT: 208px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5537731861178016978" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2mhr_Ulj7CKVKNwafqUTI_Js7KhPSauhgsVqBpvVzmK41Kc-oqZcS4QtnwbD0vZdzKXn5BsLeyQ4h14TauWNJ_zqUbRV3EYoEFYvy7ftfGe0K3UHra_zY0DsMt7qhTXZZWqLqQmcTbfc/s400/3_60_fig1.jpg" /></a><br /><div>Based on the method of images we can replace the infinite conducting plane by an infintely long charged wire of charge <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkwBISUuoWmi1tCNBbPcMQhmoBbo48r-7c1n7yQ1qGsvWsxmXBzJ593A5Ms-LDT8GBOFO_Se1F1aOjWu_MWWKuglHMvqA9XEkIiXLst0mAuIg0vRfUspYQ2XS29XfJhgoaRBnuNCvC7bw/s1600/3_60_1.jpg"><img style="WIDTH: 19px; HEIGHT: 23px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5537733135812954306" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkwBISUuoWmi1tCNBbPcMQhmoBbo48r-7c1n7yQ1qGsvWsxmXBzJ593A5Ms-LDT8GBOFO_Se1F1aOjWu_MWWKuglHMvqA9XEkIiXLst0mAuIg0vRfUspYQ2XS29XfJhgoaRBnuNCvC7bw/s400/3_60_1.jpg" /></a> per unit length. As seen from Problem <strong>3.22</strong>, the electric field due to the negively charged wire at a distance <strong>2l</strong> will be,<br /></div><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiy98jBZK8QjehFOwH1Bpzn7FFc-67X5U2spcQlx3qR90JvYxzUxDkxYGjWrX9LB6NZ7qWkTEfZxxk4_RiWOyuRmwjeIGiLdzavSCVJxBh6Y5MjKqq2WWS_J-FrJ2SX1WgYuJ0SzKlxB9c/s1600/3_60_2.jpg"><img style="WIDTH: 117px; HEIGHT: 37px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5537734235842295282" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiy98jBZK8QjehFOwH1Bpzn7FFc-67X5U2spcQlx3qR90JvYxzUxDkxYGjWrX9LB6NZ7qWkTEfZxxk4_RiWOyuRmwjeIGiLdzavSCVJxBh6Y5MjKqq2WWS_J-FrJ2SX1WgYuJ0SzKlxB9c/s400/3_60_2.jpg" /></a></div><div><br /></div><div>Consider an infintesimally small section of the positively charged wire of length <strong>dl</strong>. The charge contained in this infinitesimally small section of wire will be <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJDVT85SMTw2XvOXOiwlgohIASI1LQNu7C6v_tEctN7fQ3ogjfNQSTNjo_BoP9Jl0KpENrnpeFRNBz5HNvXIh8w2z6PrkKU68GFCehtFMXa-d3EC_i0X7mm3mMh4lQEw6DG58bt0t4Gic/s1600/3_60_3.jpg"><img style="WIDTH: 19px; HEIGHT: 17px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5537734723802379458" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJDVT85SMTw2XvOXOiwlgohIASI1LQNu7C6v_tEctN7fQ3ogjfNQSTNjo_BoP9Jl0KpENrnpeFRNBz5HNvXIh8w2z6PrkKU68GFCehtFMXa-d3EC_i0X7mm3mMh4lQEw6DG58bt0t4Gic/s400/3_60_3.jpg" /></a>. Thus, the force acting on this infinitesiamlly small section of the wire will be, </div><div><br /></div><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiITch7i6ufnVVjyFnv1LQTSipID2gVTTFfEcuuLixMsb85YgFY1_Vx_se3A0eBNjkNcRsdgz8C5mYp-qxBSomE4OCkyCm3kgunWRLGByOoS9l98KYZjwN9L6hGoklMBrOeX_xCoBvGmGA/s1600/3_60_4.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 184px; FLOAT: left; HEIGHT: 80px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5537736177199363746" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiITch7i6ufnVVjyFnv1LQTSipID2gVTTFfEcuuLixMsb85YgFY1_Vx_se3A0eBNjkNcRsdgz8C5mYp-qxBSomE4OCkyCm3kgunWRLGByOoS9l98KYZjwN9L6hGoklMBrOeX_xCoBvGmGA/s400/3_60_4.jpg" /></a><br /></div><div><br /><br /><br /> </div><div> </div><div><strong>b)<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhvwjpZGegY7eXzoFmLklLW2l3Zag-uTPunzPqMoeO1ltJsKIg-JCRO5gm-hjvRkFu5qiTP0D18hHnd3npSriWoIIa1KddEMj88Jq153uu0SBklM6nVWOL4qP9mw41b4pVcNaxP7orSSlo/s1600/3_60_fig2.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 400px; FLOAT: left; HEIGHT: 188px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5537737200229759346" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhvwjpZGegY7eXzoFmLklLW2l3Zag-uTPunzPqMoeO1ltJsKIg-JCRO5gm-hjvRkFu5qiTP0D18hHnd3npSriWoIIa1KddEMj88Jq153uu0SBklM6nVWOL4qP9mw41b4pVcNaxP7orSSlo/s400/3_60_fig2.jpg" /></a></strong></div><div><br /><br />Based on problem <strong>3.22</strong>, the electric field at a location on the conductor that is <strong>x</strong> untis from the conductor (as shown in the figure) is given by,</div><div><br /><br /><br /></div><div></div><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEib__yzmbt5uEq3bGV2TBj4dmL4IwEZ66tdIzJsHjRFRlvizpmWBoCwUyFGwRrhdD8ifeNGUDYOxDW5D7mSqjUm6c9467M1Df4TLVx5SADr5QRj47ozLEnJaTY6buhs1VcJEwZjeQ-imlM/s1600/3_60_5.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 386px; FLOAT: left; HEIGHT: 109px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5537741300727514018" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEib__yzmbt5uEq3bGV2TBj4dmL4IwEZ66tdIzJsHjRFRlvizpmWBoCwUyFGwRrhdD8ifeNGUDYOxDW5D7mSqjUm6c9467M1Df4TLVx5SADr5QRj47ozLEnJaTY6buhs1VcJEwZjeQ-imlM/s400/3_60_5.jpg" /></a><br /><br /><br /></div><div><br /></div><div><br /> </div><div></div><div>Similar to problem <strong>3.59</strong>, if the charge density induced in the conductor at this location is given by <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiI7etTBh7YVY2xwqNzM-QguFoaqqnv3GElq0RG2ch7E_yzdCld4pLv2V1qRv8FRqDtNq-zc5_u2GoxGyPb7CpbCum5dOo-7hZoyW4bV7qOPF_uCL7jTdYVdkTCh1BJn7uD-C6he8mlPfc/s1600/3_60_6.jpg"><img style="WIDTH: 24px; HEIGHT: 19px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5537742068580098482" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiI7etTBh7YVY2xwqNzM-QguFoaqqnv3GElq0RG2ch7E_yzdCld4pLv2V1qRv8FRqDtNq-zc5_u2GoxGyPb7CpbCum5dOo-7hZoyW4bV7qOPF_uCL7jTdYVdkTCh1BJn7uD-C6he8mlPfc/s400/3_60_6.jpg" /></a>, then,</div><div> </div><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhNkXqqSwAb_NkYPcR5LbuNTTon_p-QwPCadBR-kBwl_bjADmO-EmqHtstPYKU_SspkRKwXfD07IiqZ5TeuXj6r4Q8ipZKQPtASOyIX9tnuCk-eK45Yej-pEZ3c0jU4byIbuS_yBMdh-ok/s1600/3_60_7.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 115px; FLOAT: left; HEIGHT: 91px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5537743336790076898" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhNkXqqSwAb_NkYPcR5LbuNTTon_p-QwPCadBR-kBwl_bjADmO-EmqHtstPYKU_SspkRKwXfD07IiqZ5TeuXj6r4Q8ipZKQPtASOyIX9tnuCk-eK45Yej-pEZ3c0jU4byIbuS_yBMdh-ok/s400/3_60_7.jpg" /></a><br /></div><div></div></div></div></div></div></div>Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-8917124841013563709.post-17286087646855484022010-10-26T07:14:00.018+05:302010-12-05T05:48:14.746+05:30Irodov Problem 3.59<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjOHxIF6WhsJu8d0hN4mTDb5cSn9AiS9bWYcPSVzjW3up684nKDoWyPmGh4NbMp1cGd_kASce8JF1q2KaksbLHj-6DSf2EltbkumWMrEiTWY68CWW6QgQDKcMALmapCh1HxF8Z7tymHCv8/s1600/3_59_fig1.jpg"><img style="margin: 0px 10px 10px 0px; width: 370px; float: left; height: 400px;" id="BLOGGER_PHOTO_ID_5532177170632132818" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjOHxIF6WhsJu8d0hN4mTDb5cSn9AiS9bWYcPSVzjW3up684nKDoWyPmGh4NbMp1cGd_kASce8JF1q2KaksbLHj-6DSf2EltbkumWMrEiTWY68CWW6QgQDKcMALmapCh1HxF8Z7tymHCv8/s400/3_59_fig1.jpg" border="0" /></a><br /><br /><div><div><div><div><div><div><div><div>As described in problems <strong>3.54</strong> by using the <a href="http://en.wikipedia.org/wiki/Method_of_images">method of images</a>, we can place a negative charge <strong>-q</strong> at a distance <strong>2l</strong> from the charge <strong>q</strong> on the other side of the conducting plane. </div><br /><div>Consider a point on the conducting plane that is at a distance <strong>r</strong> from the perpendicular on the plane from <strong>q</strong>. The net electric field at this location will be given by,<br /><br /></div><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhICR7AHZltZHuMowieKHkajXKpV8J2hdg_TysxbMQ6MLccgI9KuXV77zquqbUzr3IR19CmEQ9pRLx6SBmLp_ojwp5PYq_0En_bCo-jK_qlEFsJsVzD0sjpW-1wWM-XVrD0NJ_Bf6W8-BY/s1600/3_59_1.jpg"><img style="width: 177px; height: 115px;" id="BLOGGER_PHOTO_ID_5532169760264817378" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhICR7AHZltZHuMowieKHkajXKpV8J2hdg_TysxbMQ6MLccgI9KuXV77zquqbUzr3IR19CmEQ9pRLx6SBmLp_ojwp5PYq_0En_bCo-jK_qlEFsJsVzD0sjpW-1wWM-XVrD0NJ_Bf6W8-BY/s400/3_59_1.jpg" border="0" /></a><br /><br />There can never be an electric field inside the body of conductor, in other words there will be no electric field just beneath the surface of the conductor. This is because, the entire conductor has to be at the same electric potential. If there were a electric field within the conductor it would mean that there would be a potential difference between two points within the conductor. Let us assume that the charge density on the conductor at a distance <strong>r</strong> from the point where the perpendicular from the charge <strong>q</strong> touches the conducting plane is <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzJw_254v_Xvr8PuyjU7fhatCyOKDA4JFsEXucqMcwTEej1i8ZoUM_EosFlFNjxQPd58kHeHbvxqhIJK8FpBSPIC-rgySVmxA45JmVfupXl6ZdtEo2EBZMLdNoVRoYivb8XsXWNlW0038/s1600/3_59_2.jpg"><img style="width: 26px; height: 24px;" id="BLOGGER_PHOTO_ID_5532171696100355778" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzJw_254v_Xvr8PuyjU7fhatCyOKDA4JFsEXucqMcwTEej1i8ZoUM_EosFlFNjxQPd58kHeHbvxqhIJK8FpBSPIC-rgySVmxA45JmVfupXl6ZdtEo2EBZMLdNoVRoYivb8XsXWNlW0038/s400/3_59_2.jpg" border="0" /></a>. </div><br />Consider an infinitesimally small cylinderical <a href="http://en.wikipedia.org/wiki/Gaussian_surface">Gaussian Pillbox</a>, of area <strong>dA</strong> and height <strong>dh </strong>(considered to be as thin as the thickness of a plane), a this location as shown in the figure below. One flat end surface of this cylinder lies within the thickness of the conductor (it just penetrates the conductor) and other flat end is just above the surface of the conductor.<br /><div><br /><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjxk7c0WlvcWfOenAwjWxc39I30rdegu37wdGSRP3zOcDcBwqpby-52OkfE3tVJXF9A3ug5m_3vvthAXE3SoZpZfJ5xAsnCzDpE4wzosznu5AodZemBv-LoHgKnaUfbttjrCae45ZCDuRo/s1600/3_59_fig2.jpg"></a></div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghb0FFHX8I-5lG4B4dAisTyJGdQMkTvf9zf5hpC_lQn7Ap1dp5nvLe31yIgjp0N_wRw59mTtqGqgbQcF3or9OqIqnFs2SnP-t6Sl1EZopT0IFZ1-arMtq1Qv11PHZOJw0umb0fiG65C98/s1600/3_59_fig2.jpg"><img style="margin: 0px 10px 10px 0px; width: 400px; float: left; height: 237px;" id="BLOGGER_PHOTO_ID_5532176836728586994" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghb0FFHX8I-5lG4B4dAisTyJGdQMkTvf9zf5hpC_lQn7Ap1dp5nvLe31yIgjp0N_wRw59mTtqGqgbQcF3or9OqIqnFs2SnP-t6Sl1EZopT0IFZ1-arMtq1Qv11PHZOJw0umb0fiG65C98/s400/3_59_fig2.jpg" border="0" /></a><br /><br /></div><div><span style="color: rgb(0, 0, 0);">The electric field <strong>E</strong> impinges perpendicularly into the Gaussian pillbox's surface into the upper plane of the cylinder as shown. The area vector of the upper surface of the cylinder is given by (<strong>dA k</strong>). There is no electric flux through the other end since there is no electric field on that other surface (that lies just below the surface). Hence,</span></div><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEidrbFKTJdA1Zscss0GrPXMBZQt4I1CWj1ZR67tk6HjfaUnn6D1oOlo_366l6s65arIrFpD6yMo4O2SzTqJklbvNjQXqbc7JbfDmbAJNoy2dbSCUGX04FFMw9qBCidxAkcc43SPab1bCp8/s1600/3_59_3.jpg"></a><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjaBD9EG0qkRIfwOQmTbbPnrlGn-KLNEQiOUNtelBycOFRMG4CHQbTGya0-WodOYYeC-zvuosj_p9_xQFd6lsnvRfMjLWaH8McRFU2P8LkV8iyLMT-rMkV1y1JIDUAWZLf-oN5lDqk9btI/s1600/3_59_3.jpg"><img style="margin: 0px 10px 10px 0px; width: 238px; float: left; height: 128px;" id="BLOGGER_PHOTO_ID_5532183057447711282" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjaBD9EG0qkRIfwOQmTbbPnrlGn-KLNEQiOUNtelBycOFRMG4CHQbTGya0-WodOYYeC-zvuosj_p9_xQFd6lsnvRfMjLWaH8McRFU2P8LkV8iyLMT-rMkV1y1JIDUAWZLf-oN5lDqk9btI/s400/3_59_3.jpg" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><div></div><br /><div> </div><div>The total charge contained within the Gaussian pillbox comes from the surface charge density of the conductor and given by <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhmlj9BSzL3pbFR7F1CXkUGmajUPtIh7pKemmk4JYOSSL24h9E7MpuV_BXrj9YvGTcGXe2Pqvq_VcaOcgGakLlCwN2VDCZmz1fW3vSVyJzsm2ein-xC4KGJJrm2mjPoKNSB3KolTkCgZvI/s1600/3_59_4.jpg"><img style="width: 44px; height: 21px;" id="BLOGGER_PHOTO_ID_5532180011616681058" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhmlj9BSzL3pbFR7F1CXkUGmajUPtIh7pKemmk4JYOSSL24h9E7MpuV_BXrj9YvGTcGXe2Pqvq_VcaOcgGakLlCwN2VDCZmz1fW3vSVyJzsm2ein-xC4KGJJrm2mjPoKNSB3KolTkCgZvI/s400/3_59_4.jpg" border="0" /></a>. Now applying Gauss's law we have,</div><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhxJ8wBIgNlLs-gt4NK122A82YJdONohIZ0jnX7OhwUNeffQ4wSVEptfwq7cChZOqlf7jNKuljtiJHNg2AMh31-xPHeFEFGAu2Qs28YjZpP4-FAGcEnaoJNcoue7-HnTUMiT2J3mrGNLKg/s1600/3_59_4.jpg"><img style="margin: 0px 10px 10px 0px; width: 201px; float: left; height: 133px;" id="BLOGGER_PHOTO_ID_5532184832841618898" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhxJ8wBIgNlLs-gt4NK122A82YJdONohIZ0jnX7OhwUNeffQ4wSVEptfwq7cChZOqlf7jNKuljtiJHNg2AMh31-xPHeFEFGAu2Qs28YjZpP4-FAGcEnaoJNcoue7-HnTUMiT2J3mrGNLKg/s400/3_59_4.jpg" border="0" /></a><br /><div></div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzXM-A8E1QZUmNUvnuODYaVY93evbQnUedqKqFMoXbMLqY9-kD9c9FqAX3w5yuvrz1HjNXpNDLgA0ir46Gm4D9kdyBxdvwI9W55vLK2K22gSIz6wl0_Q6dCbHEyT6CyCj6qmEFIUFGEYw/s1600/3_59_4.jpg"></a><br /><br /><br /><br /><div></div></div></div></div></div></div></div></div><br /><div></div><div> </div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-6768168311804051072010-10-20T08:56:00.006+05:302010-10-20T09:29:28.553+05:30Irodov Problem 3.58<div><br /><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjaiEpwhLhn7kTgRIi2VLKmDrbnR62pams-rjeJ8kaeOHS9c8wKG04bPC5rO6sj4HMw9O8_j5RGgAcQhVujzEc0Sdfh31BR3WdIZzv65G8ZOnr5tWrUXc15FUUU9aSd_562Kg4mJ7tcrR4/s1600/3.58_fig1.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 251px; FLOAT: left; HEIGHT: 400px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5529968646679727586" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjaiEpwhLhn7kTgRIi2VLKmDrbnR62pams-rjeJ8kaeOHS9c8wKG04bPC5rO6sj4HMw9O8_j5RGgAcQhVujzEc0Sdfh31BR3WdIZzv65G8ZOnr5tWrUXc15FUUU9aSd_562Kg4mJ7tcrR4/s400/3.58_fig1.jpg" /></a><br /><div><div><div>Let the dipole comprise of two charges <strong>q</strong> and <strong>-q</strong> seperated by a distance <strong>d</strong> as shown in the figure. Thus,</div><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiniwDYqsObF_LhDJ7er2Ebq7gVYXwf6ZMy7JZSdSXcqeAcxf66KrOuLmM-ApZIY0He9twhOidL39OlXSGRiCwJSVNXvQqpgXGTG1YaodqE-lW252ebkCM4qIV9tg-oNmQUv_LZ000kbe0/s1600/3.58_1.jpg"><img style="WIDTH: 82px; HEIGHT: 21px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5529968381994534018" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiniwDYqsObF_LhDJ7er2Ebq7gVYXwf6ZMy7JZSdSXcqeAcxf66KrOuLmM-ApZIY0He9twhOidL39OlXSGRiCwJSVNXvQqpgXGTG1YaodqE-lW252ebkCM4qIV9tg-oNmQUv_LZ000kbe0/s400/3.58_1.jpg" /></a><br /><br /><div><div>As described in problem <strong>3.54</strong>, using the method of images, the effect of having a dipole <strong>p</strong> at a distance <strong>l</strong> from the conducting plane is same has having another dipole <strong>-p</strong> at a distance <strong>2l</strong> from the original dipole, as illustrated in the figure.</div><br /><div>The force acting on the dipole is the sum of the forces acting on the two charges <strong>q</strong> and <strong>-q</strong> of the original (not the reflection) dipole due to the charges in the image dipole.</div><br /><div>Thus we have,</div><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBg2rSa5ySzDjRO8OXURnk09CD-GWrIASUbwpDlTmDIyk2N0bVpMQtxCT_TAKEYSYco_2-P5X_hLDhY-hyRnUAmUdOy2ze_Kl3JM0vNj1QPo0boqXY82LRQPo0hYJb8UyDr3NZrQw4yVA/s1600/3.58_2.jpg"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 383px; FLOAT: left; HEIGHT: 283px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5529972769939155650" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBg2rSa5ySzDjRO8OXURnk09CD-GWrIASUbwpDlTmDIyk2N0bVpMQtxCT_TAKEYSYco_2-P5X_hLDhY-hyRnUAmUdOy2ze_Kl3JM0vNj1QPo0boqXY82LRQPo0hYJb8UyDr3NZrQw4yVA/s400/3.58_2.jpg" /></a><br /><br /><br /><div></div><br /><br /><br /><br /><div></div><br /><br /><br /><br /><br /><div></div><br /><br /><br /><br /><br /><div></div></div></div></div></div></div>Unknownnoreply@blogger.com2tag:blogger.com,1999:blog-8917124841013563709.post-50832320729985361412010-08-22T07:16:00.006+05:302010-08-22T07:51:58.783+05:30Irodov Problem 3.57<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhukBgTTMLJ4i4rG3GKiTwFkziicf9u2hMS4tGE48JVfbuZaXQzWaXrfvUbjhKGrERq4ls8h3A9sgM6h-yD9BA3E8q1Omi2TO1NIbljqkYT9Kq8ZPeLwZ0JnOAyBy61_LI8Ak3TuGhkJOw/s1600/3.57_fig1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 400px; height: 280px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhukBgTTMLJ4i4rG3GKiTwFkziicf9u2hMS4tGE48JVfbuZaXQzWaXrfvUbjhKGrERq4ls8h3A9sgM6h-yD9BA3E8q1Omi2TO1NIbljqkYT9Kq8ZPeLwZ0JnOAyBy61_LI8Ak3TuGhkJOw/s400/3.57_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5508049875159802498" border="0" /></a>As described in Problem <span style="font-weight: bold;">3.54</span>, an infinite conducting surface can be treated as a mirror. In this problem there are two perpendicular semi-infinite conducting planes. So considering them as mirrors we will see two infinite perpendicular planes as shown in the figure. Similar to hall of mirrors, now the single positive charge will have <span style="font-weight: bold;">3</span> other image reflections as shown in the figure.<br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjsgEdWPSYMURf6487winli2aYim6rr-iAPXYCyPczsqa3oltvj6dMlxa6yPulT6cw4AxuLZiGZlKdk6GBlF2WSXHr01nXy6O2vjD9wkze-8GiY0dP_MwsVCT0JXrIfFzgvsW06DhH6MGs/s1600/3.57_fig2.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 320px; height: 272px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjsgEdWPSYMURf6487winli2aYim6rr-iAPXYCyPczsqa3oltvj6dMlxa6yPulT6cw4AxuLZiGZlKdk6GBlF2WSXHr01nXy6O2vjD9wkze-8GiY0dP_MwsVCT0JXrIfFzgvsW06DhH6MGs/s320/3.57_fig2.jpg" alt="" id="BLOGGER_PHOTO_ID_5508051629033949554" border="0" /></a><br />The net force on the positive charge is given by the forces due to the three image charges as shown in the figure as,<br /><br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiiSHsqnnEkicPUtIwWqBBaEV4wW2_GkyryjWfD2OeJRH_YPHuLfCFVhZhrm2pWCf45TYYiqasOCEbrY7vfgtsbHEqc3O84wSp6oQ0D7NPfxWpuEIue7TgpuU7lSUBMj8nN5rfRRem9Beg/s1600/3.57_1.jpg"><img style="cursor: pointer; width: 400px; height: 138px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiiSHsqnnEkicPUtIwWqBBaEV4wW2_GkyryjWfD2OeJRH_YPHuLfCFVhZhrm2pWCf45TYYiqasOCEbrY7vfgtsbHEqc3O84wSp6oQ0D7NPfxWpuEIue7TgpuU7lSUBMj8nN5rfRRem9Beg/s400/3.57_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5508053426508257842" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-15088847819808759452010-08-22T06:36:00.010+05:302010-08-22T07:14:55.171+05:30Irodov Problem 3.56<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhogkv-Auq7PtQzTtdwjpKinlHkEMIUpmsWTOmh0h7Ni2t3ywK546vm3_tOMciJyptS93J1SS7BKQjx_rug00HWKNQlGsO9gzsOjBV2TFidKxj5inXJHyH7JQIm8pI86OFRHuoKuitmnSo/s1600/3.56_fig1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 400px; height: 152px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhogkv-Auq7PtQzTtdwjpKinlHkEMIUpmsWTOmh0h7Ni2t3ywK546vm3_tOMciJyptS93J1SS7BKQjx_rug00HWKNQlGsO9gzsOjBV2TFidKxj5inXJHyH7JQIm8pI86OFRHuoKuitmnSo/s400/3.56_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5508034741281679154" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br />Following the same argument as in Problem <span style="font-weight: bold;">3.54</span>, we can treat the infinite conducting plane as a plane mirror. Hence, for each charge and equal and opposite image charge appears at the other end of the conducting plane as shown in the figure.<br /><br /><span style="font-weight: bold;">a)</span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgmbPlzgOAuSrpP_OpZJTAAAHjvE41QtAxykqSgRiSeWXNAG44HE8mPg3Ezw2_Ta3QgZL3VLV2fMTAcAVcFKXfg7YxhUfocfvqn3h1BBiJc5xzed69G9EDQMnlP97Mpm8hqnaEJlopPhn0/s1600/3.56_fig2.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 315px; height: 320px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgmbPlzgOAuSrpP_OpZJTAAAHjvE41QtAxykqSgRiSeWXNAG44HE8mPg3Ezw2_Ta3QgZL3VLV2fMTAcAVcFKXfg7YxhUfocfvqn3h1BBiJc5xzed69G9EDQMnlP97Mpm8hqnaEJlopPhn0/s320/3.56_fig2.jpg" alt="" id="BLOGGER_PHOTO_ID_5508035393001607842" border="0" /></a><br />There are three forces acting on the positive charge <span style="font-weight: bold;">q</span> as shown in the figure. The attactive forces <span style="font-weight: bold;">F2</span> and <span style="font-weight: bold;">F1</span> from the real and the image negative charges located at distances <span style="font-weight: bold;">l</span> each and the repulsive force <span style="font-weight: bold;">F3</span> from the image charge located diagonally. The net force is given by,<br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjAzS3xn1K5Z-qDdItxK8k-ZwJ5yIpMgfQdAHnBjDFdWgqsZbJigXFF1ctUqU3_oAHqPxWJG27RCntCRKl7BD-M-7l0Tydfdswwyo5vw-0b6GolLoILgm4i-DL-2zWX5_nzwxfzGerwqeE/s1600/3.56_1.jpg"><img style="cursor: pointer; width: 400px; height: 195px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjAzS3xn1K5Z-qDdItxK8k-ZwJ5yIpMgfQdAHnBjDFdWgqsZbJigXFF1ctUqU3_oAHqPxWJG27RCntCRKl7BD-M-7l0Tydfdswwyo5vw-0b6GolLoILgm4i-DL-2zWX5_nzwxfzGerwqeE/s400/3.56_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5508039430219292402" border="0" /></a><br /><br /><span style="font-weight: bold;">b)</span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjY5843zOM2xBeFtTEuCGdYVEr-0Z-ZyNLM2N16kR1nCNxNjp_EA40wmZTIMnGAbCRL0FCHC5iZoi8cCeUVCzlTI4903G8L2pQgpJ-ZQeRJyOb8JVLkD32M1KZl_Ttm_uzh89Qv_2mo_Z0/s1600/3.56_fig3.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 320px; height: 314px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjY5843zOM2xBeFtTEuCGdYVEr-0Z-ZyNLM2N16kR1nCNxNjp_EA40wmZTIMnGAbCRL0FCHC5iZoi8cCeUVCzlTI4903G8L2pQgpJ-ZQeRJyOb8JVLkD32M1KZl_Ttm_uzh89Qv_2mo_Z0/s320/3.56_fig3.jpg" alt="" id="BLOGGER_PHOTO_ID_5508041459506095570" border="0" /></a><br /><br />The electric field at the mid point of the two charges can be obtained by considering the electric field due to all the four charges - the two real charges and the two image charges. This is given by,<br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxfTR7pAep_K10beEVgiTgjf8gUnXXueG8dBOsTwLTqmPAS6Z77cc3Bcj9I9fd_0_GKkxuwq6C-ULgzeh5TTTLBYHjASc6sPYTC3qzXRY3wokRTmVi9PPS5u6pMkj6KoLaw8XUSedCUIg/s1600/3.56_2.jpg"><img style="cursor: pointer; width: 352px; height: 323px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxfTR7pAep_K10beEVgiTgjf8gUnXXueG8dBOsTwLTqmPAS6Z77cc3Bcj9I9fd_0_GKkxuwq6C-ULgzeh5TTTLBYHjASc6sPYTC3qzXRY3wokRTmVi9PPS5u6pMkj6KoLaw8XUSedCUIg/s400/3.56_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5508044182786118722" border="0" /></a>Unknownnoreply@blogger.com6tag:blogger.com,1999:blog-8917124841013563709.post-52100087427611574362010-08-21T07:25:00.008+05:302010-08-21T08:24:02.678+05:30Irodov Problem 3.55<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgr0PPaybuQnlSxAUOxETgdls6RkV0tlNVFzVyzklTR7LPsaTQ_n8mWBNKoCpC2pziKHpW5GyyQHH870VJs2zKuDFcMLcldgVA0jqqMFS_rATtb4j_aZZ6IR-gWPuhSO35rFsrNulAfu6g/s1600/3.55_fig1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 400px; height: 208px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgr0PPaybuQnlSxAUOxETgdls6RkV0tlNVFzVyzklTR7LPsaTQ_n8mWBNKoCpC2pziKHpW5GyyQHH870VJs2zKuDFcMLcldgVA0jqqMFS_rATtb4j_aZZ6IR-gWPuhSO35rFsrNulAfu6g/s400/3.55_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5507676589685607442" border="0" /></a>As discussed is Problem <span style="font-weight: bold;">3.54</span>, a positive charge located at a distance <span style="font-weight: bold;">x</span> from the infinite plane is equivalent to replacing it with a negative charge of equal magnitude at a distance <span style="font-weight: bold;">x</span> on the other side of the plane as shown in the figure.<br /><br />When the charge is at a distance <span style="font-weight: bold;">x</span> then the force acting on the charge trying it pull it towards the conducting plane is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJJk5jiLWwB3Td3dJu8S4W0f4kvT3T6nl_AJKlKfdyi4t3CEguGInxYAps2fDu6yA0GckSB1_XViwMIR9aKS5GedWlP51Vt5YmYJ6niYF0F-_ym-50038EheBU0nYJ2zvxSF8p7KOowW4/s1600/3.55_1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 130px; height: 48px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJJk5jiLWwB3Td3dJu8S4W0f4kvT3T6nl_AJKlKfdyi4t3CEguGInxYAps2fDu6yA0GckSB1_XViwMIR9aKS5GedWlP51Vt5YmYJ6niYF0F-_ym-50038EheBU0nYJ2zvxSF8p7KOowW4/s400/3.55_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5507680300146228290" border="0" /></a><br /><br /><br /><br />The work done in moving it away to infinity is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh-VF5cmt_T7p6Mdykka7BfDC3f5_3AN5n-RFQeZsRmXbC1NFFFysbaUQP7QoOQYODVkixSP4kKsRssm7E9nte6K1CfyadUJiwvOrBxWKhpI3gMwjM6ef0lgW1b39mvRToBcSUGRYLAn9E/s1600/3.55_2.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 144px; height: 154px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh-VF5cmt_T7p6Mdykka7BfDC3f5_3AN5n-RFQeZsRmXbC1NFFFysbaUQP7QoOQYODVkixSP4kKsRssm7E9nte6K1CfyadUJiwvOrBxWKhpI3gMwjM6ef0lgW1b39mvRToBcSUGRYLAn9E/s400/3.55_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5507690621043132530" border="0" /></a>Unknownnoreply@blogger.com3tag:blogger.com,1999:blog-8917124841013563709.post-61811598295858607972010-07-27T04:41:00.009+05:302010-07-27T05:44:49.269+05:30Irodov Problem 3.54<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgjvV_nTPzg7g7nIQ_pOEaoFJPQd4V3xyppXuIBhWMbOaglioVwnbWyLSTF_qkP2AhOXqUFrXTXqZ7RdsplyV1JOCQKSminxHhRh64ppzMbqGl5mTqqw5BAb_PS-2GvLuOX4gfKNdsMWyI/s1600/3.54_fig1.jpg"><img id="BLOGGER_PHOTO_ID_5498356688946072210" style="WIDTH: 400px; CURSOR: hand; HEIGHT: 226px" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgjvV_nTPzg7g7nIQ_pOEaoFJPQd4V3xyppXuIBhWMbOaglioVwnbWyLSTF_qkP2AhOXqUFrXTXqZ7RdsplyV1JOCQKSminxHhRh64ppzMbqGl5mTqqw5BAb_PS-2GvLuOX4gfKNdsMWyI/s400/3.54_fig1.jpg" border="0" /></a><br /><div><br /><div>We shall use the Method of Images to solve this problem. Since this is the first problem that uses method of images, I will try and explain the idea behind Method of images first.</div><br /><div>A perfect conductor is an equipotential surface. In other words there cannot be any potential difference between two points on a perfect conducting surface. Why is this so? Even the slighest potential difference along the surface of a perfect conductor will immideately result in a large current flow that will neutralize the potential difference. This is very similar to thermal conductors. You cannot sustain a temperature difference between any two points on a perfect thermal conductor since immideately heat will flow between the points until these two points are at the same temperature. </div><br /><div>Equipotential surface in turn implies that there cannot be any electric field along the surface of the (tangential to the surface) of the conductor. In other words, in our problem, since the infinite conductor is along the <strong>x-y</strong> plane, <strong>Ex = Ey = 0</strong>. </div><br /><div>If there were no infinite conducting plane, the electric field lines would move radially outwards from the charge <strong>q</strong>. However, the presence of the inifinite conducting plane distorts the electric field lines in such a manner that <strong>Ex = Ey = 0</strong> at all points on the <strong>x-y</strong> plane. How does this happen? What really happens is that the electric field due to the charge <strong>q</strong>, causes the electrons in the conductor to move about and distribute themelves in such a manner that so as to make <strong>Ex = Ey = 0 </strong>all along the surface of the conductor. In other words, the charge <strong>q</strong>, induces a surface charge density distribution along the conductor's surface in a manner so as to ensure that <strong>Ex = Ey = 0</strong> all along its surface. </div><div></div><div></div><div></div><div>One way to solve the problem is to determine the necessary charge distribution on the conductor's surface which results in <strong>Ex = Ey = 0</strong> and then use this to solve the problem. A simpler way to solve the problem is using the "<strong><em><a href="http://en.wikipedia.org/wiki/Method_of_image_charges">Method of Images</a></em></strong>". The idea is behind method of images is to replace the conducting surface by other imaginary charges in such a way that create the condition that tangential component of the electric field along the conducting surface is zero. This can be done by thinking of the conducting surface as a mirror. Imagine a person at the position of the charge <strong>q</strong>. What will he see? He will see another charge at the other side of the mirror that is at a distance equal to <strong>q'</strong>s distance from the mirror. In method of images, each time you reflect a charge ithe sign of the charge is also negated (think of this as similar to lateral inversion phenomenon in mirror reflections). In other words we have an image charge of <strong>-q</strong>, on the other side of the conductor as shown in the figure. </div><br /><div>Now let us examine how placing this imaginary charge helps us solve the problem. The electric filed lines between the charges <strong>q</strong> and <strong>-q</strong> are shown in the figure. The interesting thing to notice is that the field lines are perpendicular to the <strong>x-y</strong> plane at all points on the <strong>x-y</strong> plane. In other words <strong>Ex = Ey = 0</strong> all over the infinite conducting plane. Thus, by placing the imaginary charge <strong>-q</strong> instead of the conducting plane we have created the same situation as it would have been if there were an infinite conducting plane. Now we can forget about the infinite conducting plane so solve the problem thinking that there is an imaginary image charge <strong>-q</strong>.</div><br /><div>When the charge <strong>q</strong> is at a distance <strong>l</strong> from the conducting plane, the image charge <strong>-q</strong> is at a distance <strong>l</strong> from the conducting plane on the other side. The distance between the charges is thus <strong>2l</strong>. The force <strong>F</strong> acting on the charge <strong>q</strong> is given by,<br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjTq5K_LPxCoDuho823lbnLstohu73z-LekDAniVjrtKzSkUVHPSutzh2lwep5OLPUOAtADvvcdpJjWLmTsXl55fHiKmK99X1tzGOFrnPdJev8mCrHhB8ej0gw0wKUKi3ML_Ol8Ug1YbWg/s1600/3.54_1.jpg"><img id="BLOGGER_PHOTO_ID_5498369961740067170" style="WIDTH: 135px; CURSOR: hand; HEIGHT: 120px" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjTq5K_LPxCoDuho823lbnLstohu73z-LekDAniVjrtKzSkUVHPSutzh2lwep5OLPUOAtADvvcdpJjWLmTsXl55fHiKmK99X1tzGOFrnPdJev8mCrHhB8ej0gw0wKUKi3ML_Ol8Ug1YbWg/s400/3.54_1.jpg" border="0" /></a></div><br /><div></div><div>Since the string is elongated by a length x, the restoring force acting on the charge due to the string is given by <strong>kx</strong>. Since the charge is static under the action of the two forces (there is no acceleration) we have,</div><br /><div></div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiypjZiQ-YjAswRDb0dosAwijRZbw96fBMYPm_mrWDaka-5SRPNp_FhS8kOPETjAOIGgVvTBs8gR5TJ4viby5gtZ_uIdW9ZAOockfObnDOL5g3eX3eig7IaHhjvS4aYcbCnH3ZMyxTyiPg/s1600/3.54_2.jpg"><img id="BLOGGER_PHOTO_ID_5498371334755575570" style="WIDTH: 101px; CURSOR: hand; HEIGHT: 54px" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiypjZiQ-YjAswRDb0dosAwijRZbw96fBMYPm_mrWDaka-5SRPNp_FhS8kOPETjAOIGgVvTBs8gR5TJ4viby5gtZ_uIdW9ZAOockfObnDOL5g3eX3eig7IaHhjvS4aYcbCnH3ZMyxTyiPg/s400/3.54_2.jpg" border="0" /></a></div>Unknownnoreply@blogger.com14tag:blogger.com,1999:blog-8917124841013563709.post-72798699402810769432010-07-05T09:13:00.002+05:302010-07-05T09:14:38.122+05:30Irodov Problem 3.53From the problem definition,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg3a9UcC34lDmudH1etN9F1tcyUm_l-nifGgiCgzI7spBqLDj-2wE5WL5AjWhTq5HFPO_xcnzkw70qbcxtsqesdHl57oSOJcsIuMfyAv62hSGkis4F3uhT3kemexXU6D02mkV7aOsanrzM/s1600/3.53_1.jpg"><img style="cursor: pointer; width: 156px; height: 59px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg3a9UcC34lDmudH1etN9F1tcyUm_l-nifGgiCgzI7spBqLDj-2wE5WL5AjWhTq5HFPO_xcnzkw70qbcxtsqesdHl57oSOJcsIuMfyAv62hSGkis4F3uhT3kemexXU6D02mkV7aOsanrzM/s400/3.53_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5490262896805999650" border="0" /></a><br /><br />We know that,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgDeGkHjCLKuoSs4UsVQfaltTg2lDraNCY2o6kR_8MpBUEikGGSruGzdf6a2lla_mbu5TfsdtSthrx2XWjCA6Muw7-oj2IQ8Lw5JdWHagwvtrYJdpspc4z73Y6Zb3rz_pGp4UUeSm5Q31Y/s1600/3.53_2.jpg"><img style="cursor: pointer; width: 198px; height: 265px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgDeGkHjCLKuoSs4UsVQfaltTg2lDraNCY2o6kR_8MpBUEikGGSruGzdf6a2lla_mbu5TfsdtSthrx2XWjCA6Muw7-oj2IQ8Lw5JdWHagwvtrYJdpspc4z73Y6Zb3rz_pGp4UUeSm5Q31Y/s400/3.53_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5490262967697659122" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-80801727596642108182010-07-05T09:08:00.005+05:302010-07-05T09:13:02.084+05:30Irodov Problem 3.52<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhRoPVQ7zIhfzwk1YcWCru6o7jfGWDc6Cr0AztBqFurmWGPUlPILx7NHssWmRh7JvCKTdOaZVMm8h3eP8HLaQFooFLNtl8w96hYDxggtpT0gLUP8cVDETyIFWNNXB8kDBjsViCEyGDCCiI/s1600/3.52_fig1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 233px; height: 286px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhRoPVQ7zIhfzwk1YcWCru6o7jfGWDc6Cr0AztBqFurmWGPUlPILx7NHssWmRh7JvCKTdOaZVMm8h3eP8HLaQFooFLNtl8w96hYDxggtpT0gLUP8cVDETyIFWNNXB8kDBjsViCEyGDCCiI/s400/3.52_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5490261770786154354" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Let the left plate be at a higher potential than the right plate. So the direction of the electric field will be from left plate to right plate. Further let us suppose that the Electric field at the left plate is equal to zero. We know that,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigbBm2Y5ew_bvTOqTJTrJFjV-GtfKeoE_siPJ1r0V1JuCbIe7-U_E5YF42hxQOHKyDl2WRVVfGS4e_cbrJBFeuP0PIZgw2f65xDfly6ckStd7a7OZMRT_EOVDcj_45V7I2pCTCdrul8DQ/s1600/3.52_1.jpg"><img style="cursor: pointer; width: 260px; height: 135px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigbBm2Y5ew_bvTOqTJTrJFjV-GtfKeoE_siPJ1r0V1JuCbIe7-U_E5YF42hxQOHKyDl2WRVVfGS4e_cbrJBFeuP0PIZgw2f65xDfly6ckStd7a7OZMRT_EOVDcj_45V7I2pCTCdrul8DQ/s400/3.52_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5490261894398611634" border="0" /></a><br /><br />Since the parallel plates are large, the direction of electric field will be along the x-direction (from the left plate to the right one). This means that,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5X1HheXy92i3ROEbaGWg4hN_VJGrIGosnyb28DBVkrnofvjCTVCfBxZ65P1BawEqZZuETN9B6gaq5PXU5GKO37USIrSgoaDf-EChjcpUIjB-DVR6jfhY8oHiOLjAJvpg2gqI5h8pHTV4/s1600/3.52_2.jpg"><img style="cursor: pointer; width: 92px; height: 101px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5X1HheXy92i3ROEbaGWg4hN_VJGrIGosnyb28DBVkrnofvjCTVCfBxZ65P1BawEqZZuETN9B6gaq5PXU5GKO37USIrSgoaDf-EChjcpUIjB-DVR6jfhY8oHiOLjAJvpg2gqI5h8pHTV4/s400/3.52_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5490262047563166946" border="0" /></a><br /><br />Hence, we can determine the electric field as a function of distance form the left plate as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhizVHywUOuTx5QJ2gVUhI0ujI0UGGXiUjvt3KTFpHFdNtLIbJwePqa6EV_5gVTUXp3qrO1uzeuqbmuEffU8zgN9avhyphenhyphenBIkVHt0MvSp_jyaLlR-XYsJ1uWgTKo5bP5MR2zxjhdNaYZ_Quc/s1600/3.52_3.jpg"><img style="cursor: pointer; width: 116px; height: 101px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhizVHywUOuTx5QJ2gVUhI0ujI0UGGXiUjvt3KTFpHFdNtLIbJwePqa6EV_5gVTUXp3qrO1uzeuqbmuEffU8zgN9avhyphenhyphenBIkVHt0MvSp_jyaLlR-XYsJ1uWgTKo5bP5MR2zxjhdNaYZ_Quc/s400/3.52_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5490262161538227266" border="0" /></a><br /><br />At the right plate <span style="font-weight: bold;">x = d</span>, and and so the electric field at the right plate is,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjx4kyduOtnBPi1xvlq3X0bHh4X1fW9gn2IRMtzzwWwLZ5xMIR-TOSo4fBZ9PzKepVhWI4V_a2ZAyGwQS1jGkEhHeynKpZCRyBkg5-InsIzu2nC78wzce-3QfmLP5a1L9WDHNRCyctIrI8/s1600/3.52_4.jpg"><img style="cursor: pointer; width: 62px; height: 23px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjx4kyduOtnBPi1xvlq3X0bHh4X1fW9gn2IRMtzzwWwLZ5xMIR-TOSo4fBZ9PzKepVhWI4V_a2ZAyGwQS1jGkEhHeynKpZCRyBkg5-InsIzu2nC78wzce-3QfmLP5a1L9WDHNRCyctIrI8/s400/3.52_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5490262277060291122" border="0" /></a><br /><br />We also know that,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiE-AM66hNBr5MQAisArmn8fFXKqp2cYZLr6uS3u2UShPoVJkS8trOWELXxpo2p7xoZ8d6KE_5ICMSu5zc9Z96CYMi2dy3q8EwnGb0Dj2IX1G9DUAzN_Wj9pnSKLFZd2FNmYSJsOgmPEzk/s1600/3.52_5.jpg"><img style="cursor: pointer; width: 122px; height: 217px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiE-AM66hNBr5MQAisArmn8fFXKqp2cYZLr6uS3u2UShPoVJkS8trOWELXxpo2p7xoZ8d6KE_5ICMSu5zc9Z96CYMi2dy3q8EwnGb0Dj2IX1G9DUAzN_Wj9pnSKLFZd2FNmYSJsOgmPEzk/s400/3.52_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5490262417509849522" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-61221260929777752812010-07-05T09:06:00.003+05:302010-07-05T09:08:06.969+05:30Irodov Problem 3.51<span style="text-decoration: underline;">From the problem definition we have,<br /><br /></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEijaPt9kxVUZn8Bq205OoWlBUAKOhZ2I9d864fQJHsgROZKtVvReu37jcCn4-qWWwiJTsVTcXFJP8d_rO4NzKExRKvy7_6YZSzdOMIKxJYrtL9-UmYeNjLYwgxNWHhuE7ScBi2IGiOLX-s/s1600/3.51_1.jpg"><img style="cursor: pointer; width: 318px; height: 245px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEijaPt9kxVUZn8Bq205OoWlBUAKOhZ2I9d864fQJHsgROZKtVvReu37jcCn4-qWWwiJTsVTcXFJP8d_rO4NzKExRKvy7_6YZSzdOMIKxJYrtL9-UmYeNjLYwgxNWHhuE7ScBi2IGiOLX-s/s400/3.51_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5490261010390203218" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-36058515183517663492010-07-03T09:48:00.006+05:302010-07-03T09:52:26.372+05:30Irodov Problem 3.50This problem is similar to <span style="font-weight: bold;">3.48</span> and <span style="font-weight: bold;">3.49</span>, so I shall outline this solution without as much explanation.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgPhIgsGi-u0It5u5ZOZGU412283K-D2vy1v4bZHiMz_QkqqwP0DLPQfGppCKqOuF8lTf2RgyNVwpcWQLmqUs-G03nyYe7np1tl3gyRnWxUs1APd0Iu1_YEpwSkqw_Mtw7_CEdr2AccROg/s1600/3.50_1.jpg"><img style="cursor: pointer; width: 194px; height: 178px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgPhIgsGi-u0It5u5ZOZGU412283K-D2vy1v4bZHiMz_QkqqwP0DLPQfGppCKqOuF8lTf2RgyNVwpcWQLmqUs-G03nyYe7np1tl3gyRnWxUs1APd0Iu1_YEpwSkqw_Mtw7_CEdr2AccROg/s400/3.50_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5489529809007239970" border="0" /></a><br /><br />Integrating <span style="font-weight: bold;">(1a),(1b),(1c)</span> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0M4olpUhFcjGc1gUV7gCZGkEJl9UD57g6nJ5dy6LipJILqmzEgRR87yCN2j4lmheE2lvq8wP1NIb8ZQTl_ru8SThnGxb0PjkB_PylDmSZvyUZTqGMELtbRyVpNyBqfe19nM_hMWSmrF8/s1600/3.50_2.jpg"><img style="cursor: pointer; width: 222px; height: 92px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0M4olpUhFcjGc1gUV7gCZGkEJl9UD57g6nJ5dy6LipJILqmzEgRR87yCN2j4lmheE2lvq8wP1NIb8ZQTl_ru8SThnGxb0PjkB_PylDmSZvyUZTqGMELtbRyVpNyBqfe19nM_hMWSmrF8/s400/3.50_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5489530013019104306" border="0" /></a><br /><br />Subtracting <span style="font-weight: bold;">(2a)</span> and <span style="font-weight: bold;">(2b)</span> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghc2S7HAmO6OvsQ1oRO-LD3_b0joLCrHlcVvVNgEMcGGCzUkcKSntnDDYMhmNX12r8ApRgRNPnsmR8dYYvpT06s179BuqxGQ8ZqnZFE8JVX0ujfxH3OC6HeMHzSQHaBPNm3lgoQbda55g/s1600/3.50_3.jpg"><img style="cursor: pointer; width: 196px; height: 156px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghc2S7HAmO6OvsQ1oRO-LD3_b0joLCrHlcVvVNgEMcGGCzUkcKSntnDDYMhmNX12r8ApRgRNPnsmR8dYYvpT06s179BuqxGQ8ZqnZFE8JVX0ujfxH3OC6HeMHzSQHaBPNm3lgoQbda55g/s400/3.50_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5489530199641645330" border="0" /></a><br /><br />Now subtracting <span style="font-weight: bold;">(4)</span> from <span style="font-weight: bold;">(2c)</span> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiVR7HDatMmxCIZ0OpOtXBVBV217N74Cbb6kM5Vbo8rKyvrhhp7HxIInHz3nalkG9rc51ALb4RRmkShjpJzVu7-t3_VLgSPTJs0ccPXOiFbMJt0nyu42qCqMQrJyuHnRFdX0eYohV1F7GY/s1600/3.50_4.jpg"><img style="cursor: pointer; width: 154px; height: 156px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiVR7HDatMmxCIZ0OpOtXBVBV217N74Cbb6kM5Vbo8rKyvrhhp7HxIInHz3nalkG9rc51ALb4RRmkShjpJzVu7-t3_VLgSPTJs0ccPXOiFbMJt0nyu42qCqMQrJyuHnRFdX0eYohV1F7GY/s400/3.50_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5489530350810834866" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-44848650065542699842010-07-03T09:45:00.002+05:302010-07-03T09:48:10.260+05:30Irodov Problem 3.49This problem is similar to <span style="font-weight: bold;">3.48</span>, so I shall outline this solution without as much explanation as <span style="font-weight: bold;">3.48</span>.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEheqlIXqStzV8dnp6pmoYfcqWFrgxsZimgNNcWj1jJBOQH-QoPBxZN_8RftNCL4yhWrTQrX7CUd8W6ohrrcO3BqABUVOgaJ7PRZXE3Zz4FmUsRKl82IyActKePo3ACvcTryxuHa98Li8eA/s1600/3.49_1.jpg"><img style="cursor: pointer; width: 175px; height: 175px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEheqlIXqStzV8dnp6pmoYfcqWFrgxsZimgNNcWj1jJBOQH-QoPBxZN_8RftNCL4yhWrTQrX7CUd8W6ohrrcO3BqABUVOgaJ7PRZXE3Zz4FmUsRKl82IyActKePo3ACvcTryxuHa98Li8eA/s400/3.49_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5489529107936050482" border="0" /></a><br /><br />Integrating we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgIXpqD9l4VtJ_x2pTeWOM2JqJNx8KbXZr7AiOw3ky7uysXtzFyyCy29-kbO7tMuN8gxwKoDkiPuivnlxIZnfJYp_euGYMjJsQ-GxqI60Q-IuXNndA7szL4-Bp1Bm9md2euHFrhYo1ocms/s1600/3.49_2.jpg"><img style="cursor: pointer; width: 241px; height: 92px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgIXpqD9l4VtJ_x2pTeWOM2JqJNx8KbXZr7AiOw3ky7uysXtzFyyCy29-kbO7tMuN8gxwKoDkiPuivnlxIZnfJYp_euGYMjJsQ-GxqI60Q-IuXNndA7szL4-Bp1Bm9md2euHFrhYo1ocms/s400/3.49_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5489529192423916354" border="0" /></a><br /><br />Subtracting <span style="font-weight: bold;">(2a)</span> and <span style="font-weight: bold;">(2b)</span> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8ImdSALYgTS7Mbh8xT3G6Nn2SqXH0Owl13vrm5Z2g-XKqJvnz-CvUXddJ_bzxFctL006SWLfFib-wLa_WSVxAdwIMLSyTr4PJ6h7Ov2eCRoBDvK2pWll726vyBySorigHy0rbKag0CsY/s1600/3.49_3.jpg"><img style="cursor: pointer; width: 216px; height: 156px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8ImdSALYgTS7Mbh8xT3G6Nn2SqXH0Owl13vrm5Z2g-XKqJvnz-CvUXddJ_bzxFctL006SWLfFib-wLa_WSVxAdwIMLSyTr4PJ6h7Ov2eCRoBDvK2pWll726vyBySorigHy0rbKag0CsY/s400/3.49_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5489529317028106210" border="0" /></a><br /><br />Now subtracting <span style="font-weight: bold;">(4)</span> from <span style="font-weight: bold;">(2c)</span> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8tCvQ3ngPm0Neb4wDbJcjNDLnfOCaKqx0s84n1hnrmlK_8U89QuJLd-45WjuvTWgEe0bXB5IeZxNtj_szrpFozU_eW30QXEJFdiQ2wiBoV2PD_vZOL0fjmQiIufxg3nANuRNE6OPSpEM/s1600/3.49_4.jpg"><img style="cursor: pointer; width: 208px; height: 121px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8tCvQ3ngPm0Neb4wDbJcjNDLnfOCaKqx0s84n1hnrmlK_8U89QuJLd-45WjuvTWgEe0bXB5IeZxNtj_szrpFozU_eW30QXEJFdiQ2wiBoV2PD_vZOL0fjmQiIufxg3nANuRNE6OPSpEM/s400/3.49_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5489529438747562594" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-5517276443921486982010-07-03T09:37:00.006+05:302010-07-03T09:45:26.297+05:30Irodov Problem 3.48We know that,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhx55pr6ih_GCIWtfv59qP8SQEOv5lJIBWukMgySqrUyeWRVzgfTsnvu91O2lGwrugqjzsDFm38-0IafRkmiEsiYxs-GK7BTvX6zhxZ6A-o3bnTiDz8CY8B3PuYeWkBEtfznMz4qHdHbg/s1600/3.48_1.jpg"><img style="cursor: pointer; width: 196px; height: 72px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhx55pr6ih_GCIWtfv59qP8SQEOv5lJIBWukMgySqrUyeWRVzgfTsnvu91O2lGwrugqjzsDFm38-0IafRkmiEsiYxs-GK7BTvX6zhxZ6A-o3bnTiDz8CY8B3PuYeWkBEtfznMz4qHdHbg/s400/3.48_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5489526975407029250" border="0" /></a><br /><br />From the problem definition we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj2Mo4JeS7Bq4PDdByf2ZuVJiNWhsfRA63QxPkqiuB2Wd87T5mJL6FPQCT_Y2oMXBNQoLlOEgT8wbqoKLuyER_YREy96T3DKz3487EAK2dqBWET_vok7LW7J4DxoQRl1N9FgHRkEGdhz-U/s1600/3.48_2.jpg"><img style="cursor: pointer; width: 106px; height: 140px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj2Mo4JeS7Bq4PDdByf2ZuVJiNWhsfRA63QxPkqiuB2Wd87T5mJL6FPQCT_Y2oMXBNQoLlOEgT8wbqoKLuyER_YREy96T3DKz3487EAK2dqBWET_vok7LW7J4DxoQRl1N9FgHRkEGdhz-U/s400/3.48_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5489527082643348530" border="0" /></a><br /><br />Integrating both sides of each of the set of simultaneous equations <span style="font-weight: bold;">(2a), (2b) (2c) </span>we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiK7ydqZIgSCb423KrDBCltyDIYCKIe6UQQx5uU3ZbbQLDNyAYEdos6a1WY5liAfk_tKoxe1Wie41lIzqxKFYt1WZTNoVj0o6cqjtPWOoXi0PGkAGzr1tkrmrec7TgOwoKvDdC_i-7DdU/s1600/3.48_4.jpg"><img style="cursor: pointer; width: 171px; height: 93px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiK7ydqZIgSCb423KrDBCltyDIYCKIe6UQQx5uU3ZbbQLDNyAYEdos6a1WY5liAfk_tKoxe1Wie41lIzqxKFYt1WZTNoVj0o6cqjtPWOoXi0PGkAGzr1tkrmrec7TgOwoKvDdC_i-7DdU/s400/3.48_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5489527499858415698" border="0" /></a><br /><br />Now we have to find <span style="font-weight: bold;">f1(y,z), f2(x,z)</span> and <span style="font-weight: bold;">f3(x,y)</span> such that all the equations <span style="font-weight: bold;">(3a),(3b)</span> and <span style="font-weight: bold;">(3c)</span> are simultaneously satisfied.<br /><br />Subtracting <span style="font-weight: bold;">(3a)</span> and <span style="font-weight: bold;">(3b)</span> we get,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgMWNIQmD-aXpNiGzFBQmC2AbRgRv-YwlcwuLWvb_cOI1zpH92839_c5Aq-H2KOZ4w4nMzvz-lstxIaIoF0wSMozTJ24hv2kFefyKxvxNMlW8FEehoknpcK5c10KB70gw7GkKy-cfW5NHg/s1600/3.48_5.jpg"><img style="cursor: pointer; width: 142px; height: 22px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgMWNIQmD-aXpNiGzFBQmC2AbRgRv-YwlcwuLWvb_cOI1zpH92839_c5Aq-H2KOZ4w4nMzvz-lstxIaIoF0wSMozTJ24hv2kFefyKxvxNMlW8FEehoknpcK5c10KB70gw7GkKy-cfW5NHg/s400/3.48_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5489527819227958786" border="0" /></a><br /><br />Equation <span style="font-weight: bold;">(4a)</span> can only be true if<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigK1-iV7HZ4Qjz36PCQljvgmZVnVeWn7kW_swt0V79ID03RVlQr4JiSb5QGYK2ll9E_Uc9KqcDji7YdohPmnCgAgVkArvxXdHBBYvhax-RS6EtxsV57klr_J4GetJOWwHBlyHS7sMRW24/s1600/3.48_6.jpg"><img style="cursor: pointer; width: 175px; height: 24px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigK1-iV7HZ4Qjz36PCQljvgmZVnVeWn7kW_swt0V79ID03RVlQr4JiSb5QGYK2ll9E_Uc9KqcDji7YdohPmnCgAgVkArvxXdHBBYvhax-RS6EtxsV57klr_J4GetJOWwHBlyHS7sMRW24/s400/3.48_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5489527985027273122" border="0" /></a><br /><br />In other words <span style="font-weight: bold;">f1</span> and <span style="font-weight: bold;">f2</span> must be only functions of <span style="font-weight: bold;">z</span> and must be equal and opposite. Hence, now we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjgDlL2aAdUG8SvUmlCKtH4K8GAUdyt0G5xWzcnrojbprK9cAZ2DWhnnjI0QZSjN7aq9Gf5oegRhYBsNdT5Oo86DvfPum8m5zqbifAIl-jV8oAPXP9wrcGFnhORELd_g9gU3JLUO6Am9R4/s1600/3.48_7.jpg"><img style="cursor: pointer; width: 162px; height: 24px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjgDlL2aAdUG8SvUmlCKtH4K8GAUdyt0G5xWzcnrojbprK9cAZ2DWhnnjI0QZSjN7aq9Gf5oegRhYBsNdT5Oo86DvfPum8m5zqbifAIl-jV8oAPXP9wrcGFnhORELd_g9gU3JLUO6Am9R4/s400/3.48_7.jpg" alt="" id="BLOGGER_PHOTO_ID_5489528152212335506" border="0" /></a><br /><br />Now subtracting <span style="font-weight: bold;">(3c)</span> and <span style="font-weight: bold;">(5)</span> we obtain,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_lOmqS7RBF7GqdMXad2u5UtHcr6sWcZ7mPegRPd6rqxLGzARb4dzsmhccftkU9MXPNLOTphpGqWubNYxDY2BMaYf64XrzUIci6Qx-WAyi-uu_P3OsEgsRzaByhyE1gCY7S6pifhOIDdU/s1600/3.48_8.jpg"><img style="cursor: pointer; width: 181px; height: 23px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_lOmqS7RBF7GqdMXad2u5UtHcr6sWcZ7mPegRPd6rqxLGzARb4dzsmhccftkU9MXPNLOTphpGqWubNYxDY2BMaYf64XrzUIci6Qx-WAyi-uu_P3OsEgsRzaByhyE1gCY7S6pifhOIDdU/s400/3.48_8.jpg" alt="" id="BLOGGER_PHOTO_ID_5489528399031895442" border="0" /></a><br /><br />Now <span style="font-weight: bold;">(6)</span> can only be true if and only if,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjj_XVXGtCOGMNXtr-pP07p10AohjcKxhoencDFEnqi-p6_JEs3qtV3ClTMSicygevFuPndhPtPFacUCzccXgsz5Ed0aXn7fpABvMhpvmsbSvSsCwIIDGQv5nbvCLvabEM5DKIC1NcFaNA/s1600/3.48_9.jpg"><img style="cursor: pointer; width: 121px; height: 54px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjj_XVXGtCOGMNXtr-pP07p10AohjcKxhoencDFEnqi-p6_JEs3qtV3ClTMSicygevFuPndhPtPFacUCzccXgsz5Ed0aXn7fpABvMhpvmsbSvSsCwIIDGQv5nbvCLvabEM5DKIC1NcFaNA/s400/3.48_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5489528543001022914" border="0" /></a><br /><br />Hence, we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiaKF3IisUkG7y3mgqGJgs3ktFnwlGT0-_I867W2bUwPp29UNdkBfIObbm5hYN11-q-ARLwkUSl7xmVt9cxM6EbUbUtlZ4Ja1RBpWzbA_fEJY6T9KAw6Hiu19wtVPB9T7rdd7UMqH3-c9w/s1600/3.48_10.jpg"><img style="cursor: pointer; width: 84px; height: 23px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiaKF3IisUkG7y3mgqGJgs3ktFnwlGT0-_I867W2bUwPp29UNdkBfIObbm5hYN11-q-ARLwkUSl7xmVt9cxM6EbUbUtlZ4Ja1RBpWzbA_fEJY6T9KAw6Hiu19wtVPB9T7rdd7UMqH3-c9w/s400/3.48_10.jpg" alt="" id="BLOGGER_PHOTO_ID_5489528646017586258" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-75289349491225271032010-07-03T09:29:00.005+05:302010-07-03T09:36:53.332+05:30Irodov Problem 3.47<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhErN67IxwA1uK1Sy6nR-UrUFUrQWjECSTHq7ZzrO5CgwrBFEyOJaON3bDh1dIVyvV4Lzd-YA3tEutHnXHjjMMv7aLT7hihTjZcXYxWQRdSoe6-isjjwtflg7kEb5hjs6d7CDDJjdAToYs/s1600/3.47_fig1.jpg"><img style="cursor: pointer; width: 241px; height: 400px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhErN67IxwA1uK1Sy6nR-UrUFUrQWjECSTHq7ZzrO5CgwrBFEyOJaON3bDh1dIVyvV4Lzd-YA3tEutHnXHjjMMv7aLT7hihTjZcXYxWQRdSoe6-isjjwtflg7kEb5hjs6d7CDDJjdAToYs/s400/3.47_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5489525288648762850" border="0" /></a>Using the expression derived for <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEht99KsjTAPgCN6JASty7mR04koBJOHDTn0SKA2M6vBcRIe3DwV-mt_ZarYQXcuDGwP3T7CAFFMH96aiV7AKLIeGhBIWoQG1qh27UBbztMnK6dR8SoqyM2GzCfCzn_fiPvUAfsqpYM7lOs/s1600/3.47_0.jpg"><img style="cursor: pointer; width: 18px; height: 19px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEht99KsjTAPgCN6JASty7mR04koBJOHDTn0SKA2M6vBcRIe3DwV-mt_ZarYQXcuDGwP3T7CAFFMH96aiV7AKLIeGhBIWoQG1qh27UBbztMnK6dR8SoqyM2GzCfCzn_fiPvUAfsqpYM7lOs/s400/3.47_0.jpg" alt="" id="BLOGGER_PHOTO_ID_5489525563463444050" border="0" /></a> in problem <span style="font-weight: bold;">3.40</span> and setting <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgTKte7Ct9fNb1b2Tm54cBaD8ueQql8RL3f3cGTKlnZf1MbhtIOj1M7O5TUA2zhUzyCYq22aTMTqNpgn8W_3Ea0wowiS9PX1qjQer9PIs8UYFy3-k4nzmbRVVL_KZdTkHu6G0gnL6L2jgs/s1600/3.47_1.jpg"><img style="cursor: pointer; width: 40px; height: 20px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgTKte7Ct9fNb1b2Tm54cBaD8ueQql8RL3f3cGTKlnZf1MbhtIOj1M7O5TUA2zhUzyCYq22aTMTqNpgn8W_3Ea0wowiS9PX1qjQer9PIs8UYFy3-k4nzmbRVVL_KZdTkHu6G0gnL6L2jgs/s400/3.47_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5489525608740083202" border="0" /></a>, we obtain the electric field due to a dipole oriented along the <span style="font-weight: bold;">z-axis</span>, at a distance <span style="font-weight: bold;">z</span> along the <span style="font-weight: bold;">z-axis</span> as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgc2WarDj8CIRoTWLDhAznTpS0zOUKJPEYOWUddCUS3fS1WffqqRfcwKnsHAxZ2vkyoHXr7nywKN2_SIH268fsX31qR4MEE-vFZtYCZ2aJxEF9-_CSR3X3c0eUFItud66D-bI-is5IHBDo/s1600/3.47_2.jpg"><img style="cursor: pointer; width: 126px; height: 42px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgc2WarDj8CIRoTWLDhAznTpS0zOUKJPEYOWUddCUS3fS1WffqqRfcwKnsHAxZ2vkyoHXr7nywKN2_SIH268fsX31qR4MEE-vFZtYCZ2aJxEF9-_CSR3X3c0eUFItud66D-bI-is5IHBDo/s400/3.47_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5489525853843164610" border="0" /></a><br /><br />Consider another dipole with strength <span style="font-weight: bold;">p</span> and direction along the <span style="font-weight: bold;">z-axis</span>, in this field at a distance <span style="font-weight: bold;">l</span> as shown in the Figure, with charges <span style="font-weight: bold;">q</span> and <span style="font-weight: bold;">-q</span> separated by a distance <span style="font-weight: bold;">d</span>. Thus,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjZDPAaq76GBHw3KHsoqrTpXXLB6HvecY6hhyHPTqlXOCpNQYcmw1hYc7TH_WYDVuKnK_1I-pYZGZLEw0b-oAsXkSdch8OLPx2Dff8S-HnYW76ZFErJrd0GO98HiCgsb0qbizX0PAulSTY/s1600/3.47_3.jpg"><img style="cursor: pointer; width: 76px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjZDPAaq76GBHw3KHsoqrTpXXLB6HvecY6hhyHPTqlXOCpNQYcmw1hYc7TH_WYDVuKnK_1I-pYZGZLEw0b-oAsXkSdch8OLPx2Dff8S-HnYW76ZFErJrd0GO98HiCgsb0qbizX0PAulSTY/s400/3.47_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5489526345792259026" border="0" /></a><br /><br />The net force on this dipole will be,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVVWrOC01n-lEjeoh04iVJiMFy44VD5cjECyu6fPnIkXU6F1_ijfttv9oAErvgg3YooCeg1x6iP-cnNcGv_4kUUTom7GTV6IuonNQm9GMN6iG-5hnh39cxI1vl4IHYUxZizQrJxPAJe1A/s1600/3.47_4.jpg"><img style="cursor: pointer; width: 225px; height: 248px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVVWrOC01n-lEjeoh04iVJiMFy44VD5cjECyu6fPnIkXU6F1_ijfttv9oAErvgg3YooCeg1x6iP-cnNcGv_4kUUTom7GTV6IuonNQm9GMN6iG-5hnh39cxI1vl4IHYUxZizQrJxPAJe1A/s400/3.47_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5489526520482456642" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-19771351518685916242010-07-02T07:50:00.008+05:302010-07-02T07:57:16.280+05:30Irodov Problem 3.46<span style="font-weight: bold;">a) </span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEieLL13kU66Y6OfYZr1ChfvYoeB9Dzus6RIxS8HfvEdiJM5Ib04v1kwDFKIvwF4y9fVT7unnb9ApgFZO5aRruK9EksLLil1UdUC878GdKFQxl-uGGLdRUxHCob-X1Kq-6AV79S_ReEQ7aY/s1600/3.46_fig1.jpg"><img style="cursor: pointer; width: 400px; height: 289px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEieLL13kU66Y6OfYZr1ChfvYoeB9Dzus6RIxS8HfvEdiJM5Ib04v1kwDFKIvwF4y9fVT7unnb9ApgFZO5aRruK9EksLLil1UdUC878GdKFQxl-uGGLdRUxHCob-X1Kq-6AV79S_ReEQ7aY/s400/3.46_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5489128188824016210" border="0" /></a><br />The dipole moment <span style="font-weight: bold;">p</span> can be represented as two charges <span style="font-weight: bold;">q </span>and <span style="font-weight: bold;">-q</span> separated by a distance <span style="font-weight: bold;">d</span> along the <span style="font-weight: bold;">z=axis</span>, located at a distance <span style="font-weight: bold;">r</span> from the wire as shown in Figure 1. Then<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEin1B0c7VQagvYgDfX6HJkzB4r7zHXRC9BnDBPP5R4K9kTYGfFAaeR8evYJX5vj9s3f7aGOWx2q8qjs6BbS60uyZGWPB41ikIGXcxlcd5G66FM3-tq3omNtRYbBq1aznCX8H_tj1oTwAYc/s1600/3.46_2.jpg"><img style="cursor: pointer; width: 79px; height: 31px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEin1B0c7VQagvYgDfX6HJkzB4r7zHXRC9BnDBPP5R4K9kTYGfFAaeR8evYJX5vj9s3f7aGOWx2q8qjs6BbS60uyZGWPB41ikIGXcxlcd5G66FM3-tq3omNtRYbBq1aznCX8H_tj1oTwAYc/s400/3.46_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5489128779481397858" border="0" /></a><br /><br />We already know from <span style="font-weight: bold;">3.22 Eqn 1</span>, that the electric field due to a long wire at a distance <span style="font-weight: bold;">r</span> from its axis is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjXrh03fvgD_TbAGcPsM3xXXWin8xa1fNDiUNWo4S5ShSf0j9xsT1FumbgAmmmRcXSLgcSzD33OeSh2IVveo7LVl3I7Hztj6PON29ZupLty7wgjlJZ-y-kk-3ohVcEjXd3QTpI-FLXwbiQ/s1600/3.46_1.jpg"><img style="cursor: pointer; width: 98px; height: 41px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjXrh03fvgD_TbAGcPsM3xXXWin8xa1fNDiUNWo4S5ShSf0j9xsT1FumbgAmmmRcXSLgcSzD33OeSh2IVveo7LVl3I7Hztj6PON29ZupLty7wgjlJZ-y-kk-3ohVcEjXd3QTpI-FLXwbiQ/s400/3.46_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5489128922876088914" border="0" /></a><br /><br />acting radially outwards.<br /><br />The force on positive and negative electric charges in the dipole are then respectively given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjUrcQJiegub1mWNlTHQxajbmDtv_aE79uTje68ybfPhwoSb95Ua1qNiqiMoe_x8_qrW1cz1NUnlgNXfOUrGureJetfFp9-MjInDnfUl9chgICr0Ug8eazFaIvO7u54Ny4JMZA2HDh0mxs/s1600/3.46_3.jpg"><img style="cursor: pointer; width: 220px; height: 96px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjUrcQJiegub1mWNlTHQxajbmDtv_aE79uTje68ybfPhwoSb95Ua1qNiqiMoe_x8_qrW1cz1NUnlgNXfOUrGureJetfFp9-MjInDnfUl9chgICr0Ug8eazFaIvO7u54Ny4JMZA2HDh0mxs/s400/3.46_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5489129107351053170" border="0" /></a><br /><br />The net force on the dipole is then given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgH8EdNmqsQelZ8-Zl_lHR7AMUZM8Z8OgwQod-ikWpbD4yfi_wM04ND9lbiIluceYtGh2YqbkORxZ4I6v3jeNyreq01_KS0ZR9wKRJHtsYddltyuq4-lCjrXLzdeHX0RuB-LtcnBIPy0E8/s1600/3.46_4.jpg"><img style="cursor: pointer; width: 96px; height: 28px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgH8EdNmqsQelZ8-Zl_lHR7AMUZM8Z8OgwQod-ikWpbD4yfi_wM04ND9lbiIluceYtGh2YqbkORxZ4I6v3jeNyreq01_KS0ZR9wKRJHtsYddltyuq4-lCjrXLzdeHX0RuB-LtcnBIPy0E8/s400/3.46_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5489129208516533122" border="0" /></a><br /><br /><span style="font-weight: bold;">b)</span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhplnYuRYm_bniXNsZxvNmrKngTtA5O0jCD2dUqR-YPWsKEJ85pTXhcbeBrbUp5ksIh01tkwKDJGZF8a_-QcAu5TOt8l4ExUYt8GFELCNlLOlBdwC7VYjJiZ9soBMAwOG_DrMPmeZr6L4A/s1600/3.46_fig2.jpg"><img style="cursor: pointer; width: 400px; height: 289px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhplnYuRYm_bniXNsZxvNmrKngTtA5O0jCD2dUqR-YPWsKEJ85pTXhcbeBrbUp5ksIh01tkwKDJGZF8a_-QcAu5TOt8l4ExUYt8GFELCNlLOlBdwC7VYjJiZ9soBMAwOG_DrMPmeZr6L4A/s400/3.46_fig2.jpg" alt="" id="BLOGGER_PHOTO_ID_5489128109225980594" border="0" /></a><br /><br />Similar to part <span style="font-weight: bold;">a</span>, now the charges are located along the <span style="font-weight: bold;">x-axis</span>. The force is then given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8TbbQTABpF9ZMVpwqsWy8ZycJRp-JO_qssbinZ0clK78vtCVm4xZTzGwYWbB8NkeFw3QSx80VSEQlP6ty-J9XFfTk92eqbTLlhCTO0GK60_Y1_YyWynCppvLlAdpyf0AVUM-6sXYqHvY/s1600/3.46_5.jpg"><img style="cursor: pointer; width: 217px; height: 238px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8TbbQTABpF9ZMVpwqsWy8ZycJRp-JO_qssbinZ0clK78vtCVm4xZTzGwYWbB8NkeFw3QSx80VSEQlP6ty-J9XFfTk92eqbTLlhCTO0GK60_Y1_YyWynCppvLlAdpyf0AVUM-6sXYqHvY/s400/3.46_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5489129423269800674" border="0" /></a><br /><br /><span style="font-weight: bold;">c)</span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEisboaCO5L4klQIyy-isOB3nN6kiSgp7g7rOMeRLuRbi54iMTGrJ_BukUG6gKk3vIXAjEbT7hmwZoGYeZaI_0TVtfmPMx66e_Aq-BUCdlpAMory1rdNL-JF6IRWBNRzWpVx11MhR7RdXhE/s1600/3.46_fig3.jpg"><img style="cursor: pointer; width: 400px; height: 282px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEisboaCO5L4klQIyy-isOB3nN6kiSgp7g7rOMeRLuRbi54iMTGrJ_BukUG6gKk3vIXAjEbT7hmwZoGYeZaI_0TVtfmPMx66e_Aq-BUCdlpAMory1rdNL-JF6IRWBNRzWpVx11MhR7RdXhE/s400/3.46_fig3.jpg" alt="" id="BLOGGER_PHOTO_ID_5489129572254331330" border="0" /></a><br /><br />Similar to parts <span style="font-weight: bold;">a</span> and <span style="font-weight: bold;">b</span>, the charges are located along the <span style="font-weight: bold;">y-axis</span>. The force is then given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjAKHafhyRo-FFRIibqb91gdvjpN8rhKmINISu4TQLDROr9MAuj-CTVTnLzZ_cFgBzlX5ied6dmTRemCqYUYXYJqkMDLciZKppu9T7DZG05IXOXaz8ijKrSMyn4Jc13ja9lKwexI6HQGb4/s1600/3.46_6.jpg"><img style="cursor: pointer; width: 226px; height: 285px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjAKHafhyRo-FFRIibqb91gdvjpN8rhKmINISu4TQLDROr9MAuj-CTVTnLzZ_cFgBzlX5ied6dmTRemCqYUYXYJqkMDLciZKppu9T7DZG05IXOXaz8ijKrSMyn4Jc13ja9lKwexI6HQGb4/s400/3.46_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5489129721305046130" border="0" /></a><br /><span style="text-decoration: underline;"><br /><br /><br /></span>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-35175476329461358352010-06-30T19:02:00.006+05:302010-06-30T19:08:52.801+05:30Irodov Problem 3.45<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJxoYg6ZJbPm6PToTWdqWR9eoXSKXRv_nMiK2kOVJ6lnEW02cfLilz7zWDETZbBScNNjjRiFQr7OEtixQo2Ui9flcdqruTyLiWflNMXEZO8ZdvqJ-8PUF7rLzKCvHL6cCcZaPc668NMj4/s1600/3.45_fig1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 361px; height: 400px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJxoYg6ZJbPm6PToTWdqWR9eoXSKXRv_nMiK2kOVJ6lnEW02cfLilz7zWDETZbBScNNjjRiFQr7OEtixQo2Ui9flcdqruTyLiWflNMXEZO8ZdvqJ-8PUF7rLzKCvHL6cCcZaPc668NMj4/s400/3.45_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5488559257992931394" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />In Problem <span style="font-weight: bold;">3.44 Eqn (1)</span> we have already determined the electric field due a disc as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiTOeDjg4tan35_SDgAv87cKjPo-TdUmezR9INrlD9DzQ4_ce-qFSfK-IW2A5uq3zI-xPtnNbeD4lPXx6YWRKEdomZ8t19lPDvM6IsZEPJ4u94jaIShrp7UhvzNUZx9YC-bj9NtSkVmKxo/s1600/3.45_1.jpg"><img style="cursor: pointer; width: 231px; height: 58px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiTOeDjg4tan35_SDgAv87cKjPo-TdUmezR9INrlD9DzQ4_ce-qFSfK-IW2A5uq3zI-xPtnNbeD4lPXx6YWRKEdomZ8t19lPDvM6IsZEPJ4u94jaIShrp7UhvzNUZx9YC-bj9NtSkVmKxo/s400/3.45_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5488559868606972850" border="0" /></a><br />The Electric field due to a negative and positively charged discs is then given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgTAqPzkK1xGzJTsptrmF7F54T48D5Uro-_CZH25_0ff14HkaONY29EReMGzz2Hl5hXic7HG4fUPsyhgam6RrAzrqjn6sojvJu8IvLpEJpVvaPcMdKcedMOLfgkP31ZjYF8oSCAAhBRomk/s1600/3.45_2.jpg"><img style="cursor: pointer; width: 400px; height: 51px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgTAqPzkK1xGzJTsptrmF7F54T48D5Uro-_CZH25_0ff14HkaONY29EReMGzz2Hl5hXic7HG4fUPsyhgam6RrAzrqjn6sojvJu8IvLpEJpVvaPcMdKcedMOLfgkP31ZjYF8oSCAAhBRomk/s400/3.45_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5488559977458202418" border="0" /></a><br /><br />When<span style="font-weight: bold;"> l</span> is very small, using Taylor's series on <span style="font-weight: bold;">l</span> for <span style="font-weight: bold;">(2)</span> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjyBfkDrxSh0OGGENykPP7wwj_9jLYjPXsm0j3wan9oiFgmGeLHvrika0UlB_m1ay13Js9gf3QNe3MfJd5SprSOVXnXpGCOaOlg-z-3o_0rSejxtN3wDNshicedijM9jtgiEZR7pyGtCzA/s1600/3.45_3.jpg"><img style="cursor: pointer; width: 257px; height: 109px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjyBfkDrxSh0OGGENykPP7wwj_9jLYjPXsm0j3wan9oiFgmGeLHvrika0UlB_m1ay13Js9gf3QNe3MfJd5SprSOVXnXpGCOaOlg-z-3o_0rSejxtN3wDNshicedijM9jtgiEZR7pyGtCzA/s400/3.45_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5488560124203363986" border="0" /></a><br /><br />When <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj2LuE8sRy6rTDNJo0FnFCgE6LK3-DMglr_TajYUvKwW6AU2-bv8ErtEIclFgkJdA-Hw8mOYKF3LEh8YNxcAKlW2FXuPwXGG03gmcC9NeAHJvhZkpKrAbZjccR-MLVVEUQFGe01NkJJc0U/s1600/3.44_9.jpg"><img style="cursor: pointer; width: 42px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj2LuE8sRy6rTDNJo0FnFCgE6LK3-DMglr_TajYUvKwW6AU2-bv8ErtEIclFgkJdA-Hw8mOYKF3LEh8YNxcAKlW2FXuPwXGG03gmcC9NeAHJvhZkpKrAbZjccR-MLVVEUQFGe01NkJJc0U/s400/3.44_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5488560404533265618" border="0" /></a> we can further approximate <span style="font-weight: bold;">(3)</span> as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhKEOqFzSfhipiyCf-3CWYwTgk45uWENTnmYScT6BkBZ8kpcqTJo2K_QTfWTHThM-xj_ODZwX2Qwen5EN09Jpcc7ajlPHkCNho3TSqB06PirQ_a04nTu_Fux_ws1kO5C5HXzE4ygc16NJc/s1600/3.45_4.jpg"><img style="cursor: pointer; width: 113px; height: 45px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhKEOqFzSfhipiyCf-3CWYwTgk45uWENTnmYScT6BkBZ8kpcqTJo2K_QTfWTHThM-xj_ODZwX2Qwen5EN09Jpcc7ajlPHkCNho3TSqB06PirQ_a04nTu_Fux_ws1kO5C5HXzE4ygc16NJc/s400/3.45_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5488560217508441010" border="0" /></a><br /><br />We can now find the electric potential at a distance <span style="font-weight: bold;">x</span> approaching from the positive direction as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1gc7-nM9lDttVOB7vD2TCS-DGvA6iSqr5PaVSrq8nm7Ag913mUkvR7DePadncB7GKTglWjCMwmnfDmBVvkxeJAAe8J8OmFwo3de2gQC0FveMGIhKAxUHeP1L-_fZlaj6U5QJxItmBkvk/s1600/3.45_5.jpg"><img style="cursor: pointer; width: 274px; height: 128px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1gc7-nM9lDttVOB7vD2TCS-DGvA6iSqr5PaVSrq8nm7Ag913mUkvR7DePadncB7GKTglWjCMwmnfDmBVvkxeJAAe8J8OmFwo3de2gQC0FveMGIhKAxUHeP1L-_fZlaj6U5QJxItmBkvk/s400/3.45_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5488560532250738002" border="0" /></a><br /><br />Similarly the electric potential at a distance <span style="font-weight: bold;">x</span> approaching from the negative direction is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgru7nNZmAk9xwlY8_1Jjdv6GuAJRC9LW6h2vK5c6n0AnYJ9ti-0d1uqq84e3DYPG2_xG6U_pEVWIfO1QNa0yOc552bm7nhIDUVuGd2uW1zlU-n2hY1GqYPsBl5OOi7-CeQXauIkYZ5vTg/s1600/3.45_6.jpg"><img style="cursor: pointer; width: 276px; height: 132px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgru7nNZmAk9xwlY8_1Jjdv6GuAJRC9LW6h2vK5c6n0AnYJ9ti-0d1uqq84e3DYPG2_xG6U_pEVWIfO1QNa0yOc552bm7nhIDUVuGd2uW1zlU-n2hY1GqYPsBl5OOi7-CeQXauIkYZ5vTg/s400/3.45_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5488560627553807890" border="0" /></a>Unknownnoreply@blogger.com35tag:blogger.com,1999:blog-8917124841013563709.post-92172557216799463462010-06-13T10:16:00.011+05:302010-06-13T10:30:16.438+05:30Irodov Problem 3.44<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhE77LdrSyw0L5ezv-l3Vm6bsVYUhKfUrcffQr0SxCV6l2oRewi-m01Qr3WGH-JPn2EjwRe7M0fuzztSr_zYG-TNXmYLVUD32oPgzdixDdnD4pEpu9kaAUdpCrQ9XHHZzYKWKEMHlRnymM/s1600/3.44_fig1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 400px; height: 122px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhE77LdrSyw0L5ezv-l3Vm6bsVYUhKfUrcffQr0SxCV6l2oRewi-m01Qr3WGH-JPn2EjwRe7M0fuzztSr_zYG-TNXmYLVUD32oPgzdixDdnD4pEpu9kaAUdpCrQ9XHHZzYKWKEMHlRnymM/s400/3.44_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5482115180051314546" border="0" /></a><br /><br /><br />In this problem first we need to find the electric field due to an infinite thin plane sheet with a hole. For this, we shall first use the superposition principle. The idea being that the electric field due to a sheet with cavity is same as the summation of fields due to a disc with negative charge density and a solid infinite plane sheet with positive charge density as shown in figure 1.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEho925HDrzWUyR3EfilpzcskDNhWCoZ4ge8-iilT6YsjqBDJnjn8bHLHPoNOMdHTNrRj8h_CEoTk6znG4xPbyvrQ53A0nORkNK0qtctsMCmxxyD-SBeEEB5SbPkelUH9E15eowoP0EOORc/s1600/3.44_fig2.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 336px; height: 400px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEho925HDrzWUyR3EfilpzcskDNhWCoZ4ge8-iilT6YsjqBDJnjn8bHLHPoNOMdHTNrRj8h_CEoTk6znG4xPbyvrQ53A0nORkNK0qtctsMCmxxyD-SBeEEB5SbPkelUH9E15eowoP0EOORc/s400/3.44_fig2.jpg" alt="" id="BLOGGER_PHOTO_ID_5482115568993546706" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />First let us determine the electric field due to a charged disc. Consider an infinitesimally thin ring section on the disc (as shown in <span style="font-weight: bold;">Figure 2</span>) of thickness <span style="font-weight: bold;">dr</span> and radius <span style="font-weight: bold;">r</span>.The surface area of this ring will be <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFNiTvT7gEzz6ec7QLEj3yqRGlqbfmXVY0xkdbTQ1fWbAxCwmfChuWQXyhXiYOECu534J4qCqbGUhE_t6wn1Q1ecJEg40Sq3nJNK2NnGHjUd0M1qAhObTHBtJSEOoGtRusFZ6wHU3nxgw/s1600/3.44_1.jpg"><img style="cursor: pointer; width: 40px; height: 19px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFNiTvT7gEzz6ec7QLEj3yqRGlqbfmXVY0xkdbTQ1fWbAxCwmfChuWQXyhXiYOECu534J4qCqbGUhE_t6wn1Q1ecJEg40Sq3nJNK2NnGHjUd0M1qAhObTHBtJSEOoGtRusFZ6wHU3nxgw/s400/3.44_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5482115788293104898" border="0" /></a>. The charge contained in this infinitesimally thin ring will thus be <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGNqOjrPnNuJJz8VCqey0VaYEptTzURjDsLaxEcteKiv3YpjewX9NlLUXHf4LmgxXRnFWIolc_oUgDtIaGRcxpgbIKH14I0f_ClbK4xQcSyyUSujo7ueLeYa0gdirUFpUq4QcMtr8lmY0/s1600/3.44_2.jpg"><img style="cursor: pointer; width: 57px; height: 23px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGNqOjrPnNuJJz8VCqey0VaYEptTzURjDsLaxEcteKiv3YpjewX9NlLUXHf4LmgxXRnFWIolc_oUgDtIaGRcxpgbIKH14I0f_ClbK4xQcSyyUSujo7ueLeYa0gdirUFpUq4QcMtr8lmY0/s400/3.44_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5482115889500374482" border="0" /></a>. We have already derived the expression for the electric field due to a charged ring at a distance <span style="font-weight: bold;">x</span> along the axis from <span style="font-weight: bold;">Problem 3.9 Equation 4</span>. Using this expression we have the electric field <span style="font-weight: bold;">dE</span> due to this infinitesimally thin ring as<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj-BLMo6U5WiX4945_EeUWHl-Ovzc6kLe_YpKwn7J8VdxPCIgwQr5GcQjicynLzYNGWWGilfWz6NeOZJYRxH2rteAvZU7vAy5OkL5Helgdxa4zqULksDCOwTVFUeuaaTNEfkhVi51KBx3A/s1600/3.44_3.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 212px; height: 195px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj-BLMo6U5WiX4945_EeUWHl-Ovzc6kLe_YpKwn7J8VdxPCIgwQr5GcQjicynLzYNGWWGilfWz6NeOZJYRxH2rteAvZU7vAy5OkL5Helgdxa4zqULksDCOwTVFUeuaaTNEfkhVi51KBx3A/s400/3.44_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5482116425496273682" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><span style="font-weight: bold;">(1)</span> thus, gives the electric field due to a positively charged disc at a distance <span style="font-weight: bold;">x</span> along the axis of the disc.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEimCtX6u33niGzryUNzqfK58Y-if8n1eP8dpkryWSUNN_kWrkTgjtEJDEletYsWoCqSRTcMfWsxLpe_AIyoOcUBGCdK08X8hUqL-uiLXF3UpIYY4UuFg1H3n2lQuvtF9MCde5Ozlu38mg8/s1600/3.44_fig3.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 345px; height: 400px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEimCtX6u33niGzryUNzqfK58Y-if8n1eP8dpkryWSUNN_kWrkTgjtEJDEletYsWoCqSRTcMfWsxLpe_AIyoOcUBGCdK08X8hUqL-uiLXF3UpIYY4UuFg1H3n2lQuvtF9MCde5Ozlu38mg8/s400/3.44_fig3.jpg" alt="" id="BLOGGER_PHOTO_ID_5482117096068646386" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Now let us determine the elxtric field due to an infinitely large thin charged sheet. For this we can use Gauss law. Consider a Gaussian surface that is in the shape of a cylinder with its axis perpendicular to the sheet and having and area of cross-section <span style="font-weight: bold;">A</span>. The total charge contained within this Gaussian cylinder is <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgDj0xQ9FP0dA3WELdq2BDqNOzF_ev0Y4AFoCSIgasdTZylk6IuvgLQe53DhsLgxSMje2QPaQUeT3ZS31CsExkaw8KMJjBzSDN_05JHedbcEYdT4S2-hKFrvUDuT9oFIn0r7bz5ltcrShQ/s1600/3.44_4.jpg"><img style="cursor: pointer; width: 23px; height: 19px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgDj0xQ9FP0dA3WELdq2BDqNOzF_ev0Y4AFoCSIgasdTZylk6IuvgLQe53DhsLgxSMje2QPaQUeT3ZS31CsExkaw8KMJjBzSDN_05JHedbcEYdT4S2-hKFrvUDuT9oFIn0r7bz5ltcrShQ/s400/3.44_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5482116849580481730" border="0" /></a>. The electric field <span style="font-weight: bold;">E</span> will be perpendicular to the Gaussian cylinder at both its ends. Hence we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgMhaWOwEhayKGVR6wEyoVqFiJ65N5rxgrhop5whLdr0uY9d8MSHWmbNDhTALgOQLyVOEAujTx4NjjnqCYqbFUwuYc9vx2UsTGbKJhO-XlFZpHzczstyVC2j3e-eEHP2Rrqmy-pd9U2f0c/s1600/3.44_5.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 233px; height: 80px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgMhaWOwEhayKGVR6wEyoVqFiJ65N5rxgrhop5whLdr0uY9d8MSHWmbNDhTALgOQLyVOEAujTx4NjjnqCYqbFUwuYc9vx2UsTGbKJhO-XlFZpHzczstyVC2j3e-eEHP2Rrqmy-pd9U2f0c/s400/3.44_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5482116994592036706" border="0" /></a><br /><br /><br /><br /><br />The electric field due to the infinite sheet with a circular hole can now be determined as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEht1FZBO_vOvpIEfeV67QO-u_AG7KvSl3t7mUBj5PB5RDny152GCKW5aEwMZ4Ef8RrRviCPc-iGOAZDPDMJhMCtO_drBnTIu-dWQnnSMKkYBvXsoqmhc7I_ovs5aVdYPFZ-Bo-Pg7Hi70E/s1600/3.44_6.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 179px; height: 89px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEht1FZBO_vOvpIEfeV67QO-u_AG7KvSl3t7mUBj5PB5RDny152GCKW5aEwMZ4Ef8RrRviCPc-iGOAZDPDMJhMCtO_drBnTIu-dWQnnSMKkYBvXsoqmhc7I_ovs5aVdYPFZ-Bo-Pg7Hi70E/s400/3.44_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5482117306948826962" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2OkIFigG3cF0hHmcko-QrG1kczT0xHYa_NJ4903i_V0unge7hNWdn4d59SmJVwevXIx6tRu2kzEBhkh2jsvwt_ejv8Bh2R-mSOSah95lLbQH3nlfGy-78wGm1EykZ5XwqjaF7VaCzSQc/s1600/3.44_fig4.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 330px; height: 400px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2OkIFigG3cF0hHmcko-QrG1kczT0xHYa_NJ4903i_V0unge7hNWdn4d59SmJVwevXIx6tRu2kzEBhkh2jsvwt_ejv8Bh2R-mSOSah95lLbQH3nlfGy-78wGm1EykZ5XwqjaF7VaCzSQc/s400/3.44_fig4.jpg" alt="" id="BLOGGER_PHOTO_ID_5482117628301834226" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Now using <span style="font-weight: bold;">(3)</span> we can determine the electric field due to two such sheets, one negatively charged and the other positively charged separated by a distance of <span style="font-weight: bold;">l</span> from each other (as shown in <span style="font-weight: bold;">Figure 4</span>) as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3gikv89_dMZ8zHtqCF6khaGTTiVOaLwb7XfNjhMpKvG3D8lzCUZCigSmgHC1raKpt5yIZ2V_YeEn-NfquNsgZNL5B5OFdkpgSFCewlWdlIm40CX75VGSlwuzGPlgEofz7EOoE7qeYWNA/s1600/3.44_7.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 297px; height: 68px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3gikv89_dMZ8zHtqCF6khaGTTiVOaLwb7XfNjhMpKvG3D8lzCUZCigSmgHC1raKpt5yIZ2V_YeEn-NfquNsgZNL5B5OFdkpgSFCewlWdlIm40CX75VGSlwuzGPlgEofz7EOoE7qeYWNA/s400/3.44_7.jpg" alt="" id="BLOGGER_PHOTO_ID_5482117519565687170" border="0" /></a><br /><br /><br /><br /><br />When <span style="font-weight: bold;">l</span> is very small we can use Taylor's series to approximate <span style="font-weight: bold;">(4)</span> as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjMMrEiCahUJ34ndZOchNkQVGJMl1uV2M6NVOAD0y2RCnGTjjNbipj7evlhYMNOFL3b2gmt_vXnnATfYwK1tBQEgSn5hySzEzdhQn4cPpR07Dus_ZZ5W-zvLlx8ZkBhz4FKwVlz1d9HSo/s1600/3.44_8.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 169px; height: 54px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjMMrEiCahUJ34ndZOchNkQVGJMl1uV2M6NVOAD0y2RCnGTjjNbipj7evlhYMNOFL3b2gmt_vXnnATfYwK1tBQEgSn5hySzEzdhQn4cPpR07Dus_ZZ5W-zvLlx8ZkBhz4FKwVlz1d9HSo/s400/3.44_8.jpg" alt="" id="BLOGGER_PHOTO_ID_5482117852689566066" border="0" /></a><br /><br /><br /><br />When <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh33T8r1aoAMgZ4CT7X5kIR-2lq_ZhszRJ8lDpM59_m51CocroSVumYCBrkeBMai5mtzfYMWzwN7crHKS2fpztR2PoJS-YrgXwIL8g2teUNcRi1CTG3-4hmY4OZRcGcv1FfpPz9JOHX9fI/s1600/3.44_9.jpg"><img style="cursor: pointer; width: 42px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh33T8r1aoAMgZ4CT7X5kIR-2lq_ZhszRJ8lDpM59_m51CocroSVumYCBrkeBMai5mtzfYMWzwN7crHKS2fpztR2PoJS-YrgXwIL8g2teUNcRi1CTG3-4hmY4OZRcGcv1FfpPz9JOHX9fI/s400/3.44_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5482117981134350882" border="0" /></a> <span style="font-weight: bold;">(5)</span> becomes,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg5l8PJFolcFmSyu6Zv0YB2CgvSGkTSUKAbItXe8aWFumBNGE6i3E68bSU5acUX-Y4IhJ25ALMqlv8ImMWbNRb9GVPXrNBnp7jTgxsgHylpWykwyEFt6sFDce4GFz05EU17bWOHMZEYyak/s1600/3.44_10.jpg"><img style="cursor: pointer; width: 123px; height: 49px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg5l8PJFolcFmSyu6Zv0YB2CgvSGkTSUKAbItXe8aWFumBNGE6i3E68bSU5acUX-Y4IhJ25ALMqlv8ImMWbNRb9GVPXrNBnp7jTgxsgHylpWykwyEFt6sFDce4GFz05EU17bWOHMZEYyak/s400/3.44_10.jpg" alt="" id="BLOGGER_PHOTO_ID_5482118072836658802" border="0" /></a><br /><br />Now the electric field potential can be determined as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhbxRIYIHZlnce-32i_ARMmHWH7SGX2JbT3LQ-w8xl0-o19h8BQ0wcfPQ5P-tk-BM5N1rlXM0m_JN0O3rgi7VL6G1BX9Cxlxq1zVcxG-m3jgkqrh1QD06ROYykRqklFjcwvLLDk5HQGl6M/s1600/3.44_11.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 166px; height: 168px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhbxRIYIHZlnce-32i_ARMmHWH7SGX2JbT3LQ-w8xl0-o19h8BQ0wcfPQ5P-tk-BM5N1rlXM0m_JN0O3rgi7VL6G1BX9Cxlxq1zVcxG-m3jgkqrh1QD06ROYykRqklFjcwvLLDk5HQGl6M/s400/3.44_11.jpg" alt="" id="BLOGGER_PHOTO_ID_5482118239424018354" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />When <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh33T8r1aoAMgZ4CT7X5kIR-2lq_ZhszRJ8lDpM59_m51CocroSVumYCBrkeBMai5mtzfYMWzwN7crHKS2fpztR2PoJS-YrgXwIL8g2teUNcRi1CTG3-4hmY4OZRcGcv1FfpPz9JOHX9fI/s1600/3.44_9.jpg"><img style="cursor: pointer; width: 42px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh33T8r1aoAMgZ4CT7X5kIR-2lq_ZhszRJ8lDpM59_m51CocroSVumYCBrkeBMai5mtzfYMWzwN7crHKS2fpztR2PoJS-YrgXwIL8g2teUNcRi1CTG3-4hmY4OZRcGcv1FfpPz9JOHX9fI/s400/3.44_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5482117981134350882" border="0" /></a> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0w5tXy8aUYTRHZAMx6AYD46AeWyn4Sp2RUzWLyRRJtOrnMcnAP8iZph7_yWUVEwPW1HaEKbjRAJ6HgQ-gHdalXdUDWInseQ0QMFn8RZflB3xapPTyp6i6f0ncaxYgp9dmfKJMOm_aEIc/s1600/3.44_12.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 74px; height: 40px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0w5tXy8aUYTRHZAMx6AYD46AeWyn4Sp2RUzWLyRRJtOrnMcnAP8iZph7_yWUVEwPW1HaEKbjRAJ6HgQ-gHdalXdUDWInseQ0QMFn8RZflB3xapPTyp6i6f0ncaxYgp9dmfKJMOm_aEIc/s400/3.44_12.jpg" alt="" id="BLOGGER_PHOTO_ID_5482118525243705058" border="0" /></a>Unknownnoreply@blogger.com3tag:blogger.com,1999:blog-8917124841013563709.post-62890915795833418192010-06-06T21:51:00.013+05:302010-06-06T22:16:32.360+05:30Irodov Problem 3.43<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgAzDKwcuyeTOPwiAQw91rk3U4ToGJtEgmdMHMIF4QEmL5jsF8OnyzN0ooBt3sG-l8ogvTS1InIF61QvFm-3ej95Q9OYfaUhSYp9iAMn7OF3-WXkFIs_hGpZeQpWj8Bd4oW_o2FvpZRct8/s1600/3.43_fig1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 319px; height: 400px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgAzDKwcuyeTOPwiAQw91rk3U4ToGJtEgmdMHMIF4QEmL5jsF8OnyzN0ooBt3sG-l8ogvTS1InIF61QvFm-3ej95Q9OYfaUhSYp9iAMn7OF3-WXkFIs_hGpZeQpWj8Bd4oW_o2FvpZRct8/s400/3.43_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5479696682825921122" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />From Problem <span style="font-weight: bold;">3.9 Eqn 4</span>, we already know that the field due to a ring of radius <span style="font-weight: bold;">R</span> with charge <span style="font-weight: bold;">q</span> at a distance <span style="font-weight: bold;">x</span> along its axis is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjvQw6IfisL99rx7a9WLU0OKD8YgW7w2gPCACO4rFT-MctgSFWboEmhRVfsxStiydSzM0f10N4u7KWdU7skNkNATrTfGhxBdP4oGV0EBXqVZAQp85UTd_DvaPRKRwAQTmesyUSJ08_Y2qY/s1600/3.43_1.jpg"><img style="cursor: pointer; width: 135px; height: 45px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjvQw6IfisL99rx7a9WLU0OKD8YgW7w2gPCACO4rFT-MctgSFWboEmhRVfsxStiydSzM0f10N4u7KWdU7skNkNATrTfGhxBdP4oGV0EBXqVZAQp85UTd_DvaPRKRwAQTmesyUSJ08_Y2qY/s400/3.43_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5479696893662772418" border="0" /></a><br /><br />Hence the electric field due to the two rings is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiVhneEW9u6AfuAPGxTOjmORFlY1ew_OXPcOqPGAUxtx_SZrfEXX_eTnd7i1f2aAO1aXz6-zOlVpaqA0iYM95fhdI9QKxoR6hWwuyhQ00wjzk4ijrMaTdvgM8-1uAnM1U9B1S5FUBdBj78/s1600/3.43_2.jpg"><img style="cursor: pointer; width: 370px; height: 53px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiVhneEW9u6AfuAPGxTOjmORFlY1ew_OXPcOqPGAUxtx_SZrfEXX_eTnd7i1f2aAO1aXz6-zOlVpaqA0iYM95fhdI9QKxoR6hWwuyhQ00wjzk4ijrMaTdvgM8-1uAnM1U9B1S5FUBdBj78/s400/3.43_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5479696984871085698" border="0" /></a><br /><br />For <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnfqTQ53r7jBaUtXcTLOXrHSsjQFxLmNfd0GOi3sPJgxHdbVTlhfFxaGNeljsPpdFzA9-6-hEPbwuNWUJn_yE8aOqbBS_UE8OM8P0vKeJzB7H3uoFXyU41lW8F0ruL16FAMeCD-h6COEI/s1600/3.43_3.jpg"><img style="cursor: pointer; width: 41px; height: 19px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnfqTQ53r7jBaUtXcTLOXrHSsjQFxLmNfd0GOi3sPJgxHdbVTlhfFxaGNeljsPpdFzA9-6-hEPbwuNWUJn_yE8aOqbBS_UE8OM8P0vKeJzB7H3uoFXyU41lW8F0ruL16FAMeCD-h6COEI/s400/3.43_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5479697082804529458" border="0" /></a>, we can use Taylor's series with <span style="font-weight: bold;">l</span> as the variable around zero and approximate the electric field as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjz8fGG5d63lshfDN_uxMhTaVNTLLb4QkaTGovkQluFLiWLNCVveXawiGEcBSfyptyaJVJTWTYYDgXIWKpo2bkW5wGGFYppxRsIEJanzZAiSHGmK2QjoXbtAAKMhWT-y6097-iiwA_vfNI/s1600/3.43_8.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 293px; height: 111px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjz8fGG5d63lshfDN_uxMhTaVNTLLb4QkaTGovkQluFLiWLNCVveXawiGEcBSfyptyaJVJTWTYYDgXIWKpo2bkW5wGGFYppxRsIEJanzZAiSHGmK2QjoXbtAAKMhWT-y6097-iiwA_vfNI/s400/3.43_8.jpg" alt="" id="BLOGGER_PHOTO_ID_5479700090043401794" border="0" /></a><br /><br /><br /><br /><br /><br /><br />When <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjscYOAZViUmNwr9XcSKotJj7rDozTroVYqmEkBDBcA3p5BkyRpChr3II8IL9QA_hB5zmID3Nn1HZZVHktw0hkbT4eIb_GyFEANAOgPg6lt0ZTMuxHiKC1Bjy3RolqD5RUGOW-fDpEdhc/s1600/3.43_5.jpg"><img style="cursor: pointer; width: 50px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjscYOAZViUmNwr9XcSKotJj7rDozTroVYqmEkBDBcA3p5BkyRpChr3II8IL9QA_hB5zmID3Nn1HZZVHktw0hkbT4eIb_GyFEANAOgPg6lt0ZTMuxHiKC1Bjy3RolqD5RUGOW-fDpEdhc/s400/3.43_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5479697314749127794" border="0" /></a> the electric field becomes,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9TNKZ0i61XKiReBkp9btH0k0agssB5abjM5dhRXA3h-iinLAWE142Efk-CuZ9N2ilKDaUu6xyRWsAUUaZ4iMhD-A67wgB0NtxfOF9XtZ13arPoZK0tw9uRN7VFHUTsxjDTQKJTyPwH-o/s1600/3.43_4.jpg"><img style="cursor: pointer; width: 88px; height: 50px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9TNKZ0i61XKiReBkp9btH0k0agssB5abjM5dhRXA3h-iinLAWE142Efk-CuZ9N2ilKDaUu6xyRWsAUUaZ4iMhD-A67wgB0NtxfOF9XtZ13arPoZK0tw9uRN7VFHUTsxjDTQKJTyPwH-o/s400/3.43_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5479697220307304738" border="0" /></a><br /><br />We also know that the potential due to a ring of radius <span style="font-weight: bold;">R</span> with charge <span style="font-weight: bold;">q</span> at a distance <span style="font-weight: bold;">x</span> along the axis is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZkwWCXkeJNOku-_h-u_JddQvCoGhwkqDxlnMz5L7EiWiOPuJRMJ4uDFvrl2IkjYJ7CbPjUVI503ge2w3HmvrvTBj-rLzPKkYaeUasvnz7Yzt9lxGhMxSZ4uUjEzqX6QT9lUWZdDqvLR8/s1600/3.43_6.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 141px; height: 50px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZkwWCXkeJNOku-_h-u_JddQvCoGhwkqDxlnMz5L7EiWiOPuJRMJ4uDFvrl2IkjYJ7CbPjUVI503ge2w3HmvrvTBj-rLzPKkYaeUasvnz7Yzt9lxGhMxSZ4uUjEzqX6QT9lUWZdDqvLR8/s400/3.43_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5479700575342850914" border="0" /></a><br /><br /><br /><br />Hence, the potential due to the two rings is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgEHhThGEm4uDEhzDvpBC2fRF0AdOx1wFrbvm3zouYcvbLRVa1IwGnxDCnFYKKW30QSgwqmvdz9o4_G4kAOpN8DsNTjM6kdMewhNRcNzg5ZSxyylmjuO1-8GQNfKL41hTAbY_Vvgv2bEhY/s1600/3.43_7.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 369px; height: 54px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgEHhThGEm4uDEhzDvpBC2fRF0AdOx1wFrbvm3zouYcvbLRVa1IwGnxDCnFYKKW30QSgwqmvdz9o4_G4kAOpN8DsNTjM6kdMewhNRcNzg5ZSxyylmjuO1-8GQNfKL41hTAbY_Vvgv2bEhY/s400/3.43_7.jpg" alt="" id="BLOGGER_PHOTO_ID_5479700847596972146" border="0" /></a><br /><br /><br /><br />For <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjscYOAZViUmNwr9XcSKotJj7rDozTroVYqmEkBDBcA3p5BkyRpChr3II8IL9QA_hB5zmID3Nn1HZZVHktw0hkbT4eIb_GyFEANAOgPg6lt0ZTMuxHiKC1Bjy3RolqD5RUGOW-fDpEdhc/s1600/3.43_5.jpg"><img style="cursor: pointer; width: 50px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjscYOAZViUmNwr9XcSKotJj7rDozTroVYqmEkBDBcA3p5BkyRpChr3II8IL9QA_hB5zmID3Nn1HZZVHktw0hkbT4eIb_GyFEANAOgPg6lt0ZTMuxHiKC1Bjy3RolqD5RUGOW-fDpEdhc/s400/3.43_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5479697314749127794" border="0" /></a>, we can use Taylor's series with <span style="font-weight: bold;">l</span> as the variable around zero and approximate the potential as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgwIOO7P0U4xRMKN2g3JA-kJq0jb1y-faHoZgKqMnMdB_NUgbsRurjLIF1HDveiGYU5vXQic05wXRdzwJOKliZAo99CglGduAIQDtsWZewth51I3tF035TyqWgRlf7onwyqxip1wpB6RXI/s1600/3.43_9.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 142px; height: 153px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgwIOO7P0U4xRMKN2g3JA-kJq0jb1y-faHoZgKqMnMdB_NUgbsRurjLIF1HDveiGYU5vXQic05wXRdzwJOKliZAo99CglGduAIQDtsWZewth51I3tF035TyqWgRlf7onwyqxip1wpB6RXI/s400/3.43_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5479701646402998882" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8917124841013563709.post-16053803730249140392010-06-06T21:47:00.005+05:302010-06-06T21:51:16.422+05:30Irodov Problem 3.42<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizHIEIhJ30nkE_oVI0-8svb_gimNy0PEUwB5ZoBxO9qbeGtgpZoqX0PPt49QLGB31FUPgGJ3TlwYvyQqmL1IoZqyGNzoPqSfcsiuX7li7D4_Vfin_EjmV8TgbkvCn7FWuTDGVAae8iGnA/s1600/3.42_fig1.jpg"><img style="float: left; margin: 0pt 10px 10px 0pt; cursor: pointer; width: 400px; height: 291px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizHIEIhJ30nkE_oVI0-8svb_gimNy0PEUwB5ZoBxO9qbeGtgpZoqX0PPt49QLGB31FUPgGJ3TlwYvyQqmL1IoZqyGNzoPqSfcsiuX7li7D4_Vfin_EjmV8TgbkvCn7FWuTDGVAae8iGnA/s400/3.42_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5479695634606832866" border="0" /></a><br /><br /><br /><br /><br /><br />As shown in the figure the two infinitely long wires are perpendicular to the screen (coming out of the screen). For <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfAoxlSlxgR3MFAM5RkoJJfMgaxQD6rY7qfVQD-Spue3rh8Hw_cAVR7zAQJeyeNE73hjkvZerUHag0t_d8kcQAPpIGF4JuQ2MAYCWALLLzF8gY9hDM_ft4PsV7OWNQJ_OyVNx5P8mivJk/s1600/3.42_1.jpg"><img style="cursor: pointer; width: 28px; height: 17px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfAoxlSlxgR3MFAM5RkoJJfMgaxQD6rY7qfVQD-Spue3rh8Hw_cAVR7zAQJeyeNE73hjkvZerUHag0t_d8kcQAPpIGF4JuQ2MAYCWALLLzF8gY9hDM_ft4PsV7OWNQJ_OyVNx5P8mivJk/s400/3.42_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5479695797132755154" border="0" /></a>, the distances of the positive and negative wires from <span style="font-weight: bold;">X</span> are given by <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiCqo7XQCTU3opfwVrXxinpipOUK6aq339IbvPVRy_j0xfbACs9TjI0CBtVFkOaMiDj9TUT80wx2NshjA2pqBnhvJrJ5-DWUHJ4fC2f6HcGnvrNIvpOylZ7IrIXCDhU7VhiXbZE6eYPAXs/s1600/3.42_2.jpg"><img style="cursor: pointer; width: 82px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiCqo7XQCTU3opfwVrXxinpipOUK6aq339IbvPVRy_j0xfbACs9TjI0CBtVFkOaMiDj9TUT80wx2NshjA2pqBnhvJrJ5-DWUHJ4fC2f6HcGnvrNIvpOylZ7IrIXCDhU7VhiXbZE6eYPAXs/s400/3.42_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5479695903066183602" border="0" /></a> and <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhihNAKZxMpI4823DnSY5ZdSum4pjGE-sd3VM-1fNdd4cfXSKswtFe9XPPM0-jcMnrAHg6vZec4KfqYkBHc8Kt909kmURHabv94cePbWaUwVIjKAnSQh7JH7ceCE4SaBAX3ldk7gUD24Y/s1600/3.42_3.jpg"><img style="cursor: pointer; width: 82px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhihNAKZxMpI4823DnSY5ZdSum4pjGE-sd3VM-1fNdd4cfXSKswtFe9XPPM0-jcMnrAHg6vZec4KfqYkBHc8Kt909kmURHabv94cePbWaUwVIjKAnSQh7JH7ceCE4SaBAX3ldk7gUD24Y/s400/3.42_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5479696030125500018" border="0" /></a> respectively.<br /><br />We know that the electric field due to a single long wire at a distance d acts radially outwards and is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiKUsLtRWuB2XCrKB97DFkMyHKZkQQOfoGKuKnYl55Wof1SR46LV4sZmAMf6bhzLOamBgMx7PHWionGzy8A4G2ECO4CDKKftIcRtcmC6RwZfzjoau-d3E8RGGwyEQYpvzcg8yLTebYvIw0/s1600/3.42_4.jpg"><img style="cursor: pointer; width: 69px; height: 33px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiKUsLtRWuB2XCrKB97DFkMyHKZkQQOfoGKuKnYl55Wof1SR46LV4sZmAMf6bhzLOamBgMx7PHWionGzy8A4G2ECO4CDKKftIcRtcmC6RwZfzjoau-d3E8RGGwyEQYpvzcg8yLTebYvIw0/s400/3.42_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5479696239164107858" border="0" /></a><br /><br />Hence the total electric field for two wires is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhIsFaxQqpNUIEsZP9B9evTb7BO9t4f3MboMqRA6zxc_OCsmPbssjHW0LDSTdPgNGEG644WeJFwIQZsNnGxbFKLVGlJ97xF4UHEi7rwmpc4aExbyOrCdRcKiFh8e8RHmdZCYO4IV5AKxFA/s1600/3.42_5.jpg"><img style="cursor: pointer; width: 305px; height: 109px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhIsFaxQqpNUIEsZP9B9evTb7BO9t4f3MboMqRA6zxc_OCsmPbssjHW0LDSTdPgNGEG644WeJFwIQZsNnGxbFKLVGlJ97xF4UHEi7rwmpc4aExbyOrCdRcKiFh8e8RHmdZCYO4IV5AKxFA/s400/3.42_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5479696353317286098" border="0" /></a><br /><br />The direction of the electric field being radially outwards.<br /><br />The electric potential can now evalauted as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhNoIjboRSgcTjmE0wsV4D9CKAfErxl_CsSTlPs08_UEHfV4h7CsEw1jA_GdkAZCUJKza8QXNHnHOiaexWD-s1FJe_RZ0xzONXwxDuoKrtVd0q5NULKbudfILKzwoMdkn8DKLkHvSXF0mM/s1600/3.42_6.jpg"><img style="cursor: pointer; width: 222px; height: 105px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhNoIjboRSgcTjmE0wsV4D9CKAfErxl_CsSTlPs08_UEHfV4h7CsEw1jA_GdkAZCUJKza8QXNHnHOiaexWD-s1FJe_RZ0xzONXwxDuoKrtVd0q5NULKbudfILKzwoMdkn8DKLkHvSXF0mM/s400/3.42_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5479696458303197122" border="0" /></a>Unknownnoreply@blogger.com5