tag:blogger.com,1999:blog-8917124841013563709.post7823899902870741003..comments2023-12-17T17:11:15.939+05:30Comments on Solutions to I E Irodov - Electrodynamics: Irodov Problem 3.29Unknownnoreply@blogger.comBlogger1125tag:blogger.com,1999:blog-8917124841013563709.post-18507385079430454832012-05-01T18:33:44.386+05:302012-05-01T18:33:44.386+05:30Electric Field Intensity
It is the force experienc...Electric Field Intensity<br />It is the force experienced by a unit positive charge placed at a point in an electric field.<br /><br />Due to a point charge<br /> <br />Due to linear distribution of charge<br /><br /> λ = linear charge density of rod<br /> <br /> <br />(i) At a point on its axis.<br /> <br />(ii) At a point on the line perpendicular to one end<br /> <br /> <br /> <br />Due to ring of uniform charge distribution<br />At a point on its axis<br /> <br /> <br />Due to uniformly charged disc.<br />At a point on its axis.<br /> <br />Thin spherical shell<br /> <br />1. Non conducting solid sphere having uniform volume distribution of charge<br />(i) Outside point<br /> <br />(ii) Inside point <br /> q = total charge<br /> ρ = volume density of charge<br /> <br />Cylindrical Conductor of Infinite length<br /><br />Outside point<br /> <br />Inside point λ = linear density of charge<br /> E = 0 (as charges reside only on the surface)<br />Non-conducting cylinder having uniform volume density of charge<br /> <br />Infinite plane sheet of charge<br /> <br /> λ = surface charge density<br />Two oppositely charged sheets (Infinite)<br />(i) Electric field at points outside the charged sheets<br /> EP = ER = 0<br />(ii) Electric field at point in between the charged sheets<br /> <br />Example 3. Find electric field intensity due to long uniformly charged wire. <br /> (charge per unit length is λ)DHARMENDRA SINGH SISODIAhttps://www.blogger.com/profile/07936793199831171717noreply@blogger.com