Consider an infinitesimally small section of the ring that subtends an angle
at the center. The charge contained in this infinitesimally small section of the ring is 
. The force exerted by the charge q0 on this infinitesimally small section of the ring is given by,
Let T be the tension in the ring induced due to the electrostatic force resulting from charge q0. The rest of ring exerts this force T on each of the two ends of the infinitesimally small segment of the ring. The component of this force T in the direction of dFe (as shown in the diagram) is given by
. Since the infinitesimally thing ring section is in equilibrium (no accelerating) using Newton's second law we have,
the final answer is having dθ, but it was canceled in the step just above it!
ReplyDeleteIt's been a mistake
Deleteit is just a writing error. final answer won't contain d@. rest of the answer is perfectly correct
ReplyDeleteChange in Tension is required but it is only the final Tension
ReplyDeletelook shubham try to reakise that the tension in the ring initially is zero due to no F external so change in the force will be this only.i hope you will try to understand......................................................................................................................................................................................
DeleteIt is not 0. We have to find change in tension. The intial tension in the ring is already balancing the forces by the ring electrostatic repulsion.As we can see by symmetry that its direction will be radially outward (same as that of the new repulsion)so the increase will happen only due the central charge force. Hence we are assuming initially 0 which is just a reference point.
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ReplyDeleteHow did the charge in the small section of the ring = qd@/2π
ReplyDeletecharge on the whole ring was q. we know that a circle subtends an angle 2π at the center. it means that the part that subtends unit angle on the center has a charge of q/2π (unitary method).
Deleteso, now the part that subtends d@ angle will have a charge of (q/2π)x d@= qd@/2π
Why have we not considered the electrostatic force due to remaining part of ring?...please reply
ReplyDeleteWhy have we not considered the electrostatic force due to remaining part of ring?...please reply
ReplyDeleteWe have to find change in tension. The intial tension in the ring is already balancing the forces by the ring electrostatic repulsion.As we can see by symmetry that its direction will be radially outward (same as that of the new repulsion)so the increase will happen only due the central charge force.
Delete