Question

A ring of radius R carries a charge +q.A test charge -qo is released on its axis at a distance under root 3R from its center .How much kinetic energy will be acquired by the test charge when it reaches the center of the ring?

Answer 1

Answer: Find potential at initial point and at final point. Difference of potential energy will turn into kinetic energy.

Potential at centre of ring = kq/R

Potential at dist root(3)R from centre =kq/root(R^2 + (root(3))^2) = kq/2R

Initial Potential energy = -kqqo/2R

Final Potential Energy =-kqqo/R

So, Initial Potential energy = Final Potential Energy + kinetic energy

-kqqo/2R +kqqo/R = kqqo/2R =kinetic energy