Question

A ring of radius R is having two charges q and 2q

distributed on its two half parts. The electric
potential at a point on its axis at a distance of 2/2
R from its centre is k = 4€
* 3R
(C) R
(d)​

Answer 1

Given :

Radius of the ring = R

Charge 1 (q₁) = q

Charge 2 (q₂) = 2q

To Find :

The electric potential at distance of 2√2R from the axis

Solution :

• We can find the distance between the point and charges present on the ring using Pythagoras theorem

           $r = \sqrt{R^{2}+(2\sqrt{2R}) ^{2} }$

           $r = \sqrt{R^{2}+8R^{2} }$

           $r = 3R$

• Electric potential (V)  =  $\frac{kq_{net} }{r}$

           $V=\frac{k(3q)}{3R}$

           $V=\frac{kq}{R}$

The electric potential at a distance of 2√2R from the axis is (kq)/R