A charge Q is uniformly on a ring of radius R. Find the electric potential on the axis of ring at distance x from the centre of the ring:

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$\frac{1}{4 {πε}_{0}} \frac{2 Q}{(R^{2} + x^{2})}$

$\frac{1}{4 {πε}_{0}} \frac{Q}{(R^{2} + x^{2})}$

$\frac{1}{4 {πε}_{0}} \frac{Q}{{(R^{2} + x^{2})}^{1 / 2}}$

$\frac{1}{2 {πε}_{0}} \frac{Q}{{(R^{2} + x^{2})}^{1 / 2}}$

answer is A.

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Detailed Solution

Small potential at P due to dq

$dV = \frac{1}{4 {πϵ}_{0}} \cdot \frac{dq}{\sqrt{R^{2} + x^{2}}}$

$V_{P} = ΣdV = \frac{1}{4 {πϵ}_{0}} \cdot \frac{Σdq}{\sqrt{R^{2} + x^{2}}}$

$, = \frac{1}{4 {πϵ}_{0}} \cdot \frac{Q}{\sqrt{R^{2} + x^{2}}}$

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