A ring with a uniform charge $Q$ and radius $R$ , is placed in the $y z$ plane with its center at the origin $O$ of the coordinates. Let $V_{O} a n d E_{O}$ be the potential and the field at the origin, $V$ and $E$ be the potential and field at all points on the positive $x$ axis, other than the origin, then

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$V = \frac{1}{4 π ε_{0}} \frac{Q}{\sqrt{x^{2} + R^{2}}}$

$V_{O} = \frac{1}{4 π ε_{0}} \frac{Q}{R}$

$E_{O} = 0$

$E = \frac{1}{4 π ε_{0}} \frac{Q x}{{(R^{2} + x^{2})}^{\frac{3}{2}}} a n d i s m a x i m u m a t x = \frac{R}{\sqrt{2}}$

answer is A, B, C, D.

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Detailed Solution

The electric potential due to the ring at a distance x from the origin on the x axis of ring is $E = k \frac{Q}{{(x^{2} + R^{2})}^{1 / 2}}$ At center, $x = 0, V_{0} = \frac{Q}{4 π \in_{0} R}$ .

As $E = - \frac{d V}{d z},$ the electric potential due to the ring at a point on the axis of ring is E= $k \frac{Q x}{{(x^{2} + R^{2})}^{\frac{3}{2}}}$