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Q.

For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h form its centre. Then $+ v e$ value of h is

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a

$R \sqrt{2}$

b

$R$

c

$\frac{R}{\sqrt{2}}$

d

$\frac{R}{\sqrt{5}}$

answer is C.

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Detailed Solution

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Electric field on axis of ring
$E = \frac{kQh}{{(h^{2} {+R}^{2})}^{3/2}}$
For maximum electric field $\frac{dE}{dh} =0$
$\begin{array}{l} \frac{dE}{dh} =KQ [\frac{{(R^{2} {+h}^{2})}^{3/2} - h× \frac{3}{2} {(R^{2} {+h}^{2})}^{1/2} ×2h}{{(R^{2} {+h}^{2})}^{3}}] \\ \Rightarrow R^{2} {+h}^{2} {-3h}^{2} =0 \Rightarrow h = \pm \frac{R}{\sqrt{2}} . \end{array}$

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