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Q.

Two equal point charges are fixed at x = - $α$ and x = + $α$ on the X-axis . Another point charge Q is placed at the origin. The change in electrical potential energy of Q when it is displaced by a small distance x along the X-axis, is approximately proportional to

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a

X²

b

1/X

c

X³

d

X

answer is B.

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Detailed Solution

When Q is at position x, then
$U = \frac{qQ}{4 {πε}_{0}} [\frac{1}{(a + x)} + \frac{1}{(a - x)}]$
or $U = \frac{2 qQa}{4 {πε}_{0} (a^{2} - x^{2})}$
$\begin{array}{l} ∴ \frac{dU}{dx} = \frac{2 aqQ}{4 {πε}_{0}} [\frac{2 x}{{(a^{2} - x^{2})}^{2}}] \\ = \frac{2 aqQ}{4 {πε}_{0}} \times \frac{2 x}{a^{4}} (∵ (a^{2} - x^{2}) \approx a^{2}) \end{array}$
Now $dU \propto x dx or U \propto x^{2} .$

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