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

Question

Dr. Abubacker · Accepted Answer

(b) Initially according to figure (i) potential energy of $Q$ is
$U_{i} = \frac{2kqQ}{a}$ …(i)
According to figure (ii) when charge $Q$ is displaced by small distance $x$ then it’s potential energy now
$U_{f} = kqQ\left\lbrack \frac{1}{(a + x)} + \frac{1}{(a - x)} \right\rbrack = \frac{2kqQa}{(a^{2} - x^{2})}$ …(ii)
Hence change in potential energy
$\mathrm{\Delta}U = U_{f} - U_{i} = 2kqQ\left\lbrack \frac{a}{a^{2} - x^{2}} - \frac{1}{a} \right\rbrack = \frac{2kqQx^{2}}{(a^{2} - x^{2})a}$
Since $x < < a$ so $\mathrm{\Delta}U = \frac{2kqQx^{2}}{a^{3}} \Rightarrow \mathrm{\Delta}U \propto x^{2}$

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

Explanation

Reviewed by Dr. Abubacker

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