Two equal point charges are fixed at x = -a a | vedclass

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(B) Initially,according to figure $(i)$,the potential energy of $Q$ is $U_i = \frac{2kqQ}{a}$ ... $(i)$According to figure $(ii)$,when charge $Q$ is d

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$x^2$

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(B) Initially,according to figure $(i)$,the potential energy of $Q$ is $U_i = \frac{2kqQ}{a}$ ... $(i)$According to figure $(ii)$,when charge $Q$ is displaced by a small distance $x$,its potential energy becomes:$U_f = kqQ \left[ \frac{1}{a+x} + \frac{1}{a-x} \right] = kqQ \left[ \frac{(a-x) + (a+x)}{a^2 - x^2} \right] = \frac{2kqQa}{a^2 - x^2}$ ... $(ii)$Hence,the change in potential energy is:$\Delta U = U_f - U_i = 2kqQ \left[ \frac{a}{a^2 - x^2} - \frac{1}{a} \right] = 2kqQ \left[ \frac{a^2 - (a^2 - x^2)}{a(a^2 - x^2)} \right] = \frac{2kqQx^2}{a(a^2 - x^2)}$Since $x $\Delta U \approx \frac{2kqQx^2}{a(a^2)} = \frac{2kqQ}{a^3} x^2$Thus,$\Delta U \propto x^2$.

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

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