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
IIT 2002, Diffcult
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(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[ {\frac{1}{{(a + x)}} + \frac{1}{{(a - x)}}} \right]$$ = \frac{{2kqQa}}{{({a^2} - {x^2})}}$ .......$(ii)$
Hence change in potential energy
$\Delta U = {U_f} - {U_i} = 2kqQ\,\left[ {\frac{a}{{{a^2} - {x^2}}} - \frac{1}{a}} \right]$$ = \frac{{2kqQ{x^2}}}{{({a^2} - {x^2})}}$
Since $x << a$ so $\Delta U = \frac{{2kqQ{x^2}}}{{{a^2}}} \Rightarrow \Delta U \propto {x^2}$
According to figure $(ii)$ when charge $Q$ is displaced by small distance $x$ then it’s potential energy now
${U_f} = kqQ\,\left[ {\frac{1}{{(a + x)}} + \frac{1}{{(a - x)}}} \right]$$ = \frac{{2kqQa}}{{({a^2} - {x^2})}}$ .......$(ii)$
Hence change in potential energy
$\Delta U = {U_f} - {U_i} = 2kqQ\,\left[ {\frac{a}{{{a^2} - {x^2}}} - \frac{1}{a}} \right]$$ = \frac{{2kqQ{x^2}}}{{({a^2} - {x^2})}}$
Since $x << a$ so $\Delta U = \frac{{2kqQ{x^2}}}{{{a^2}}} \Rightarrow \Delta U \propto {x^2}$
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