- Home
- Standard 12
- Physics
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
$x$
${x^2}$
${x^3}$
$1/x$
Solution
(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 $\Delta U = \frac{{2kqQ{x^2}}}{{{a^2}}} \Rightarrow \Delta U \propto {x^2}$
