Question

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

• A. $x$
• B. $x^2$
• C. $x^3$
• D. $\frac{1}{x}$

Answer 1

The correct option is B $x^2$

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
$U_f=kqQ[\frac{1}{(a+x)}+\frac{1}{(a-x)}]=\frac{2kqQa}{(a^2-x^2)}$ ......... (ii)
Hence change in potential energy
$\Delta U=U_f-U_i=2kqQ[\frac{a}{a^2-x^2}-\frac{1}{a}]=\frac{2kqQa}{(a^2-x^2)}$
Since x<<a so $\Delta U=\frac{2kqQ{x}^2}{a^3}\Rightarrow \Delta U\propto x^2$