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 displaced by small distance x then it's potential energy now
$U_f=kqQ\left[\frac{1}{(a+x)+\frac{1}{(a-x)}}\right]=\frac{2kqQa}{(a^2-x^2)}$ .......... (i)

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