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 electrical potential energy $Q$ 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. 1$/x$

Answer 1

The correct option is A x

$\begin{array}{c}ui=\frac{KQq}{a}\times 2=\frac{2\times Qq}{a}\\ uf=\frac{KQq}{\left(a+r\right)}+\frac{KQq}{\left(a-r\right)}=KQq\left[\frac{a-x+q+x}{a^2-x^2}\right]\\ =\frac{2KQqa}{a^2\left(1-\frac{x^2}{a^2}\right)}\\ =\frac{2KQq}{a\left(1-\frac{x^2}{a^2}\right)}\\ \approx \frac{2KQq}{a}\left(1+\frac{2x}{a}\right)\\ \therefore \Delta u=u_f-u_i\propto x\end{array}$

$\therefore$ option $A$ is correct .