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 $x^2$

Potential energy $=\frac{1}{4\pi {ϵ}_o}q.\frac{Q}{r}$

$|\Delta P|=\frac{1}{4\pi {ϵ}_o}q.\frac{Q}{xdx}-\frac{1}{4\pi {ϵ}_o}\frac{qQ}{x+dx}$

$li{m}_{d}x\to 0=\frac{1}{4\pi {ϵ}_o}.\frac{Q.2x}{x^2-(dx{)}^2}$

Where $dx=0$

so the answer is $x^2$