Question

A system of three charges are placed as shown in the figure.


If $D>>d$, the potential energy of the system is best given by

• A. $\frac{1}{4\pi {ϵ}_0}[\frac{-q^2}{d}-\frac{qQd}{2D^2}]$
• B. $\frac{1}{4\pi {ϵ}_0}[-\frac{q^2}{d}-\frac{qQd}{D^2}]$
• C. $\frac{1}{4\pi {ϵ}_0}[+\frac{q^2}{d}+\frac{qQd}{D^2}]$
• D. $\frac{1}{4\pi {ϵ}_0}[\frac{-q^2}{d}+\frac{2qQd}{D^2}]$

Answer 1

The correct option is B $\frac{1}{4\pi {ϵ}_0}[-\frac{q^2}{d}-\frac{qQd}{D^2}]$

Potential energy due to combination of charges is $\sum \frac{q_1{q}_2}{r^2}$


$U=\frac{1}{4\pi {ϵ}_0}[\frac{q(-q)}{d}+\frac{qQ}{(D+\frac{d}{2})}+\frac{(-q)Q}{(D-\frac{d}{2})}]$

$U=\frac{1}{4\pi {ϵ}_0}[\frac{-q^2}{d}+\frac{qQ(D-\frac{d}{2})-qQ(D+\frac{d}{2})}{D^2-\frac{d^2}{4}}]$

$U=\frac{1}{4\pi {ϵ}_0}[-\frac{q^2}{d}-\frac{qQd}{D^2}],\because \frac{d^2}{4}<<D$