Question

Three charges are placed as shown in figure. If the electric potential energy of the system is zero, then $Q:q$ is :

• A. $\frac{Q}{q}=\frac{-2}{1}$
• B. $\frac{Q}{q}=\frac{2}{1}$
• C. $\frac{Q}{q}=\frac{-1}{2}$
• D. $\frac{Q}{q}=\frac{1}{4}$

Answer 1

The correct option is D $\frac{Q}{q}=\frac{1}{4}$


Total potential energy of the given system
$U=\frac{-KqQ}{r}+\frac{K{q}^2}{2r}-\frac{KQq}{r}$
Given that, $U=0$
$\Rightarrow \frac{2KqQ}{r}=\frac{K{q}^2}{2r}$
or $\frac{Q}{q}=\frac{1}{4}$