Question

A certain charge Q is divided into two parts q and $Q-q$. How the charge Q and q must be related so that when q and $(Q-q)$ is placed at a certain distance apart experience maximum electrostatic repulsion?

• A. $Q=2q$
• B. $Q=3q$
• C. $Q=4q$
• D. $Q=4q+c$

Answer 1

Answer: (A)

The electrostatic force of repulsion between the charge q and $(Q-q)$ at separation r is given by
$F=\frac{1}{4\pi {\epsilon }_0}$ $\frac{q(Q-q)}{r^2}$ $=\frac{1}{4\pi {\epsilon }_0}$ $\frac{qQ-q^2}{r^2}$
If F is maximum, then $\partial F/\partial q=0$
i.e., $\frac{1}{4\pi {\epsilon }_0}\cdot \frac{(Q-2q)}{r^2}=0$
As $\frac{1}{4\pi {\epsilon }_0{r}^2}$ is constant, therefore
$Q-2q=0$ or $Q=2q$