Question

A certain charge $Q$ is to be divided into two parts $q$ and $Q-q$. The relationship between $Q$ and $q$ placed at a certain distance apart to have the maximum Coulomb repulsion is

• A. $q=\frac{Q}{2}$
• B. $q=\frac{Q}{3}$
• C. $q=\frac{2Q}{2}$
• D. $q=\frac{Q}{4}$

Answer 1

The correct option is A $q=\frac{Q}{2}$

Let the divided parts be $q$ and $Q-q$, here $q$ is the variable part and total charge $Q$ is a fixed quantity.
The coulombic force between charges is given by,
$F=\frac{1}{4\pi {ϵ}_0}\frac{q(Q-q)}{r^2}$
$\Rightarrow$ For the force to be maximum $\frac{dF}{dq}=0$ and $\frac{d^{2}F}{d{q}^2}<0$, the charge at which $\frac{dF}{dq}$ is zero.

$\Rightarrow \frac{dF}{dq}=\frac{1}{4\pi {ϵ}_0{r}^2}\frac{d}{dq}[Qq-q^2]=0$
$\Rightarrow \frac{d}{dq}[Qq-q^2]=0$
$\Rightarrow Q-2q=0$
$\Rightarrow q=\frac{Q}{2}$
also, $\frac{d^{2}F}{d{q}^2}=\frac{1}{4\pi {ϵ}_0{r}^2}\frac{d}{dq}(Q-2q)$
$\Rightarrow \frac{d^{2}F}{d{q}^2}=-\frac{2}{4\pi {ϵ}_0{r}^2}$
Thus, $q=\frac{Q}{2}$ is a point of maxima.
$\therefore$ Force will be maximum at $q=\frac{Q}{2}$.

$\begin{array}{c}\text{Why this question ?}\\ \text{Caution: When you are dividing the charge into}\\ \text{q and Q-q, the variable part is q only.}\end{array}$