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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?
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The correct answer is:
$Q=2 q$
The electrostatic force of repulsion between the charge $q$ and $(Q-q)$ at separation $r$ is given by
$F=\frac{1}{4 \pi \varepsilon_0} \frac{q(Q-q)}{r^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q Q-q^2}{r^2}$
If $F$ is maximum, then $\partial F / \partial q=0$ i.e. $\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(Q-2 q)}{r^2}=0$
As $\frac{1}{4 \pi \varepsilon_0 r^2}$ is constant, therefore
$Q-2 q=0$ or $Q=2 q$
$F=\frac{1}{4 \pi \varepsilon_0} \frac{q(Q-q)}{r^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q Q-q^2}{r^2}$
If $F$ is maximum, then $\partial F / \partial q=0$ i.e. $\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(Q-2 q)}{r^2}=0$
As $\frac{1}{4 \pi \varepsilon_0 r^2}$ is constant, therefore
$Q-2 q=0$ or $Q=2 q$
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