Q120 / 180 · JIPMER 2017 · Physics

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?

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

Solution

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 $Q=2 q$

Hence, the correct answer is option: 'A'.

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