Q125 / 160 · TS EAMCET 2014 · Physics

A charge $Q$ is divided into two charges $q$ and $Q-q$. The value of $q$ such that the force between them is maximum, is

  • A
    $Q$
  • B
    $\frac{3 Q}{4}$
  • Correct answer
    C
    $\frac{Q}{2}$
  • D
    $\frac{Q}{3}$

Solution

By Coulomb's law, When charge $Q$ is divided into two charges $q$ and $Q-q$ $$ F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q \cdot(Q-q)}{r^2} $$ The value of $q$ $$ F_{\max }=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\frac{Q}{2}\left(Q-\frac{Q}{2}\right)}{r^2} $$ (Putting $q=\frac{Q}{0}$ for the maximum force) $$ \begin{aligned} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\frac{Q}{2}\left(\frac{Q}{2}\right)}{r^2} \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\frac{Q^2}{4}}{r^2} \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{4 r^2} \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\left(\frac{Q}{2}\right)^2}{r^2} \end{aligned} $$ So, option (3) is correct.

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

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