Q84 / 160 · JIPMER 2009 · Physics

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges in them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ but uncharged, is brought in contact with $B$, then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is

  • A
    $\frac{F}{4}$
  • B
    $\frac{3 F}{4}$
  • C
    $\frac{F}{8}$
  • Correct answer
    D
    $\frac{3 F}{8}$

Solution

Let the spherical conductors $B$ and $C$ have same charge as $q$. The electric force between them is $F=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2}$ $r$, being the distance between them. When third uncharged conductor $A$ is brought in contact with $B$, then charge on each conductor $q_A=q_B=\frac{q_A+q_B}{2}=\frac{0+q}{2}=\frac{q}{2}$ When this conductor $A$ is now brought in contact with $C$, then charge on each conductor $q_A=q_C=\frac{q_A+q_C}{2}=\frac{(q / 2)+q}{2}=\frac{3 q}{4}$ Hence, electric force acting between $B$ and $C$ is $\begin{aligned} F^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{q_B q_C}{r^2} & =\frac{1}{4 \pi \varepsilon_0} \frac{(q / 2)(3 q / 4)}{r^2} \\ & =\frac{3}{8}\left[\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2}\right]=\frac{3 F}{8} \end{aligned}$

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

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