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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 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-
$\frac{F}{4}$
$\frac{3F}{4}$
$\frac{F}{8}$
$\frac{3F}{8}$
Solution
Let the spherical conductors $B$ and $C$ have same charge as $q$.
The electric force between them is
$\mathrm{F}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$
$r,$being the distance between them. When third uncharged conductor $A$ is brought
in contact with $\mathrm{B}$, 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
$q_{A}=q_{C}=\frac{q_{A}+q_{C}}{2}=\frac{(q / 2)+q}{2}=\frac{3 q}{2}$
Hence, electric force acting between $\mathrm{B}$ and $\mathrm{C}$ is
$\mathrm{F}^{\prime}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{B} q_{C}}{r^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(q / 2)(3 q+4)}{r^{2}}$
$=\frac{3}{8}\left[\frac{q}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}\right]=\frac{3 F}{8}$
