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Two identical conducting spheres carry identical charges. If the spheres are set at a certain distance apart, they repel each other with a force $F$. A third conducting sphere identical to the other two, but initially uncharged is touched to one sphere and then to the other before being removed. The force between the original two spheres is now
$\frac{F}{2}$
$\frac{F}{4}$
$\frac{3 F}{4}$
$\frac{3 F}{8}$
Solution
(d) Initial condition is
So, initial value of force between spheres is $F=k q^2 / r^2$.
When another identical and uncharged sphere $C$ is touched to $A$, charge on each of $A$ and $C$ will be $\frac{q}{2}$.
When $C$ is touched with $B$, then charge on each of them will be mean value of total charge.
So, charge on $B=$ charge on $C$
$=\frac{q+\frac{q}{2}}{2}=\frac{3 q}{4}$
Now, the final condition is
So, final value of force between spheres is
$\frac{F^{\prime}-k q_1 \cdot q_2}{r^2}$
$=\frac{k \frac{9}{2} \times \frac{3 q}{4}}{r^2}=\frac{3}{8} \frac{k q^2}{r^2}$
or $F^{\prime}=\frac{3}{8} F$
