Question

Two identical conducting spheres $A$ and $B$ carry equal charge. They are initially separated by a distance much larger than their diameters and the force between them is $F.A$ third identical conducting sphere $C$ is uncharged. Sphere $C$ is first touched to $A$ , then to $C$ and removed. As a result, the force between $A$ and $B$ now is

• A. $F/16$
• B. $F/4$
• C. $3F/8$
• D. $F/2$

Answer 1

Answer: (C)

$r$ = distance between two identical spheres,
and $Q$ = charge on spheres $A$ and $B$
$F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{r^2}\dots \left(i\right)$
When a third identical sphere 'C' is first touches with $A$, then the value of charges on $A$ and $C$ will be $\frac{Q}{2}$
Again, when $C$ touches with $B$, then value of charge on $B$ and $C=\frac{\frac{Q}{2}+Q}{2}=\frac{3Q}{4}$
Now, the electrostatic force between spheres $A$ and $B$
$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\frac{Q}{2}\cdot \frac{3Q}{4}}{r^2}=\frac{3}{8}\cdot \frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{r^2}$
$=\frac{3}{8}F$ [from Eq. $\left(i\right)$]