Question

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

• A. $\frac{3F}{4}$
• B. $\frac{F}{2}$
• C. $F$
• D. $\frac{3F}{8}$

Answer 1

The correct option is D $\frac{3F}{8}$

Initially charges on both sphere, $q$

$F=\frac{K{q}^2}{d^2}$ .............(1)

When sphere C will get touched with sphere A, then final charges on both will become $\frac{q}{2}$.

Now, when this sphere C will get touched to sphere B, then final charges on both of them will be

$q_c=q_d=\frac{q/2+q}{2}=\frac{3q}{4}$

Now force between A and B will be

$F^{\prime }=\frac{k\times \frac{q}{2}\times \frac{3q}{4}}{d^2}$

$F^{\prime }=\frac{3F}{8}$