Question

Two identical conducting spheres $A$ and $B$, carry equal charge. They are 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 $B$, and then removed. As a result, the force between $A$ and $B$ would be equal to :

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

Answer 1

The Correct Option is C

Let the charge on each sphere $A$ and $B$ be $q$ and the separation is $d$.
Therefore, force between spheres $A$ and $B$ is

$F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{d^{2}}$...(1)
When spheres $A$ and $C$ are touched and then separated, charge on each will be $\frac{q+0}{2}$, i.e. $\frac{q}{2} .$

Now sphere $B$ is touched with sphere $C$, charge on each will be $\left[\frac{q+\frac{q}{2}}{2}\right]=\frac{3 q}{4}$.

Now force between sphere $A$ and sphere $B$ will be

$F'=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\frac{q}{2} \cdot \frac{3 q}{4}}{d^{2}}=\frac{3}{8} \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{d^{2}}$
$\Rightarrow F'=\frac{3}{8} F$