Physics - Electrostatics Question with Solution | TestHub

Let charge on $A$ and $B$ be $q_A=q_B=q$.

Then force acting between them is $F=\frac{K{q}^2}{r^2}$.

When $C$ is placed in contact with $A$, charge on $A$ and $C$ will be $=\frac{q}{2}$

Now $C$ is placed in contact with $B$, charge on $B$  and $C$ will be $=\frac{q+\frac{q}{2}}{2}=\frac{3q}{4}$

Now, force experienced by sphere $C$ is 

$$\begin{gathered} F^{\text{'}}=F_2-F_1 \\ =\frac{K9q^2}{16\times {\displaystyle \frac{r^2}{4}}}-\frac{K3q^2}{8\times {\displaystyle \frac{r^2}{4}}} \\ =\frac{\left(K{\displaystyle \frac{3q}{4}}-K{\displaystyle \frac{q}{2}}\right)}{{\displaystyle \frac{r^2}{4}}}\times \frac{3q}{4} \end{gathered}$$

$=\frac{3K{q}^2}{4r^2}=\frac{3 F}{4}$