Question

Force between two identical spheres charged with same charge is $F$. If $50\%$ charge of one sphere is transfered to the other sphere keeping the distance between them same, then the new force between spheres will be

• A. $\frac{3}{4}F$
• B. $\frac{3}{8}F$
• C. $\frac{3}{2}F$
• D. None of these

Answer 1

The correct option is A $\frac{3}{4}F$

Given initial condition


Magnitude of force acting between both spheres is,
$F=\frac{k{q}^2}{r^2}$

After $50\%$ charge of $S_1$ being transferred to sphere $S_2$ the new charges on spheres have been shown in figure given below.


Therefore magnitude of new elctrostatic force between spheres is given by,
$F^{\prime }=\frac{k\left(q/2\right)\left(3q/2\right)}{r^2}=\frac{3}{4}\frac{k{q}^2}{r^2}$
$\Rightarrow F^{\prime }=\frac{3F}{4}$
$$\begin{array}{c}\text{Why this problem?}\\ \text{To revise the charge dependence for electrostatic force.}\\ \text{Tip: Due to symmetry of sphere, we can consider the whole}\\ \text{charge of sphere at its centre. Thus we can apply}\\ \text{coulomb's law by taking separation between charges as r}\end{array}$$