Question

Two unlike charges equal in magnitude are placed at some distance and force of F Newton acts between them. If $75\%$ of the charge from one is transferred to another. Then, how much would the value of force become :

• A. $\frac{F}{16}$
• B. $\frac{7F}{16}$
• C. $\frac{9F}{16}$
• D. $\frac{15F}{16}$

Answer 1

The correct option is B $\frac{7F}{16}$

$F=\frac{k{q}^2}{r^2}$ ........(1)

When the charge is transferred

$F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}=\frac{k{q}_1{q}_2}{r^2}$ .......(2)

The transferred charge

$75\%$ of $q=\frac{75}{100}\times q=\frac{3q}{4}$

$\therefore Charge{q}_1=q-\frac{3}{4}q=\frac{q}{4}$

$q_2=q+\frac{3}{4}q=\frac{7q}{4}$

The net force between them

$F^{\prime }=\frac{k{q}_1{q}_2}{r^2}=\frac{k\frac{q}{4}\times \frac{7q}{4}}{r^2}$

$F^{\prime }=\frac{7k}{16}\frac{q_2}{r^2}=\frac{7F}{16}$