Question

Two point charge <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{A}$ and $\text{B}$ having charges $+Q$ and $-Q$ respectively, are placed at certain distance apart and force acting between them is $F$. If $25\%$ charge of <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{A}$ is transfered to $\text{B}$, then force between the charges becomes

• A. $F$
• B. $\frac{4}{3}F$
• C. $\frac{9F}{16}$
• D. $\frac{16F}{9}$

Answer 1

The correct option is C $\frac{9F}{16}$

Given, initially

$q_A=+Q{q}_B=-Q$

finally
$q_A^{{}^{\prime }}=Q-\frac{25}{100}Q=\frac{3Q}{4}{q}_B^{{}^{\prime }}=-Q+\frac{25}{100}Q=-\frac{3Q}{4}$


$F\propto q_A{q}_B.......\left(1\right)$

$F^{{}^{\prime }}\propto q_A^{{}^{\prime }}{q}_B^{{}^{\prime }}.......\left(2\right)$

On dividing (2) $÷$ (1)

$\frac{F^{{}^{\prime }}}{F}=\frac{q_A^{{}^{\prime }}{q}_B^{{}^{\prime }}}{q_A{q}_B}=\frac{\frac{3Q}{4}(-\frac{3Q}{4})}{Q(-Q)}=\frac{9}{16}$

$F^{{}^{\prime }}=\frac{9F}{16}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (c) is the correct answer.