Question

Two equal and opposite charges are placed at a certain distance apart and force of attraction between them is F. If $75\%$ charge of one is transferred to another, then the force between the charge becomes

• A. $\frac{7F}{16}\left(attraction\right)$
• B. $\frac{F}{16}\left(attraction\right)$
• C. $\frac{7F}{16}\left(repulsion\right)$
• D. $\frac{7F}{16}\left(repulsion\right)$

Answer 1

The correct option is B $\frac{F}{16}\left(attraction\right)$

$F_{a{H}_{-}}^1\alpha \frac{q_1{q}_2}{d^2}\to \left(1\right)q_1=q_2$
$F_{a{H}_{-}}^1\alpha \frac{q_1{q}_2}{44d^2}\to \left(2\right)q_1^1=\frac{q_1}{4};q_2^1=-q_1+\frac{3}{4}{q}_1{q}_2^1=-\frac{q}{4}$
$\frac{\left(2\right)}{\left(1\right)}given{F}_{a{H}_{-}}^1=\frac{F}{16}\left(attractive\right)$