Question

Two point charges A and B having charges $+q$ and $–q$ respectively are placed at certain distance apart and force acting between them is $F$. If 25% charge of A is transferred to B, force between the charges becomes

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

Answer 1

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


$F=\frac{-K{q}^2}{r^2}$
25% charge from A is transferred to B


New force $F\text{'}=\frac{K\left(\frac{3q}{4}\right)\left(\frac{-3q}{4}\right)}{r^2}=-\frac{9}{16}\frac{K{q}^2}{r^2}=\frac{9F}{16}$