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 $B$, then force between the charges becomes

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

Answer 1

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


Magnitude of electrostatic force between the two point charges $A$ and $B$

$F=\frac{K{Q}^2}{r^2}$ ______(1)

Now $25$% charge of $A$ is transfered to $B$


In this case electrostatic force becomes

$F_1=\frac{K(\frac{3Q}{4}{)}^2}{r^2}$ ______(2)

$F_1=\frac{9}{16}\frac{K{Q}^2}{r^2}$

from equation (1) $F=\frac{K{Q}^2}{r^2}$

$\therefore F_1=\frac{9}{16}F$

So, (b) is the correct option