Question

Two point charges $3\times {10}^{-6}C$ and $8\times {10}^{-6}C$ repel each other by a force of $6\times {10}^{-3}N$. If each of them is given an additional charge $-6\times {10}^{6}C$, the force between them will be

• A. $2.4\times {10}^{-3}N$ (attractive)
  
• B. $2.4\times {10}^{-9}N$ (attractive)
  
• C. $1.5\times {10}^{-3}N$ (repulsive)
  
• D. $1.5\times {10}^{-3}N$ (attractive)

Answer 1

The correct option is D $1.5\times {10}^{-3}N$ (attractive)

The electrostatic force between the charges is given by,

$F=\frac{k{Q}_1{Q}_2}{r^2}$
$\therefore F\propto Q_1{Q}_2$

$\Rightarrow \frac{F_1}{F_2}=\frac{Q_1{Q}_2}{Q_1^{\prime }{Q}_2^{\prime }}$

$=\frac{3\times {10}^{-6}\times 8\times {10}^{-6}}{(3\times {10}^{-6}-6\times {10}^{-6})(8\times {10}^{-6}-6\times {10}^{-6})}=\frac{3\times 8}{-3\times 2}=-\frac{4}{1}$

$\Rightarrow F_2=-\frac{F_1}{4}=-\frac{6\times {10}^{-3}}{4}=-1.5\times {10}^{-3}N$

Since now $Q_1^{\prime }$ is negative and $Q_2^{\prime }$ is positive therefore the electrostatic force between them becomes attractive.