Question

Two equal charges are placed at a distance of $3$ m. A repulsive force of $1.6$ N acts on it. The value of each charge will be :

• A. $2\mu C$
• B. $4\mu C$
• C. $40\mu C$
• D. $80\mu C$

Answer 1

The correct option is C $40\mu C$

Given $r=3$ metre

$F=1.6N$ $q_1=q_2=q$

$\therefore 1.6=\frac{1}{4\pi ϵ}.\frac{q\times q}{3^2}$

$\Rightarrow 1.6=9\times {10}^9\times \frac{q^2}{9}$

$\Rightarrow q^2=\frac{1.6}{{10}^9}=\frac{16}{{10}^{10}}\Rightarrow q=\frac{4}{{10}^5}C$

$\therefore q=4\times {10}^{-5}C\Rightarrow q=40\mu C$