Question

A charge of $1\mu C$ is divided into two parts such that their charges are in the ratio of $2:3$. These two charges are kept at a distance 1m apart in vaccum. Then, the electric force between them (in newton) is :

• A. $0.216$
• B. $0.00216$
• C. $0.0216$
• D. $2.16$

Answer 1

Answer: (B)

$0.00216$

Let the charges be

$2q$ and $3q$

$2q+3q=1\mu C$

$5q=1\mu C$

$q=\frac{1}{5}\mu C$

$\therefore chargesare\frac{2}{5}\mu Cand\frac{3}{5}\mu C$

Force $=\frac{K{q}_1{q}_2}{r^2}$

$=\frac{9\times {10}^9\times }{1^2}\frac{2}{5}\times {10}^{-6}\times \frac{3}{5}\times {10}^{-6}$

$=\frac{9\times 6\times {10}^{-3}}{25}$

$=2.16\times {10}^{-3}$

$=0.00216$