Question

A change 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 1 m 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

The correct option is B 0.00216

Given that,

$1\mu C$ charge is divided in ratio $2:3$ and kept at distance.

$d=1m$ apart. to find electric force between them.

so, two parts have charges

$\frac{2}{5}\times 1\mu Cand\frac{3}{5}\times 1\mu Cie,q_1=\frac{2}{5}\mu Cand\frac{3}{5}\mu C=q_{2}d=1m$

so; coulomb's force b/w these 2 parts is

$f=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{d^2}=\frac{9\times {10}^9}{{\left(1\right)}^2}\times \frac{2}{5}\times \frac{3}{5}\times {10}^{-12}=2.16\times {10}^{-3}=0.00216N$

so the answer is option (b).