Question

The electrostatic attracting force on a small sphere of charge $0.2\mu C$ due to another small sphere of charge $-0.4\mu C$ in air is $0.4N$. The distance between the two spheres is

• A. $43.2\times {10}^{-6}m$
• B. $42.4\times {10}^{-3}m$
• C. $18.1\times {10}^{-3}m$
• D. $19.2\times {10}^{-6}m$

Answer 1

The correct option is B $42.4\times {10}^{-3}m$

Here, $q_1=0.2\mu C=0.2\times {10}^{-6}C,q_2=.-0.4\mu C=-0.4\times {10}^{-6}C,F=-0.4N$

As $F=\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}^2}$

$r^2=\frac{q_1{q}_2}{4\pi {\epsilon }_{0}F}=\frac{0.2\times {10}^{-6}\times 0.4\times {10}^{-6}\times 9\times {10}^9}{0.4}=1.8\times {10}^{-3}$

$\therefore r={(1.8\times {10}^{-3})}^{1/2}=0.0424m=42.4\times {10}^{-3}m$