Question

The electrostatic attracting force on a small sphere of charge $0.2\mu C$ due to another small sphere of charge $-4\mu C$ in air is $0.4N$ . The distance between the two spheres must have been

• A. $43.2\times 10^{-6}m$
• B. $42.4\times 10^{-3}m$
• C. $18.1\times 10^{-3}m$
• D. $192\times 10^{-6}m$

Answer 1

Answer: (B)

Here, $q_1=0.2\mu C=0.2\times 10^{-6}C$
$q_2=-0.4\mu C=-0.4\times 10^{-6}C,F=-0.4N$
$AsF=\frac{q_1{q}_2}{4\pi {ϵ}_0{r}^2}$
$\therefore r^2=\frac{q_1{q}_2}{4\pi {ϵ}_{0}F}=\frac{0.2\times 10^{-6}\times 0.4\times 10^{-6}\times 9\times 10^9}{0.4}$
$=1.8\times 10^{-3}$
$\therefore r={\left(1.8\times {\left(10\right)}^{-3}\right)}^{1\backslash 2}=0.0424m=42.4\times {\left(10\right)}^{-3}m$