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=-0.4 \, N$
$As \, F=\frac{q_{1} q_{2}}{4 \pi \epsilon _{0} r^{2}}$
$\therefore \, \, r^{2}=\frac{q_{1} q_{2}}{4 \pi \epsilon _{0} F}=\frac{0.2 \times 10^{- 6} \times 0.4 \times 10^{- 6} \times 9 \times 10^{9}}{0.4}$
$=1.8\times 10^{- 3}$
$\therefore \, \, \, r=\left(1.8 \times \left(10\right)^{- 3}\right)^{1 \backslash 2}=0.0424m=42.4\times \left(10\right)^{- 3}m$