Question

The electrostatic force on a small sphere of charge $0.4\mu C$ due to another small sphere of charge $-0.8\mu C$ in air is 0.2 N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

Answer 1

(a)

Electrostatic force on the first sphere, $F=0.2N$

Charge on this sphere, $q_1=0.4\mu C=0.4\times {10}^{-6}C$

Charge on the second sphere, $q_2=-0.8\mu C=-0.8\times {10}^{-6}C$

Electrostatic force between the spheres is given by the relation,

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

And, $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^{-2}$

Where, ${\epsilon }_0=$ Permittivity of free space

$r^2=\frac{q_1{q}_2}{4\pi {\epsilon }_{0}F}$

$=\frac{04\times {10}^{-6}\times 8\times {10}^{-6}\times 9\times {10}^9}{0.2}$

$=144\times {10}^{-4}$

$r=\sqrt{144\times {10}^{-4}}$

$=0.12m$

The distance between the two spheres is 0.12 m.

(b) Equal and Opposite Force acts on the other sphere (By newton's Third Law). Hence 0.2 N attractive.