Question

The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge − 0.8 µC 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 = 0.2 N
Charge on this sphere, $q_1$ = 0.4 µC = $0.4 × 10^{−6}C$
Charge on the second sphere, $q_2$ = − 0.8 µC = $− 0.8 × 10^{−6}C$
Electrostatic force between the spheres is given by the relation
$F = \frac{1 }{ 4πε_0}. \frac{q_1q_2}{r^2}$
Where, $ε_0$ = Permittivity of free space and\frac{1 }{ 4πε_0}$$= 9 × 10^9 Nm^2C^{-2}
Therefore,
$r^2 = \frac{1 }{4πε0}. \frac{q_1q_2}{F}$
$= \frac{0.4 × 10^{-6} × 8 × 10^{-6} × 9 × 10^9 }{0.2 }= 144 × 10^{-4}$
$r = \sqrt{144 × 10^{ -4}} = 12 × 10^{-2} = 0.12\,m$
The distance between the two spheres is 0.12 m.

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(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.