Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the

Question

Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 	imes 10^{-7}$ C? (The radii of A and B are negligible compared to the distance of separation.)

Accepted Answer

The force of repulsion between two point charges is given by Coulomb's Law: $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$.
Given:
$q_1 = q_2 = 6.5 	imes 10^{-7}$ C
$r = 50$ cm $= 0.5$ m
$k = 9 	imes 10^9$ N m$^2$ C$^{-2}$

Calculation:
$F = \frac{9 	imes 10^9 	imes (6.5 	imes 10^{-7}) 	imes (6.5 	imes 10^{-7})}{(0.5)^2}$
$F = \frac{9 	imes 42.25 	imes 10^{-5}}{0.25}$
$F = 9 	imes 169 	imes 10^{-5}$
$F = 1521 	imes 10^{-5}$ N
$F = 1.52 	imes 10^{-2}$ N.
(See NCERT Physics Class 12, Exercise 1.12).

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