Question

Two point charges, A and B, are separated by a distance of 23.0 cm23.0 cm . The magnitude of the charge on A is twice that of the charge on B. If each charge exerts a force of magnitude 48.0 N48.0 N on the other, find the magnitudes of the charges.

Answer 1

Let, the magnitude of the charge on B be q.

Hence, the magnitude of the charge on A is 2q.

These charges are d = 23 cm = 0.23 m apart.

Hence, force exerted by each charge on the other one = F = ( 1 / 4_o ) ( q x 2q / d² ).

Given, F = 48 N.

Hence, ( 1 / 4_o ) ( q x 2q / d² ) = 48

or, 9 x 10⁹ x 2q² / 0.23² = 48

or, 3.4 x 10¹¹ x q² = 48

or, q = { 48 / ( 3.4 x 10¹¹ ) } C = 1.2 x 10^-5 C.

So, 2q = 2 x 1.2 x 10^-5 C = 2.4 x 10^-5 C.

Hence, the magnitude of the charge on A is 2.4 x 10^-5 C, and, that of the charge on B is 1.2 x 10^-5 C.