Question

Two identical small conducting spheres are separated by 0.60 m. The spheres carry different amounts of charge and each sphere experiences an attractive electric force of 10.8N. The total charge on the two spheres is -24 ?C. The two spheres are now connected by a slender conducting wire, which is then removed. The electric force on each sphere is closest to

Answer 1

Electrostatic force between two sphere is given by

$F = \frac{kq_1q_2}{r^2}$

here we know that

r = 0.60 m

F = 10.8 N

$q_1 + q_2 = -24 \mu C$

now we will have

$10.8 = \frac{9 \times 10^9 q_1q_2}{0.60^2}$

since the two charges are attracting each other so here the two charges must be opposite in nature

$q_1q_2 = -4.32 \times 10^{-10}$

now we will use

$q_1 - \frac{4.32 \times 10^{-10}}{q_1}= -24 \times 10^{-6}$

$q_1^2 + 24 \times 10^{-6} q_1 - 4.32 \times 10^{-10} = 0$

by solving above equation we will have

$q_1 = \pm 36\mu C$

$q_2 = \mp 12\mu C$

now the force between two sphere when connected by wire is to be determine

so here we have net charge on each sphere is half of total charge

so we will have

$q_1 = q_2 = -12\mu C$

so force now is given as

$F = \frac{9\times 10^9 (-12\times 10^{-6})^2}{0.60^2}$

$F = 3.6 N$

this is repulsive force between two charges