Question

Two small, identical conducting spheres repel each other with a force of 0.045 N when they are 0.15 m apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of 0.070 N. What is the original charge on each sphere?

Answer 1

Answer:

$q_1 = \pm 1.68 \times 10^{-7} C$

$q_2 = \pm 6.68 \times 10^{-7} C$

Explanation:

As we know that the force between two small spheres is given as

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

here we know that

$q_1 , q_2$ = charges on two small spheres

r = distance between two spheres = 0.15 m

now the force between them is given as

$0.045 = \frac{(9\times 10^9)(q_1q_2)}{0.15^2}$

$q_1q_2 = 1.125 \times 10^{-13}$

now when two spheres are connected together then the charge on them is equally divided

$q = \frac{q_1+q_2}{2}$

now the force between them is given as

$F = \frac{k(\frac{q_1+q_2}{2})^2}{0.15^2}$

$0.070 = \frac{(9\times 10^9)(\frac{q_1+q_2}{2})^2}{0.15^2}$

$q_1 + q_2 = 8.37\times 10^{-7}$

so here we have

$q_1 = \pm 1.68 \times 10^{-7} C$

$q_2 = \pm 6.68 \times 10^{-7} C$