Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.2589 N when separated by 50 cm, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.3456 N. What were the initial charges on the spheres? Since one is negative and you cannot tell which is positive or negative, there are two solutions. Take the absolute value of the charges and enter the smaller value here. Enter the larger value here.

Answer 1

Answer:

$q_1 = 7.19 \times 10^{-6} C$

$q_2 = -1.0 \times 10^{-6} C$

Explanation:

Let the initial charge on the two spheres are

$q_1, -q_2$

now we know that the force between them is given as

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

$0.2589 = \frac{kq_1q_2}{0.5^2}$

$q_1q_2 = 7.19 \times 10^{-12}$

now when two spheres are connected then final charge on them is given as

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

now the force between them is given as

$0.3456 = \frac{k(q_1 - q_2)^2}{4(0.5)^2}$

now we have

$q_1 - q_2 = 6.19 \times 10^{-6}$

So by solving above two equations we have

$q_1 = 7.19 \times 10^{-6} C$

$q_2 = -1.0 \times 10^{-6} C$