Question

Two small, identical conducting spheres repel each other with a force of 0.030 N when they are 0.65 m apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of 0.075 N. What is the original charge on each sphere? (Enter the magnitudes in C.)

Answer 1

Note that the methods applied in solving this question is the appropriate method. Check the parameters you gave in the question if you did not expect a complex number for the charges. Thanks

Answer:

$q_1 = 0.00000119 + j0.00000145 C \\q_2 = 0.00000119 - j0.00000145 C$

Explanation:

Note: When a conducting wire was connected between the spheres, the same charge will flow through the two spheres.

The two charges were 0.65 m apart. i.e. d = 0.65 m

Force, F = 0.030 N

The force or repulsion between the two charges can be calculated using the formula:

$F = \frac{kq^2}{d^2} \\\\0.030 = \frac{9 * 10^9 * q^2}{0.65^2}\\\\q = 1.19 * 10^{-6} C$

Due to the wire connected between the two spheres, $q_1 = q_2 = 1.19 * 10^{-6} C$

The sum of the charges on the two spheres = $q_1 + q_2 = 2.38 * 10^{-6} C$

Note: When the conducting wire is removed, the two spheres will no longer contain similar charges but will rather share the total charge unequally

Let charge in the first sphere = $q_1$

Charge in the second sphere, q₂ = $2.38 * 10^{-6} - q_1$

Force, F = 0.075 N

$F = \frac{k q_1 q_2}{r^2} \\\\0.075 = \frac{9*10^9 * q_1 * (2.38*10^{-6} -q_1 )}{0.65^2}\\\\3.52 * 10^{-12} = q_1 * (2.38*10^{-6} -q_1 )\\\\3.52 * 10^{-12} = 2.38*10^{-6} q_1 - q_1^2\\\\q_1^2 - (2.38*10^{-6}) q_1 + (3.52 * 10^{-12}) = 0$

$q_1 = 0.00000119 + j0.00000145 C \\q_2 = 0.00000119 - j0.00000145 C$