Question

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually diminishes on both spheres by leaking off.

1. When each of the spheres has lost half its initial charge, what will be the magnitude of the electrostatic force on each one?
A) 1/16 F
B) 1/8 F Coulomos coulombs Law
C) 1/4 F
D) 1/2 F
E) 1/√ 2 F

Answer 1

C) 1/4 F

If we assume the small chargued spheres as a point chargues, we can use the Coulomb's law. According to this law, the magnitude of the electrostatic force between two point chargues is directly proportional to the product of the charges and inversely proportional to the square of the distance between them:

`F=k(q_1q_2)/(d^2)`

In this case, we have:
`q_1'=(q_1)/(2)`,
`q_2'=(q_2)/(2)` and
`d'=d`. Thus, the new magnitude of the electrostatic force on each one is:

`F'=k(q_1'q_2')/(d'^2)\\F'=k(((q_1)/(2))((q_2)/(2)))/(d^2)\\F'=k(q_1q_2)/(4d^2)\\F'=(1)/(4)k(q_1q_2)/(d^2)\\F'=(1)/(4)F`