Two identical small charged spheres are a certain distance apart,and there is an electrostatic force of magnitude F that they exert on each other.With time,charge gradually diminishes on both spheres by leaking off.When each of the spheres has lost half its initial charge,what will be the magnitude of the electrostatic force exerted on each one?

Question

Quizplus · Accepted Answer

The electrostatic force of attraction or repulsion between two charged spheres is given by the formula,

F = (1/4πε) * (q1 * q2) / r^2

Where ε is the permittivity of the medium between the spheres, q1 and q2 are the charges on the spheres, and r is the distance between the spheres.

As the spheres are identical, they will lose charge at the same rate, and when each sphere has lost half of its initial charge, the charge on each sphere becomes q/2, where q is the initial charge.

Therefore, the new force of attraction between the spheres can be calculated using the same formula as,

F' = (1/4πε) * [(q/2) * (q/2)] / r^2

F' = (1/4πε) * (q^2/4) / r^2

F' = (1/16) * [(1/4πε) * (q^2/r^2)]

F' = (1/16) * F

Therefore, the magnitude of the electrostatic force exerted on each sphere is reduced to 1/4 of the original force, and the best choice is C, 1/4 F.

Two Identical Small Charged Spheres Are a Certain Distance Apart,and 2\sqrt { 2 }2​

Two Identical Small Charged Spheres Are a Certain Distance Apart,and $\sqrt { 2 }$