Question

Coulomb's Law: Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F F due to the 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 on each one?

Answer 1

Answer:

The magnitude of electrostatic force on each charge is quarter of the magnitude of initial electrostatic force. ( ¹/₄ F)

Explanation:

The electrostatic force between two charges is given by Coulomb's law;

$F = \frac{kQ_1Q_2}{r^2}$

where;

Q₁ and Q₂ are the magnitude of the charges

r is the distance between the charges

k is Coulomb's constant

Since the charges are identical;

Q₁ = Q

Q₂ = Q

the electrostatic force experienced by each charge is given by;

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

When each of the spheres has lost half of its initial charge;

Q₁ = Q/2

Q₂ = Q/2

$F_2 = \frac{k(Q/2)(Q/2)}{r^2}\\\\ F_2 = \frac{k(Q)(Q)}{4r^2}\\\\F_2 = \frac{1}{4} (\frac{kQ^2}{r^2} )\\\\F_2 = \frac{1}{4} (F)$

Therefore, the magnitude of electrostatic force on each charge is quarter of the magnitude of initial electrostatic force.