Coulomb’s Law with Dielectric – Rankers Physics
Topic: Electrostatics
Subtopic: Coulomb's Law

Coulomb’s Law with Dielectric

Two point charges exert a force \(F_0\) on each other when placed in vacuum. Now the charges are increased to four times, separation between them is doubled and the system is placed is an insulating medium. Now they experience the same force. What should be the dielectric constant of the medium?
3
4
2
5

Solution:

Initial force in vacuum \(F_0 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\) . New force in medium \(F' = \frac{1}{4\pi\epsilon_0 K} \frac{(4q_1)(4q_2)}{(2r)^2} = \frac{16}{4K} \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\) . Since \(F' = F_0\), we have \(frac{4}{K} = 1\), so \(K = 4\).

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