Two charges q1 and q2 are placed in vacuum at a distance d and the force acting between them is F. If a medium of dielectric constant 4 is introduced around them, the force now will be4F2FF/2F/4
Question
Two charges q1 and q2 are placed in vacuum at a distance d and the force acting between them is F. If a medium of dielectric constant 4 is introduced around them, the force now will be4F2FF/2F/4
Solution
The force between two charges in a vacuum is given by Coulomb's law:
F = k * q1 * q2 / d^2
where k is Coulomb's constant.
When a medium with a dielectric constant (K) is introduced, the force becomes:
F' = k * q1 * q2 / (K * d^2)
Given that the dielectric constant K is 4, the new force is:
F' = k * q1 * q2 / (4 * d^2)
This is one fourth of the original force, so the answer is F/4.
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