Coulomb's Law: Practice Problem & Solution
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?
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Coulomb's Law. Understanding the underlying formula is key to arriving at the correct answer below:
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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