Question:
The electrostatic force between two charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum is $F$. If a dielectric medium of dielectric constant $K$ is introduced between them, the new force is
The electrostatic force between two charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum is $F$. If a dielectric medium of dielectric constant $K$ is introduced between them, the new force is
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Remember that introducing a dielectric medium with dielectric constant \( K \) reduces the electrostatic force by a factor of \( K \).
Updated On: May 31, 2026
- $F / K$
- $K F$
- $F / K^2$
- $K^2 F$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The electrostatic force between two point charges in a vacuum is given by Coulomb's law. When a dielectric medium with a dielectric constant \( K \) is introduced, the force between the charges changes due to the polarization of the dielectric.
Step 2: Meaning
A dielectric material reduces the electric field within it compared to that in a vacuum because the charges are partially neutralized by induced opposite charges on the dielectric. This results in a reduction of the electrostatic force between the original charges.
Step 3: Analysis
Coulomb's law states that the electrostatic force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) in vacuum is given by: \[F = k_e \frac{q_1 q_2}{r^2}\] where \( k_e \) is the Coulomb constant. When a dielectric medium with dielectric constant \( K \) is introduced, the effective force between the charges decreases. The new force \( F' \) can be derived by considering that the electric field within the dielectric is reduced to \( \frac{1}{K} \) of its value in vacuum. Therefore, the force becomes: \[F' = k_e \frac{q_1 q_2}{r^2 K} = \frac{F}{K}\] This shows that introducing a dielectric medium with dielectric constant \( K \) reduces the electrostatic force by a factor of \( K \).
Step 4: Conclusion
The new force between the charges when a dielectric medium is introduced is reduced to \( \frac{1}{K} \) of the original force. Final Answer: (A)
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