MP Board · Class 12 · Physics · Electric Charges and FieldsThe electrostatic force between two point charges placed in air at a distance $d$ apart is $F$. If a dielectric medium of dielectric constant $K = 4$ is introduced between them, the new electrostatic force will be:
According to Coulomb's Law, the electrostatic force in vacuum/air is: $$F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$$ \nWhen a medium with dielectric constant $K$ is introduced, the new force $F'$ is given by: $$F' = \frac{F}{K}$$ \nGiven $K = 4$: $$F' = \frac{F}{4}$$ \nThus, the force reduces to one-fourth of its initial value in air. The correct option is $\frac{F}{4}$.
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electric Charges and Fields.
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electric Charges and Fields.
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electric Charges and Fields.