Question

The force between two point charges placed in a material medium of dielectric constant $\varepsilon_r$ is $F$. If the material is removed, then the force between them becomes

• A. $\varepsilon_r F$
• B. $\varepsilon F$
• C. $\frac{F}{\varepsilon_r}$
• D. $\frac{\varepsilon}{F}$
• E. $\varepsilon_0 F$

Hint:

Dielectric reduces force → removing it increases force by factor $\varepsilon_r$.

Answer 1

The Correct Option is A

Concept: The force between two point charges depends on the medium in which they are placed. According to Coulomb’s Law: \[ F = \frac{1}{4\pi \varepsilon} \cdot \frac{q_1 q_2}{r^2} \] where: \[ \varepsilon = \varepsilon_0 \varepsilon_r \] Thus in a medium: \[ F_{\text{medium}} = \frac{1}{4\pi \varepsilon_0 \varepsilon_r} \cdot \frac{q_1 q_2}{r^2} \] In vacuum (or air approximately): \[ F_{\text{vacuum}} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \]

Step 1: Compare the two forces: \[ F_{\text{medium}} = \frac{F_{\text{vacuum}}}{\varepsilon_r} \]

Step 2: Rearranging: \[ F_{\text{vacuum}} = \varepsilon_r \cdot F_{\text{medium}} \]

Step 3: Interpretation:
• Dielectric medium reduces the force
• Removing the medium increases force
• Increase factor = $\varepsilon_r$

Step 4: Physical explanation: When a dielectric is present:
• Molecules get polarized
• Internal electric field opposes original field
• Net force reduces When removed:
• No polarization effect
• Full electric field restored
• Force increases

Step 5: Final result: \[ F_{\text{new}} = \varepsilon_r F \] Final Answer: \[ \varepsilon_r F \]