Question

The capacitance of a parallel plate capacitor with air as medium is \( 3 \, \mu \text{F} \) . With the introduction of a dielectric medium between the plates, the capacitance becomes \( 15 \, \mu \text{F} \) . The permittivity of the medium in SI units is \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \).

• A. 15
• B. \( 8.845 \times 10^{-11} \)
• C. \( 0.4425 \times 10^{-10} \)
• D. 5

Hint:

The capacitance increases with the introduction of a dielectric medium, and the permittivity is related to the dielectric constant.

Answer 1

The Correct Option is C

Step 1: Formula for capacitance.
The capacitance with a dielectric is given by: \[ C = C_0 \times \kappa \] where \( C_0 \) is the capacitance without the dielectric and \( \kappa \) is the dielectric constant. We are given that \( C_0 = 3 \, \mu \text{F} \) and \( C = 15 \, \mu \text{F} \), so: \[ \kappa = \frac{C}{C_0} = \frac{15}{3} = 5 \]
Step 2: Finding the permittivity of the medium.
The permittivity \( \epsilon \) is related to \( \kappa \) and \( \epsilon_0 \) by: \[ \epsilon = \kappa \times \epsilon_0 \] Substituting the values: \[ \epsilon = 5 \times (8.85 \times 10^{-12}) = 4.425 \times 10^{-11} \, \text{F/m} \]
Step 3: Conclusion.
The correct answer is (C) \( 0.4425 \times 10^{-10} \, \text{F/m} \).