Question

The space between the plates of a parallel plate capacitor is halved and a dielectric medium of relative permittivity \( 10 \) is introduced between the plates. The ratio of the final and initial capacitances of the capacitor is:

• A. \( 20 \)
• B. \( 10 \)
• C. \( \frac{1}{10} \)
• D. \( \frac{1}{20} \)

Hint:

For a parallel plate capacitor, capacitance increases when the plate separation decreases or a dielectric is introduced. The new capacitance is given by \( C' = \kappa \frac{\varepsilon_0 A}{d} \).

Answer 1

The Correct Option is A

Step 1: Initial Capacitance Formula
The capacitance of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 A}{d} \] where: - \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the plate area,
- \( d \) is the separation between the plates.
Step 2: Effect of Halving the Plate Separation
When the plate separation \( d \) is halved:
\[ C' = \frac{\varepsilon_0 A}{d/2} = 2C \] Step 3: Effect of Introducing Dielectric
Introducing a dielectric of relative permittivity \( \kappa = 10 \) modifies the capacitance:
\[ C'' = \kappa C' = 10 \times 2C = 20C \] Step 4: Final Ratio
The ratio of final to initial capacitance is: \[ \frac{C''}{C} = \frac{20C}{C} = 20 \] Step 5: Conclusion
Thus, the ratio of final and initial capacitance is \( 20 \).