Question

A parallel plate capacitor has capacitance $C$. If a dielectric slab of dielectric constant $K$ and thickness equal to half the plate separation is introduced parallel to the plates, the new capacitance is:

• A. $\dfrac{2K}{K+1}\,C$
• B. $\dfrac{K+1}{2K}\,C$
• C. $\dfrac{K}{K+1}\,C$
• D. $KC$

Hint:

Think of the system as two capacitors in series: $C_{air} = \dfrac{\varepsilon_0 A}{d/2}$ and $C_{dielectric} = \dfrac{K\varepsilon_0 A}{d/2}$. Their series combination gives $\dfrac{2K}{K+1}C$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Introducing a dielectric slab of thickness \( t \) into a capacitor of gap \( d \) creates two capacitors in series: one with air and one with dielectric.
Key Formula or Approach:
Capacitance with slab: \( C' = \frac{\epsilon_0 A}{d - t + t/K} \).
Original capacitance: \( C = \frac{\epsilon_0 A}{d} \).
Step 2: Detailed Explanation:
Given thickness \( t = d/2 \).
Substitute \( t \) into the formula:
\[ C' = \frac{\epsilon_0 A}{d - d/2 + (d/2)/K} \] \[ C' = \frac{\epsilon_0 A}{d/2 + d/2K} = \frac{\epsilon_0 A}{\frac{dK + d}{2K}} \] \[ C' = \frac{2K \epsilon_0 A}{d(K+1)} \] Since \( \frac{\epsilon_0 A}{d} = C \):
\[ C' = \frac{2K}{K+1} C \] Step 3: Final Answer:
The new capacitance is \( \frac{2K}{K+1} C \).
This matches Option (A).