Question

Find the capacitance of the Parallel plate capacitor shown in the figure.
(A = area of each plate, \(\varepsilon _{0}\) = permittivity of free space)

• A. \(\frac{2KA\varepsilon _{0}}{(K+1)d}\)
• B. \(\frac{2KA\varepsilon _{0}}{d}\)
• C. \(\frac{(K+1)A\varepsilon _{0}}{2d}\)
• D. \(\frac{2KA\varepsilon _{0}}{(K^{2}+1)d}\)

Hint:

The dielectric slab and the air gap are two capacitors in series.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The slab of thickness \(\frac{d}{2}\) and constant \(K\) fills the full plate area, leaving an air gap of thickness \(\frac{d}{2}\). The two layers are in series.

Step 2: Capacitance of Each Layer
\(C_1=\frac{K\varepsilon_0A}{d/2}=\frac{2K\varepsilon_0A}{d}\) and \(C_2=\frac{\varepsilon_0A}{d/2}=\frac{2\varepsilon_0A}{d}\).

Step 3: Series Combination
\(\frac{1}{C}=\frac{d}{2K\varepsilon_0A}+\frac{d}{2\varepsilon_0A}=\frac{d(1+K)}{2K\varepsilon_0A}\).

Step 4: Result
\(C=\frac{2KA\varepsilon_0}{(K+1)d}\), option (A).

Final Answer:
The capacitance is 2K A epsilon-naught over (K + 1) d. \[ \boxed{\text{(A) }\dfrac{2KA\varepsilon_{0}}{(K+1)d}} \]