Question

The equivalent capacitance between plates ' A ' and ' B ' ( A -area of each plate, d-separation between plates) ( \(\varepsilon_0\) - permittivity of free space) is

• A. \(\frac{\mathrm{A}\epsilon _{0}}{\text{ }\mathrm{d}}\)
• B. \(\frac{2\text{ }\mathrm{A}\epsilon _{0}}{\text{ }\mathrm{d}}\)
• C. \(\frac{4\text{ }\mathrm{A}\epsilon _{0}}{\text{ }\mathrm{d}}\)
• D. \(\frac{8\text{ }\mathrm{A}\epsilon _{0}}{\text{ }\mathrm{d}}\)

Hint:

Left half is one capacitor; right half has two layers in series; the two halves are in parallel.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The two halves of the plate area are in parallel. The right half has two dielectric layers of thickness \(\dfrac{d}{2}\) in series.

Step 2: Left half
\(C_1=\dfrac{K_1\epsilon_0(A/2)}{d}=\dfrac{4\epsilon_0A}{2d}=\dfrac{2\epsilon_0A}{d}\)

Step 3: Right half
\(C_2=\dfrac{3\epsilon_0(A/2)}{d/2}=\dfrac{3\epsilon_0A}{d}\), \(C_3=\dfrac{6\epsilon_0A}{d}\)
Series: \(\dfrac{C_2C_3}{C_2+C_3}=\dfrac{18}{9}\dfrac{\epsilon_0A}{d}=\dfrac{2\epsilon_0A}{d}\)

Step 4: Total
\(C_{eq}=\dfrac{2\epsilon_0A}{d}+\dfrac{2\epsilon_0A}{d}=\dfrac{4\epsilon_0A}{d}\)

Final Answer:
The equivalent capacitance is 4 A epsilon naught over d. \[ \boxed{\text{(C) }\dfrac{4A\epsilon_{0}}{d}} \]