Question

A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $ \epsilon_1 $ and $ \epsilon_2 $, as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are $ C_1 $ and $ C_2 $ respectively, then $ \frac{C_1}{C_2} $ is:

• A. \( \frac{\epsilon_1 \epsilon_2}{(\epsilon_1 + \epsilon_2)^2} \)
• B. \( \frac{4\epsilon_1 \epsilon_2}{(\epsilon_1 + \epsilon_2)^2} \)
• C. \( \frac{\epsilon_1 \epsilon_2}{\epsilon_1 + \epsilon_2} \)
• D. \( \frac{\epsilon_0 (\epsilon_1 + \epsilon_2)}{2} \)

Hint:

Use the formula for capacitance in series and parallel configurations. Remember to define a common term \( C_0 \) to simplify the calculations.

Answer 1

The Correct Option is B

Let \( C_0 = \frac{\epsilon_0 A}{d} \) 

First Configuration: Area of plate is A. 

Then \( C = \frac{\epsilon_2 \epsilon_0 A}{d/2} = \frac{2\epsilon_2 \epsilon_0 A}{d} = 2\epsilon_2 C_0 \) \( C' = \frac{\epsilon_1 \epsilon_0 A}{d/2} = \frac{2\epsilon_1 \epsilon_0 A}{d} = 2\epsilon_1 C_0 \) C and C' are in series. 

\( C_1 = \frac{CC'}{C+C'} = \frac{4\epsilon_1 \epsilon_2 C_0^2}{2C_0 (\epsilon_2 + \epsilon_1)} \) 

\( C_1 = \frac{2\epsilon_2 \epsilon_1 C_0}{(\epsilon_2 + \epsilon_1)} \) 

Second Configuration: 

Here \( C = \frac{\epsilon_1 \epsilon_0 A}{2d} = \frac{\epsilon_1 C_0}{2} \) 

\( C' = \frac{\epsilon_2 C_0}{2} \) C and C' are in parallel. 

\( C_2 = C' + C = (\epsilon_1 + \epsilon_2) \frac{C_0}{2} \) 

Thus \( \frac{C_1}{C_2} = \frac{2\epsilon_1 \epsilon_2 C_0}{(\epsilon_2 + \epsilon_1)} \times \frac{2}{(\epsilon_1 + \epsilon_2) C_0} \) \( \frac{C_1}{C_2} = \frac{4\epsilon_1 \epsilon_2}{(\epsilon_2 + \epsilon_1)^2} \)