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

Answer 2

The problem asks for the ratio of the equivalent capacitances, \( \frac{C_1}{C_2} \), for two different configurations of a parallel plate capacitor filled with two dielectric materials of constants \( \varepsilon_1 \) and \( \varepsilon_2 \).

Concept Used:

The capacitance of a parallel plate capacitor filled with a dielectric material is given by the formula:

\[ C = \frac{K \varepsilon_0 A'}{d'} \]

where \( K \) is the dielectric constant (here denoted by \( \varepsilon \)), \( \varepsilon_0 \) is the permittivity of free space, \( A' \) is the area of the plates, and \( d' \) is the distance between them.

The two configurations can be modeled as combinations of two separate capacitors:

1. Series Combination: When the dielectrics are stacked such that the distance between the plates is divided, the arrangement is equivalent to two capacitors connected in series. The equivalent capacitance \( C_{eq} \) is given by:

\[ \frac{1}{C_{eq}} = \frac{1}{C_a} + \frac{1}{C_b} \quad \text{or} \quad C_{eq} = \frac{C_a C_b}{C_a + C_b} \]

2. Parallel Combination: When the dielectrics are placed side-by-side such that the area of the plates is divided, the arrangement is equivalent to two capacitors in parallel. The equivalent capacitance \( C_{eq} \) is given by:

\[ C_{eq} = C_a + C_b \]

Step-by-Step Solution:

Step 1: Calculate the capacitance for the first configuration (\(C_1\)).

In the first configuration, the dielectrics are stacked, dividing the distance \( d \) into two equal halves (\( d/2 \)), while the area \( A \) remains the same for both parts. This corresponds to a series combination of two capacitors.

The capacitance of the part with dielectric \( \varepsilon_1 \) is:

\[ C_{1a} = \frac{\varepsilon_1 \varepsilon_0 A}{d/2} = \frac{2\varepsilon_1 \varepsilon_0 A}{d} \]

The capacitance of the part with dielectric \( \varepsilon_2 \) is:

\[ C_{1b} = \frac{\varepsilon_2 \varepsilon_0 A}{d/2} = \frac{2\varepsilon_2 \varepsilon_0 A}{d} \]

Since these are in series, the equivalent capacitance \( C_1 \) is:

\[ \frac{1}{C_1} = \frac{1}{C_{1a}} + \frac{1}{C_{1b}} = \frac{d}{2\varepsilon_1 \varepsilon_0 A} + \frac{d}{2\varepsilon_2 \varepsilon_0 A} = \frac{d}{2\varepsilon_0 A} \left( \frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2} \right) \] \[ \frac{1}{C_1} = \frac{d}{2\varepsilon_0 A} \left( \frac{\varepsilon_2 + \varepsilon_1}{\varepsilon_1 \varepsilon_2} \right) \]

Therefore, the capacitance \( C_1 \) is:

\[ C_1 = \frac{2\varepsilon_0 A}{d} \left( \frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1 + \varepsilon_2} \right) \]

Step 2: Calculate the capacitance for the second configuration (\(C_2\)).

In the second configuration, the dielectrics are placed side-by-side, dividing the area \( A \) into two equal halves (\( A/2 \)), while the distance \( d \) remains the same. This corresponds to a parallel combination of two capacitors.

The capacitance of the part with dielectric \( \varepsilon_1 \) is:

\[ C_{2a} = \frac{\varepsilon_1 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_1 \varepsilon_0 A}{2d} \]

The capacitance of the part with dielectric \( \varepsilon_2 \) is:

\[ C_{2b} = \frac{\varepsilon_2 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_2 \varepsilon_0 A}{2d} \]

Since these are in parallel, the equivalent capacitance \( C_2 \) is the sum of the individual capacitances:

\[ C_2 = C_{2a} + C_{2b} = \frac{\varepsilon_1 \varepsilon_0 A}{2d} + \frac{\varepsilon_2 \varepsilon_0 A}{2d} = \frac{\varepsilon_0 A}{2d} (\varepsilon_1 + \varepsilon_2) \]

Final Computation & Result:

Step 3: Calculate the ratio \( \frac{C_1}{C_2} \).

Now, we divide the expression for \( C_1 \) by the expression for \( C_2 \):

\[ \frac{C_1}{C_2} = \frac{\frac{2\varepsilon_0 A}{d} \left( \frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1 + \varepsilon_2} \right)}{\frac{\varepsilon_0 A}{2d} (\varepsilon_1 + \varepsilon_2)} \]

We can cancel the common term \( \frac{\varepsilon_0 A}{d} \) from the numerator and denominator:

\[ \frac{C_1}{C_2} = \frac{2 \left( \frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1 + \varepsilon_2} \right)}{\frac{1}{2} (\varepsilon_1 + \varepsilon_2)} \]

Simplifying the expression gives:

\[ \frac{C_1}{C_2} = \frac{4 \varepsilon_1 \varepsilon_2}{(\varepsilon_1 + \varepsilon_2)(\varepsilon_1 + \varepsilon_2)} = \frac{4 \varepsilon_1 \varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2} \]

The required ratio is \( \frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2} \). This corresponds to option (2).