The capacity of a parallel plate capacitor formed by the plates of same area A is `0.02muF` with air as dielectric. Now one plate is replaced by a plate of area 2A and dielectric (k=2) is introduced between the plates, the capacity is

Options

A. `0.04muF`
B. `0.08muF`
C. `0.01muF`
D. `2muF`

Answer

To solve the problem step by step, we will follow the principles of capacitors, specifically focusing on the formulas for capacitance in parallel plate capacitors.

### Step 1: Understand the initial conditions
We know that the initial capacitance \( C_1 \) of a parallel plate capacitor with area \( A \) and air as the dielectric is given as:
\[ C_1 = 0.02 \, \mu F \]

### Step 2: Write the formula for capacitance
The formula for the capacitance of a parallel plate capacitor is:
\[ C = k \frac{\varepsilon_0 A}{d} \]
where:
- \( k \) is the dielectric constant,
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the distance between the plates.

### Step 3: Calculate \( C_1 \)
For the first capacitor with air as the dielectric (where \( k_1 = 1 \)):
\[ C_1 = k_1 \frac{\varepsilon_0 A}{d} = 1 \cdot \frac{\varepsilon_0 A}{d} \]
Thus, we have:
\[ C_1 = \frac{\varepsilon_0 A}{d} \]

### Step 4: Analyze the new configuration
In the new configuration, one plate is replaced with a plate of area \( 2A \) and a dielectric with \( k_2 = 2 \) is introduced. The new capacitance \( C_2 \) can be expressed as:
\[ C_2 = k_2 \frac{\varepsilon_0 A_2}{d} \]
where \( A_2 = 2A \).

### Step 5: Substitute the values into the formula for \( C_2 \)
Substituting \( k_2 \) and \( A_2 \):
\[ C_2 = 2 \cdot \frac{\varepsilon_0 (2A)}{d} = 4 \cdot \frac{\varepsilon_0 A}{d} \]

### Step 6: Relate \( C_2 \) to \( C_1 \)
From our earlier expression for \( C_1 \):
\[ C_1 = \frac{\varepsilon_0 A}{d} \]
Thus,
\[ C_2 = 4 \cdot C_1 \]

### Step 7: Calculate \( C_2 \)
Now substituting the value of \( C_1 \):
\[ C_2 = 4 \cdot 0.02 \, \mu F = 0.08 \, \mu F \]

### Final Answer
The capacitance \( C_2 \) after replacing one plate and introducing the dielectric is:
\[ C_2 = 0.08 \, \mu F \]

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