The area of the plates of a capacitor is `50cm^(2)` each and the separation between them is 1mm. The space between the plates is filled with a material of dielectric constant 4. the capacitor, initially uncharged, is connected to a battery of EMF 50 V. the work done by the battery until the capacitor gets completely charged is (in `muJ`).

Options

A. 0.22
B. 0.44
C. 0.88
D. 1.76

Explanation

To solve the problem step by step, we need to calculate the capacitance of the capacitor and then determine the work done by the battery when charging the capacitor. Here’s how to do it:

### Step 1: Convert the given values to standard units
- Area \( A = 50 \, \text{cm}^2 = 50 \times 10^{-4} \, \text{m}^2 = 5 \times 10^{-3} \, \text{m}^2 \)
- Separation \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Dielectric constant \( K = 4 \)
- EMF of the battery \( V = 50 \, \text{V} \)

**Hint for Step 1:** Always convert units to SI units for consistency in calculations.

### Step 2: Calculate the capacitance \( C \)
The formula for the capacitance of a parallel plate capacitor filled with a dielectric is given by:

\[ C = \frac{K \epsilon_0 A}{d} \]

Where:
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) (permittivity of free space)

Substituting the values:

\[ C = \frac{4 \times (8.85 \times 10^{-12}) \times (5 \times 10^{-3})}{1 \times 10^{-3}} \]

Calculating this gives:

\[ C = \frac{4 \times 8.85 \times 5 \times 10^{-15}}{1 \times 10^{-3}} = 4 \times 8.85 \times 5 \times 10^{-12} = 1.77 \times 10^{-11} \, \text{F} \]

### Step 3: Calculate the work done \( W \)
The work done by the battery in charging the capacitor is given by:

\[ W = \frac{1}{2} C V^2 \]

Substituting the values of \( C \) and \( V \):

\[ W = \frac{1}{2} \times (1.77 \times 10^{-11}) \times (50)^2 \]

Calculating this gives:

\[ W = \frac{1}{2} \times (1.77 \times 10^{-11}) \times 2500 = 2.2125 \times 10^{-8} \, \text{J} \]

### Step 4: Convert work done to microjoules
To convert joules to microjoules:

\[ W = 2.2125 \times 10^{-8} \, \text{J} = 22.125 \, \mu\text{J} \]

### Final Answer
The work done by the battery until the capacitor gets completely charged is approximately \( 22.125 \, \mu\text{J} \).

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