A parallel plate capacitor of plate separation d = 1 mm and area 50 `cm^2` is connected to a battery of 50 V.
If a dielectric medium, k = 4 is filled between plates, then the value of C is:

Options

A. 177 `muF`
B. `17.7 muF`
C. 177 pF
D. 17.7 pF

Explanation

To find the capacitance \( C \) of a parallel plate capacitor filled with a dielectric medium, we can use the formula:

\[ C = \frac{\varepsilon_0 k A}{d} \]

where:
- \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \),
- \( k \) is the dielectric constant,
- \( A \) is the area of the plates,
- \( d \) is the separation between the plates.

### Step 1: Convert the given values to SI units

- Plate separation \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Area \( A = 50 \, \text{cm}^2 = 50 \times 10^{-4} \, \text{m}^2 = 5 \times 10^{-2} \, \text{m}^2 \)
- Dielectric constant \( k = 4 \)

### Step 2: Substitute the values into the capacitance formula

Now, substituting the values into the formula:

\[ C = \frac{(8.85 \times 10^{-12} \, \text{F/m}) \times 4 \times (5 \times 10^{-2} \, \text{m}^2)}{1 \times 10^{-3} \, \text{m}} \]

### Step 3: Calculate the numerator

Calculating the numerator:

\[ 4 \times 5 = 20 \]

Thus, the numerator becomes:

\[ 8.85 \times 10^{-12} \times 20 = 177 \times 10^{-12} \, \text{F} \]

### Step 4: Divide by the plate separation

Now, divide by the plate separation:

\[ C = \frac{177 \times 10^{-12}}{1 \times 10^{-3}} = 177 \times 10^{-12 + 3} = 177 \times 10^{-9} \, \text{F} = 177 \, \text{pF} \]

### Final Result

Thus, the value of the capacitance \( C \) is:

\[ C = 177 \, \text{pF} \]