A parallel plate capacitor has rectangular plates of `400 cm^2` and are separated by a distance of 2 mm with air as medium. What charge will appear on the plates. If a 200 volt potential difference is applied across the condenser

Options

A. `3.54 xx 10^(-6)C`
B. `3.54 xx 10^(-8)C`
C. `3.54 xx 10^(-10)C`
D. `1770.8 xx 10^(-13)C`

Answer

To solve the problem, we need to find the charge that appears on the plates of a parallel plate capacitor when a potential difference is applied. Here are the steps to arrive at the solution:

### Step 1: Convert the Area and Distance to Standard Units
The area of the plates is given as \(400 \, \text{cm}^2\) and the separation distance is \(2 \, \text{mm}\).

- Convert area from cm² to m²:
\[ A = 400 \, \text{cm}^2 = 400 \times 10^{-4} \, \text{m}^2 = 0.04 \, \text{m}^2 \]

- Convert distance from mm to m:
\[ D = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} = 0.002 \, \text{m} \]

### Step 2: Calculate the Capacitance of the Parallel Plate Capacitor
The formula for the capacitance \(C\) of a parallel plate capacitor is given by:
\[ C = \frac{A \epsilon_0}{D} \]
where \(\epsilon_0\) (the permittivity of free space) is approximately \(8.85 \times 10^{-12} \, \text{F/m}\).

Substituting the values:
\[ C = \frac{0.04 \, \text{m}^2 \times 8.85 \times 10^{-12} \, \text{F/m}}{0.002 \, \text{m}} \]
Calculating this gives:
\[ C = \frac{0.04 \times 8.85 \times 10^{-12}}{0.002} = 1.77 \times 10^{-12} \, \text{F} \]

### Step 3: Calculate the Charge on the Plates
The charge \(Q\) on the plates of the capacitor can be calculated using the formula:
\[ Q = C \times V \]
where \(V\) is the potential difference applied across the capacitor, which is given as \(200 \, \text{V}\).

Substituting the values:
\[ Q = 1.77 \times 10^{-12} \, \text{F} \times 200 \, \text{V} \]
Calculating this gives:
\[ Q = 3.54 \times 10^{-10} \, \text{C} \]

### Final Answer
The charge that will appear on the plates is:
\[ Q = 3.54 \times 10^{-10} \, \text{C} \]