A parallel plate capacitor of plate area `A = 600 cm^(2)` and plate separation d = 2.0 mm. is connected to a d.c. source of 200 V.
What is the charge density `sigma` on any one of the two plates?

Options

A. `sigma=8.85 xx 10^(-7)C//m^(2)`
B. `sigma=8.85 xx 10^(-9)C//m^(2)`
C. `sigma=4.45 xx 10^(-7) C//m^(2)`
D. `sigma = 4.45 xx 10^(-9)C//m^(2)`

Explanation

To find the charge density \(\sigma\) on one of the plates of a parallel plate capacitor, we can follow these steps:

### Step 1: Convert the given values to standard units
- The area \(A\) is given as \(600 \, \text{cm}^2\). We need to convert this to square meters:
\[ A = 600 \, \text{cm}^2 = 600 \times 10^{-4} \, \text{m}^2 = 0.06 \, \text{m}^2 \]
- The plate separation \(d\) is given as \(2.0 \, \text{mm}\). We convert this to meters:
\[ d = 2.0 \, \text{mm} = 2.0 \times 10^{-3} \, \text{m} \]

### Step 2: Calculate the electric field \(E\)
The electric field \(E\) between the plates of a capacitor is given by the formula:
\[ E = \frac{V}{d} \]
where \(V\) is the potential difference. Given \(V = 200 \, \text{V}\) and \(d = 2.0 \times 10^{-3} \, \text{m}\), we can calculate \(E\):
\[ E = \frac{200 \, \text{V}}{2.0 \times 10^{-3} \, \text{m}} = 100000 \, \text{V/m} = 10^5 \, \text{V/m} \]

### Step 3: Use the relationship between electric field and charge density
The relationship between the electric field \(E\) and charge density \(\sigma\) is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \(\epsilon_0\) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\).

### Step 4: Rearranging to find charge density \(\sigma\)
We can rearrange the formula to solve for \(\sigma\):
\[ \sigma = E \cdot \epsilon_0 \]
Substituting the values we have:
\[ \sigma = (10^5 \, \text{V/m}) \cdot (8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2) \]

### Step 5: Calculate \(\sigma\)
Now we perform the multiplication:
\[ \sigma = 10^5 \cdot 8.85 \times 10^{-12} = 8.85 \times 10^{-7} \, \text{C/m}^2 \]

### Final Answer
Thus, the charge density \(\sigma\) on one of the plates is:
\[ \sigma \approx 8.85 \times 10^{-7} \, \text{C/m}^2 \]