The surface density of charge on the surface of a charged conductor in the air is `26.5 muCm^(-2)` . The the outward force per unit area of the charged conductor is `(epsilon_0=8.85xx10^-12C^2N^-1m^-2)`

Options

A. (a)`3.5 N m ^(-2)`
B. (b)`17.7 N m ^(-2)`
C. (c)`8.85 N m ^(-2)`
D. (d)`39.7 N m ^(-2)`

Answer

To solve the problem, we need to find the outward force per unit area on a charged conductor given its surface charge density and the permittivity of free space.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Surface charge density, \( \sigma = 26.5 \, \mu C/m^2 = 26.5 \times 10^{-6} \, C/m^2 \)
- Permittivity of free space, \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \)

2. **Formula for Force per Unit Area:**
The outward force per unit area \( F/A \) on a charged conductor can be calculated using the formula:
\[ F/A = \frac{1}{2} \frac{\sigma^2}{\epsilon_0} \]

3. **Substitute the Values:**
Substitute the values of \( \sigma \) and \( \epsilon_0 \) into the formula:
\[ F/A = \frac{1}{2} \frac{(26.5 \times 10^{-6})^2}{8.85 \times 10^{-12}} \]

4. **Calculate \( \sigma^2 \):**
First, calculate \( (26.5 \times 10^{-6})^2 \):
\[ (26.5 \times 10^{-6})^2 = 7.0225 \times 10^{-10} \, C^2/m^4 \]

5. **Calculate the Force per Unit Area:**
Now, substitute this value back into the equation:
\[ F/A = \frac{1}{2} \frac{7.0225 \times 10^{-10}}{8.85 \times 10^{-12}} \]
Calculate the division:
\[ \frac{7.0225 \times 10^{-10}}{8.85 \times 10^{-12}} \approx 79.4 \, N/m^2 \]
Now, multiply by \( \frac{1}{2} \):
\[ F/A \approx \frac{79.4}{2} \approx 39.7 \, N/m^2 \]

6. **Final Answer:**
The outward force per unit area on the charged conductor is approximately:
\[ F/A \approx 39.7 \, N/m^2 \]

### Conclusion:
The correct answer is \( 39.7 \, N/m^2 \).