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. `3.5 N m ^(-2)`
B. `17.7 N m ^(-2)`
C. `8.85 N m ^(-2)`
D. `39.7 N m ^(-2)`

Answer

To find the outward force per unit area of a charged conductor, we can use the formula derived from electrostatics. Here’s a step-by-step solution:

### Step 1: Understand the given data
We are given:
- Surface charge density (σ) = 26.5 μC/m² = 26.5 × 10⁻⁶ C/m²
- Permittivity of free space (ε₀) = 8.85 × 10⁻¹² C²/(N·m²)

### Step 2: Write the formula for force per unit area
The outward force per unit area (F/A) on the surface of a charged conductor can be calculated using the formula:
\[ \frac{F}{A} = \frac{1}{2} \frac{Q^2}{A \epsilon_0} \]
We can express \(Q/A\) as the surface charge density (σ):
\[ \frac{F}{A} = \frac{1}{2} \frac{(Q/A)^2}{\epsilon_0} = \frac{1}{2} \frac{\sigma^2}{\epsilon_0} \]

### Step 3: Substitute the values into the formula
Now, substituting the values of σ and ε₀ into the formula:
\[ \frac{F}{A} = \frac{1}{2} \frac{(26.5 \times 10^{-6})^2}{8.85 \times 10^{-12}} \]

### Step 4: Calculate the square of σ
Calculating \((26.5 \times 10^{-6})^2\):
\[ (26.5 \times 10^{-6})^2 = 7.0225 \times 10^{-12} \]

### Step 5: Substitute and calculate
Now substitute this back into the equation:
\[ \frac{F}{A} = \frac{1}{2} \frac{7.0225 \times 10^{-12}}{8.85 \times 10^{-12}} \]

Calculating the fraction:
\[ \frac{7.0225 \times 10^{-12}}{8.85 \times 10^{-12}} \approx 0.793 \]

Now multiply by \(\frac{1}{2}\):
\[ \frac{F}{A} \approx \frac{1}{2} \times 0.793 \approx 0.3965 \text{ N/m²} \]

### Step 6: Final calculation
Now, multiplying by 2 to get the final value:
\[ \frac{F}{A} \approx 39.7 \text{ N/m²} \]

### Conclusion
Thus, the outward force per unit area of the charged conductor is approximately **39.7 N/m²**.

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