A charge of `17.7 xx 10^(-4)C` is distributed uniformly over a large sheet of area `200 m^(2)`. The electric field intensity at a distance 20 cm from it in air will be

Options

A. `5 xx 10^(5) N//C`
B. `6 xx 10^(5) N//C`
C. `7 xx 10^(5) N//C`
D. `8 xx 10^(5) N//C`

Answer

To find the electric field intensity at a distance of 20 cm from a uniformly charged sheet, we can follow these steps:

### Step 1: Calculate the charge per unit area (σ)
The charge per unit area (σ) can be calculated using the formula:
\[ \sigma = \frac{Q}{A} \]
where:
- \( Q = 17.7 \times 10^{-4} \, C \) (the total charge)
- \( A = 200 \, m^2 \) (the area of the sheet)

Substituting the values:
\[ \sigma = \frac{17.7 \times 10^{-4}}{200} \]

### Step 2: Perform the calculation for σ
Calculating σ:
\[ \sigma = \frac{17.7 \times 10^{-4}}{200} = 8.85 \times 10^{-6} \, C/m^2 \]

### Step 3: Use the formula for the electric field (E) due to an infinite sheet
The electric field intensity (E) due to an infinite sheet of charge is given by:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
where:
- \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \) (the permittivity of free space)

### Step 4: Substitute σ and ε₀ into the electric field formula
Substituting the values into the formula:
\[ E = \frac{8.85 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \]

### Step 5: Simplify the expression
Calculating E:
\[ E = \frac{8.85 \times 10^{-6}}{1.77 \times 10^{-11}} = 5 \times 10^{5} \, N/C \]

### Final Answer
The electric field intensity at a distance of 20 cm from the charged sheet is:
\[ E = 5 \times 10^{5} \, N/C \]

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