An injfinite large sheet has charg density `sigmaC//m^(2)` Find electric field at a distance d perendicular to the sheet.

Options

A. `E=(sigma)/(2epsi_(0))`
B. `E=(sigma)/(epsi_(0))`
C. `E=(2sigma)/(epsi_(0))`
D. None of these

Explanation

To solve the problem of finding the electric field at a distance \( d \) perpendicular to an infinite sheet with charge density \( \sigma \) (in C/m²), we can follow these steps:

### Step 1: Understand the Configuration
We have an infinite sheet with a uniform charge density \( \sigma \). The electric field produced by such a sheet is uniform and directed away from the sheet if the charge is positive and towards the sheet if the charge is negative.

**Hint:** Visualize the infinite sheet and the direction of the electric field lines.

### Step 2: Apply Gauss's Law
According to Gauss's Law, the electric flux \( \Phi \) through a closed surface is equal to the charge enclosed \( Q \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \frac{Q}{\epsilon_0} \]

**Hint:** Remember that the electric flux through a surface is the product of the electric field \( E \) and the area \( A \) of the surface.

### Step 3: Choose a Gaussian Surface
For an infinite sheet, we can choose a Gaussian surface in the form of a cylindrical pillbox that straddles the sheet. The area of the top and bottom faces of the pillbox is \( A \).

**Hint:** The electric field is constant and perpendicular to the surface of the sheet, so the flux through the curved surface of the cylinder is zero.

### Step 4: Calculate the Electric Flux
The electric flux through the top and bottom faces of the pillbox is given by:
\[ \Phi = E \cdot A + E \cdot A = 2EA \]
where \( E \) is the magnitude of the electric field at distance \( d \) from the sheet.

**Hint:** Since the electric field is uniform, you can treat it as constant over the area \( A \).

### Step 5: Determine the Charge Enclosed
The charge enclosed by the Gaussian surface is:
\[ Q = \sigma A \]

**Hint:** The charge density \( \sigma \) is the amount of charge per unit area.

### Step 6: Set Up the Equation Using Gauss's Law
Substituting the expressions for electric flux and charge into Gauss's Law gives:
\[ 2EA = \frac{\sigma A}{\epsilon_0} \]

**Hint:** Notice that the area \( A \) cancels out from both sides.

### Step 7: Solve for the Electric Field
Now, we can solve for \( E \):
\[ 2E = \frac{\sigma}{\epsilon_0} \]
\[ E = \frac{\sigma}{2\epsilon_0} \]

**Hint:** This result shows that the electric field is independent of the distance \( d \) from the sheet.

### Final Answer
Thus, the electric field at a distance \( d \) perpendicular to the infinite sheet is:
\[ E = \frac{\sigma}{2\epsilon_0} \]