Question

Using Gauss’s law, obtain an expression for the electric field at a point due to a uniformly charged infinite plane sheet.

Hint:

Electric field of an infinite plane sheet is \textbf{uniform} and does not depend on distance from the sheet. \[ E = \frac{\sigma}{2\varepsilon_0} \] This is a standard Gauss’s law result.

Answer 1

Concept Overview: 
Gauss’s law states that the total electric flux through a closed surface equals the charge enclosed divided by the permittivity of free space. \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] For an infinite plane sheet of charge: - The electric field is perpendicular to the sheet. 
- Its magnitude is uniform at all points due to symmetry. 

Step 1: Select a Suitable Gaussian Surface
Choose a cylindrical Gaussian surface (pillbox) such that: - One flat face lies above the sheet. 
- The other flat face lies below the sheet. 
- The axis of the cylinder is perpendicular to the sheet. 

Step 2: Evaluate the Electric Flux
The electric field is: - Perpendicular to the flat faces. 
- Parallel to the curved surface. Therefore, flux through the curved surface is zero. Flux only passes through the two flat circular faces. \[ \Phi = EA + EA = 2EA \] 

Step 3: Determine the Enclosed Charge
If the surface charge density is \( \sigma \), then the charge enclosed within area \( A \) is: \[ Q_{\text{enc}} = \sigma A \] 

Step 4: Apply Gauss’s Law
\[ 2EA = \frac{\sigma A}{\varepsilon_0} \] Cancel \( A \) from both sides: \[ 2E = \frac{\sigma}{\varepsilon_0} \] \[ E = \frac{\sigma}{2\varepsilon_0} \] 

Step 5: Direction of the Electric Field
- Directed away from the sheet if the charge is positive. 
- Directed toward the sheet if the charge is negative.