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:
Gauss’s law states that the total electric flux through a closed surface is equal to the charge enclosed divided by permittivity of free space. \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] For an infinite plane sheet, the electric field is: - Perpendicular to the sheet - Same magnitude everywhere (symmetry)
Step 1: Choose a Gaussian surface
Take a cylindrical Gaussian surface (pillbox) with: - One face above the sheet - One face below the sheet - Axis perpendicular to the sheet
Step 2: Electric flux through the surface
Electric field is perpendicular to the plane and parallel to the curved surface. So flux through curved surface = 0. Flux only passes through the two flat faces: \[ \Phi = EA + EA = 2EA \]
Step 3: Charge enclosed
If surface charge density = \(\sigma\), then charge enclosed: \[ Q_{\text{enc}} = \sigma A \]
Step 4: Apply Gauss’s law \[ 2EA = \frac{\sigma A}{\varepsilon_0} \] Cancel \(A\): \[ 2E = \frac{\sigma}{\varepsilon_0} \] \[ E = \frac{\sigma}{2\varepsilon_0} \]
Step 5: Direction of electric field
- Away from sheet if charge is positive - Towards sheet if charge is negative