Question

(i) Using Gauss's law, show that the electric field \( E \) at a point due to a uniformly charged infinite plane sheet is given by \( E = \frac{\sigma}{2\epsilon_0} \), where symbols have their usual meanings.

Hint:

The electric field due to an infinite sheet does not depend on the distance from the sheet, unlike a point charge or a line charge.

Answer 1

Derivation of Electric Field Due to an Infinite Plane Sheet
\includegraphics[]{q32 or.PNG} Step 1: Gauss’s Law Statement
According to Gauss’s Law: \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \] where: - \( \oint \mathbf{E} \cdot d\mathbf{A} \) = Total electric flux, - \( Q_{\text{enc}} \) = Enclosed charge,
- \( \epsilon_0 \) = Permittivity of free space.

Step 2: Choosing a Gaussian Surface
- Consider an infinite charged plane with surface charge density \( \sigma \).
- The charge is uniformly distributed over the plane.
- We use a Gaussian cylinder (pillbox) that extends equally on both sides of the plane.

Step 3: Applying Gauss’s Law
- The flux is perpendicular to the surface.
- The total flux through the two flat surfaces of the pillbox is: \[ \oint \mathbf{E} \cdot d\mathbf{A} = E A + E A = 2E A \] - The enclosed charge is: \[ Q_{\text{enc}} = \sigma A \] - Applying Gauss's Law: \[ 2E A = \frac{\sigma A}{\epsilon_0} \] \[ E = \frac{\sigma}{2\epsilon_0} \] Thus, the electric field due to an infinite plane sheet is: \[ E = \frac{\sigma}{2\epsilon_0} \]