Question

State Gauss's law in electrostatics. On the basis of it, obtain the formula for the electric field produced due to a plane charged plate.

Hint:

For a charged plane, the electric field is uniform and does not depend on the distance from the plane. This is different from the electric field due to a point charge, which decreases with the square of the distance.

Answer 1

Gauss's law in electrostatics states that the total electric flux \( \Phi_E \) through any closed surface is equal to the net charge enclosed within the surface divided by the permittivity of free space \( \varepsilon_0 \). Mathematically, Gauss's law is expressed as: \[ \Phi_E = \oint \vec{E} . d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}, \] where: - \( \vec{E} \) is the electric field,
- \( d\vec{A} \) is the differential area element on the closed surface,
- \( Q_{\text{enc}} \) is the total charge enclosed within the surface,
- \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} . \text{m}^2 \).
Now, to find the electric field due to a plane charged plate, we consider a uniformly charged infinite plane with surface charge density \( \sigma \). We will apply Gauss's law using a Gaussian surface in the form of a cylindrical pillbox that intersects the charged plane. The electric field produced by the charged plate is symmetric and uniform, and it points perpendicular to the surface of the plate.
1. Gauss's Law Application: The Gaussian surface consists of two faces, one above and one below the charged plane. The electric field is perpendicular to the surface of the plate and has the same magnitude on both sides of the plane. The flux through the curved surface of the pillbox is zero since the electric field is parallel to this surface. Therefore, the total flux is given by the flux through the two flat faces of the pillbox.
2. Flux Calculation: The flux through one face of the pillbox is \( E . A \), where \( A \) is the area of the face. The total flux through the two faces is: \[ \Phi_E = 2EA. \] 3. Charge Enclosed: The charge enclosed by the Gaussian surface is the surface charge density \( \sigma \) times the area \( A \) of the pillbox: \[ Q_{\text{enc}} = \sigma A. \] 4. Applying Gauss's Law: Using Gauss's law, we equate the flux to the charge enclosed divided by \( \varepsilon_0 \): \[ 2EA = \frac{\sigma A}{\varepsilon_0}. \] 5. Solving for Electric Field: Simplifying the equation, we get the electric field due to the uniformly charged plane: \[ E = \frac{\sigma}{2\varepsilon_0}. \] Thus, the electric field produced by an infinite plane of charge with surface charge density \( \sigma \) is: \[ E = \frac{\sigma}{2\varepsilon_0}. \] Key Notes: - The electric field is directed perpendicular to the surface of the charged plane.
- The electric field due to a uniformly charged plane is constant in magnitude and does not depend on the distance from the plane.