Field of An Infinite Plane Sheet of Charge

Use Gauss’s law to find the electric field caused by a thin, flat, infinite sheet with a uniform positive surface charge density $\sigma$.

We found that the field E of a uniformly charged infinite sheet is normal to the sheet, and that its magnitude is independent of the distance from the sheet. To take advantage of these symmetry properties, we use a cylindrical Gaussian surface with ends of area A and with its axis perpendicular to the sheet of charge (Fig. 1).

Fig. 1: A cylindrical Gaussian surface is used to find the field of an infinite plane sheet of charge.

The flux through the cylindrical part of our Gaussian surface is zero because $\mathbf{E}.\hat{\mathbf{n}}=0$ everywhere. The flux through each flat end of the surface is + EA because $\mathbf{E}.\hat{\mathbf{n}}=E_{\perp}=E$ everywhere, so the total flux through both ends and hence the total flux $\Phi_E$ through the Gaussian surface is +2EA. The total enclosed charge is $Q_{encl} = \sigma A$ and so from Gauss’s law,

$2EA=\frac{\sigma A}{\epsilon_0}$ and

$E=\frac{\sigma}{2\epsilon_0}$ (field of an infinite sheet of charge)

If $\sigma$ is negative, E is directed toward the sheet, the flux through the Gaussian surface in Fig. 1 is negative, and $\sigma$ in the expression $E=\frac{\sigma}{2\epsilon_0}$ denotes the magnitude (absolute value) of the charge density.

Example 1

A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 2. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

Fig.2

Solution:

To solve the problem, we need to calculate the electric flux through the square, given that it is part of a cube with edge length 10 cm. Here's how to proceed:

Given:

• Charge, $\$q\; =\; +10\; \backslash mu\; C\; =\; 10\; \backslash times\; 10^\{-6\}\; \backslash \; C\$$
• The square is part of a cube with edge length $a = 10 \, \text{cm} = 0.1 \, \text{m}$.
• The charge is located $5 \, \text{cm} = 0.05 \, \text{m}$ above the center of the square.

Approach:

1. Using Gauss's Law: The total electric flux through a closed surface is given by:

   $${\mathrm{\$\backslash Phi\_\{total\}=\backslash frac\{q\}\{\backslash epsilon\_0\}\$}}_{}$$

   where $\epsilon_0$ is the permittivity of free space $\epsilon_0=8.8854 \times 10^{-12} \ C^2/(N.m^2)$ 

2. Cube's symmetry:

• The point charge is symmetrically positioned above the cube.
• By symmetry, the total flux through the entire cube is uniformly distributed among its six faces.
• Flux through one face (the square) is:

$\Phi_{square}=\frac{\Phi_{total}}{6}=\frac{q}{6\epsilon_0}$​

Calculation: Substitute the values into the formula:

$\Phi_{square}=\frac{q}{6\epsilon_0}=\frac{10\times 10^{-6}}{6 \times 8.854 \times 10^{-12}}$​

Let me compute the numerical result. 

The magnitude of the electric flux through the square is approximately 

$\Phi_{square}=1.88 \times 10^5 \ N.m^2/C$

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