Gauss’s law
1. Introduction
The electric field of a given charge distribution can in principle be calculated using Coulomb's law. But the actual calculations can become quite complicated.
2. Gauss's Law
An alternative method to calculate the electric field of a given charge distribution relies on a theorem called Gauss's law. Gauss' law states that
" If the volume within an arbitrary closed mathematical surface holds a net electric charge Q, then the electric flux Φ[Phi] though its surface is Q/ε[epsilon]0 "
Gauss's law can be written in the following form:
Figure 1. Electric flux through surface area a.
The electric flux Φ[Phi] through a surface is defined as the product of the area A and the magnitude of the normal component of the electric field E:
Where θ [theta] is the angle between the electric field and the normal of the surface (see Figure 1). To apply Gauss' law one has to obtain the flux through a closed surface. This flux can be obtained by integrating the second equation over all the area of the surface. The convention used to define the flux as positive or negative is that the angle θ[theta] is measured with respect to the perpendicular erected on the outside of the closed surface: field lines leaving the volume make a positive contribution, and field lines entering the volume make a negative contribution.
Example 1: Field of point charge.
The field generated by a point charge q is spherical symmetric, and its magnitude will depend only on the distance r from the point charge. The direction of the field is along the direction (see Figure 2). Consider a spherical surface centered around the point charge q (see Figure 2). The direction of the electric field at any point on its surface is perpendicular to the surface and its magnitude is constant. This implies that the electric flux Φ[Phi] through this surface is given by
Figure 2. Electric field generated by point charge q.
Using Gauss's law we obtain the following expression
or
which is Coulomb's law.
Example 2: Problem 16
Charge is uniformly distributed over the volume of a large slab of plastic of thickness d. The charge density is ρ[rho] C/m3. The mid-plane of the slab is the y-z plane (see Figure 3). What is the electric filed at a distance x from the mid-plane?
Figure 3. Problem 16.
As a result of the symmetry of the slab, the direction of the electric field will be along the x-axis (at every point). To calculate the electric field at any given point, we need to consider two separate case: - d/2 < x < d/2 and x > d/2 or x < -d/2. Consider surface 1 shown in Figure 3. The flux through this surface is equal to the flux through the planes at x = x1 and x = - x1. Symmetry arguments show that
The flux Φ[Phi]1 through surface 1 is therefore given by
The amount of charge enclosed by surface 1 is given by
Applying Gauss' law to these equations we obtain
or
Note: this formula is only correct for - d/2 < x1 < d/2.
The flux Φ[Phi]2 through surface 2 is given by
The charge enclosed by surface 2 is given by
This equation shows that the enclosed charge does not depend on x2. Applying Gauss's law one obtains
or
