Flux and Gauss's Law

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Flux and Gauss's Law

For a given charge distribution, the electric field

can be calculated using Coulomb's Law. There are two possible methods: Calculate the field vectors due to all the charges and then add them vectorially, or first calculate the electric potential V due to these charges and take its negative gradient.

For more information on this subject, see the Electric Potential module.
There exists an alternate method to finding the electric field, which is provided by Gauss's Law. This "new" method is much more elegant and powerful when there is a certain symmetry to the physical situation--for example, when a charge distribution is spherically symmetric.

Suppose we construct a closed surface around parts of a given charge distribution. This enclosed surface, called a Gaussian Surface, can be of any shape and size we decide. (Note that a Gaussian surface is not a physical surface, such as a plastic bag, but is a mathematical construct.)

Gauss's Law relates the electric fields on the Gaussian surface to the net charge inside it.

For instance, take the above illustration. The electric field (shown by the vectors

is proportional to the enclosed charges only. Therefore, only charges q₂, q₃, and q₅ affect the net electric field through the Gaussian surface. The other charges have no effect!

Before we can formulate the exact statement of Gauss's Law, we need to introduce and discuss the concept of flux in a general way. Flux plays an important role not only in electrostatics, but also in areas such as fluid mechanics.

Gauss's Law for a Single Charge

Gauss's Law applies to any charge contribution, but let us apply it now to the simplest case, that of a single point charge q.

Construct a spherical Gaussian surface of radius r around the charge q. Take a small area on the Gaussian surface. The area vector

points radially outward, as does the electric field vector

at this point. Therefore, the electric flux through this small area is

From the spherical symmetry, all of such small area elements contribute equally to the total.

According to Gauss's Law,

because

= q. Solving for the electric field gives

which is just Coulomb's Law!

Symmetry and Gauss's Law

In applying Gauss's Law to find the electric field for a single point charge q, we used a spherical Gaussian surface (Fig. a)


Fig. a	Fig. b

Can we use other Gaussian surfaces to obtain the electric field at the same point P? For example, a cubic Gaussian surface as shown in Figure b? A little thought will convince us that this choice of a Gaussian surface will not enable us to calculate the electric field as easily as the spherical Gaussian surface. The statement of Gauss's Law

is still true for the cubic Gaussian surface, but there is no easy way to untangle the electric field

from the integral. This is because

is different at each point on the cubic surface, both in magnitude and direction.

Compare this to the situation in Figure a, where the electric field vector

is directed radially outward and has the same magnitude at every point on the spherical Gaussian surface. Therefore, since the magnitude of the electric field was a constant over the surface, we were allowed to take the electric field out of the integral,

and

This illustrates an important point: that symmetry (in our example, spherical symmetry) plays a key role in applying Gauss's Law to calculate the electric field.

Application: Charged Conductors in Equilibrium

Imagine that we start with a conductor of an arbitrary shape, with no charge on it. Now we give this conductor a net charge
q--for example, by touching it momentarily with a charged rod.

The charge q will spread throughout the conductor and will eventually settle down to its equilibrium distribution.

Where does the charge q reside in the conductor when it is in equilibrium? We will use Gauss's Law to prove an important property of a charged conductor in equilibrium:

In electrostatic equilibrium, the electric charge on an isolated conductor resides entirely on its surface. No charge is found within the body of the conductor.

Proof using Gauss's Law:

Construct a Gaussian surface just inside the surface of the conductor. By definition, an equilibrium situation is reached when the charges are arranged so that they no longer move within the conductor. That is, there is no electric field within the conductor at equilibrium. (If there were an electric field, it would cause the charges to move.) Therefore, the electric flux

about this Gaussian surface. But Gauss's Law says that

, so the enclosed charge q_enc = 0. One can make the Gaussian surface approach as close to the conductor surface as possible, and the enclosed charge is still zero. Therefore, the only place that the given charge q can reside is on the conductor's surface

A Uniformly Charged Sphere

An insulating sphere of radius R has a total charge q uniformly distributed throughout its volume.
We want to find the electric field everywhere, that is, at an inside point r < R, and also at an outside point r > R.

For the case where r < R:

We construct a spherical Gaussian surface of radius r and apply Gauss's Law.

We already know that for a spherically symmetric case,

. We have to be careful about q_enc, though. q_enc is the net charge within the Gaussian surface, not the total charge q of the entire sphere. For this reason, we need to multiply the total charge by the ratio of the volume of the Gaussian surface to the volume of the entire sphere:

Therefore,

For the case where r > R:

We construct a spherical Gaussian surface through an outside point r > R, and apply Gauss's Law. In this case,

The graph below is of the magnitude of the electric field due to a uniformly charged sphere, plotted as a function of distance from the center of the sphere.

Gauss's Law in Differential Form

The equation

expresses Gauss's Law in integral form.

Notice that the charge enclosed by the Gaussian surface can be expressed as

where r is the charge density within the Gaussian surface. Therefore,

The left-hand side of this equation involves a surface integral, while the right hand side involves a volume integral.

Can the surface integral be converted to a volume integral? The answer is yes, and it is provided by the Divergence Theorem:

Where the divergence of

,

, is a scalar function defined as

in Cartesian (rectangular) coordinates.

Gauss's Law can now be written as

or

This is the statement of Gauss's Law in differential form. It constitutes one of the four Maxwell's Equations.

Community Contributions - Articles by goIITians

Flux and Gauss's Law

Gauss's Law for a Single Charge

Symmetry and Gauss's Law

Application: Charged Conductors in Equilibrium

A Uniformly Charged Sphere

Gauss's Law in Differential Form