The Basics of Electric Flux and Gauss’s Law

In physics, a system’s symmetry properties are an important tool for simplifying problems. A cylindrical body rotating around its axis doesn’t look any different if observed from any side and a charged metal sphere will also look the same while rotating about any axis through its center. In this tutorial, we will use symmetry properties and introduce a new principle, Gauss’s Law, to simplify electric-field calculations.

Consider the expressions for the field due to a line charge (at equatorial or axial point), derived by using some fairly strenuous integrations.

With Gauss’s Law, we can obtain the same expressions in a few steps. And more than just a way to make certain calculations easier, Gauss’s Law allows us to understand the fundamental relationship between electric charges and electric fields.

A way to conceptualize Gauss’s Law would be: Given any general charge distribution, we imagine a (Gaussian) surface that encloses the charge. Then we observe the electric field at different points on this imaginary surface. Gauss’s Law is a relationship between the field at all the points on the surface and the total charge enclosed within the surface. For now this may sound indirect, but this will prove to be a tremendously useful relationship. We will frequently use the insights that Gauss’s Law provides into the character of electric fields.

In electromagnetism, electric flux measures the amount of electric field passing through a given surface area in unit time. The concept of electric flux will sound more useful as our understanding of Gauss' Law improves. Mathematically, it is defined as the electric field times the component of the area perpendicular to the field.

Consider a box that may or may not contain an electric charge. We can let this box represent an imaginary surface that may or may not enclose some charge. Since it completely holds a volume, we’ll refer to the box as a closed surface. Knowing that a charge distribution produces an electric field, we can measure on the surface of the box to determine what is inside the box.

Recall that the electric field is radially outward from a positive charge and radially in toward a negative point charge. In Figure 2a, there is a single positive point charge inside the box, and the electric field points out of the box. Figure 2b shows a negative charge inside the box, and the electric field points into the box.

The electric field does not actually “flow,” but in Figure 2a, since the electric-field points out of the surface, we say there is an outward electric flux. In Figure 2b, the vectors point to the surface, and the electric flux is inward.

Let’s find out what happens if there is zero charge inside the box. In Figure 3a, the box is empty, there is no charge and hence everywhere. Thus, there is no electric flux into or out of the box. In Figure 3b, the net charge inside is zero since the positive and negative charges (equal magnitude) cancel each other out. There is an electric field, but the same amount that “flows into” the box also “flows out of” the box. Again there is no net electric flux into or out of the box. The box is empty in Figure 3c. but there is charge outside the box. One end of the box is parallel to a uniformly charged infinite sheet, and this sheet produces a uniform electric field perpendicular to its surface. On one side of the box, points into the box; on the opposite side, E points out of the box; on the other four sides, is parallel to the surface and points neither into nor out of the box. As in Figure 3b, the inward electric flux on one side exactly compensates for the outward electric flux on the other side. Therefore, in all of the cases shown in Figure 3, no net charge is enclosed in the box and there is no net electric flux through the surface of the box.

We have just discussed the connection between the sign (positive, negative, or zero) of the net charge enclosed by a closed surface and the direction (outward, inward, or none) of the net electric flux through the surface. There is also a correlation between the magnitude of the net charge within the closed surface and the strength of the electric flux over the surface. If the magnitude of the charge is twice as great as in Figure 2, then is everywhere twice as great in magnitude. This suggests that the net electric flux through the surface of the box is directly proportional to the magnitude of the net charge enclosed by the box. This observation is independent of the size of the box.

In Figure 4, the point charge +q is enclosed by a box with twice the linear dimensions of the box in Figure 2a. The magnitude of the electric field of a point charge decreases with distance (1/r²) , so the average magnitude of on each face of the large box in Figure 4 is just 1/4 of the average magnitude on the corresponding face in Figure 2a. However, note that the area of a face of the large box is exactly four times the area of the corresponding face of the small box.

Hence, we expect the same result for the two boxes when we calculate the outward electric flux: For each face of the box, take the product of the average perpendicular component of and the area of that face; then add up the results from all faces of the box. In conclusion, the net electric flux due to a net charge inside the box is independent of the size of the box and depends only on the net charge enclosed by the box.

To summarize our observations for a closed rectangular box and charge distributions made up of point charges or infinite charged sheets, we have found:

• The sense or direction of the net electric flux (inward or outward) depends on the sign of the enclosed charge.
• Charges outside the enclosed volume do not produce a net electric flux through the surfaces.
• The net electric flux is directly proportional to the net amount of charge enclosed within the surface but is independent of the size of the closed surface.

These statements collectively and qualitatively describe Gauss’s Law, and they hold true for other configurations of charges and closed surfaces of any shape. If the area is not planar, then the evaluation of the flux generally requires an area integral considering the angle between and will be continually changing.

The electric flux through a planar area is defined as the electric field times the component of the area perpendicular to the field, or where is the angle makes with

The SI unit for electric flux is which is also equal to V m (volt-meters). Note that if the area is “parallel” to the field, are perpendicular, and the flux is zero (Figure 5c). We represent the direction of the area vector by using a unit vector perpendicular to the area; stands for “normal.” Then A surface has two sides, so there are two possible directions for and We must always specify which direction we choose. With a closed surface, we always choose the direction of to be outward; and speak of the flux out of a closed surface. Thus what we call “outward electric flux” corresponds to a positive value of and what we call “inward electric flux” corresponds to a negative value of

If the area is not planar, the evaluation of the flux generally requires an area integral since the angle is continually changing.

When the area is used in a vector operation like this, it is understood that the magnitude of the vector is equal to the area, and the direction of the vector is perpendicular to the area.

Example: Electric flux through a cube

An imaginary cubical surface of side s is in a region of uniform electric field . Find the electric flux through each face of the cube and the total flux through the cube when the cube is turned by an angle about a vertical axis.

Solution: The field is directed into faces 1 and 3, so the fluxes through them are negative; is directed out of faces 2 and 4, so the fluxes through them are positive. Solving for the flux of each individual face,

where is the normal vector of the face i.

The total flux through the surface of the cube is again zero. A uniform electric field has zero net flux through a closed surface containing no electric charge.

Example: Electric flux through a sphere

A point charge is surrounded by an imaginary sphere of radius centered on the charge. Find the resulting electric flux through the sphere.

Solution: The surface is not flat, and the electric field is not uniform, so to calculate the electric flux, we must use the integral equation or the general definition,

Because the sphere is centered on the point charge, at any point on the spherical surface, is directed out of the sphere perpendicular to the surface and is, therefore, in the same direction as The positive direction for both n and E is outward. The flux through a surface portion is Hence the integral becomes

At any point on the sphere of radius r, the electric field has the same magnitude Therefore can be taken outside the integral, which becomes

where is the total surface area of the sphere: The total flux through the sphere is

The radius of the sphere cancels out of the result for We came to the same conclusion as when we considered rectangular closed surfaces of two different sizes enclosing a point charge. There we found that the flux was independent of the size of the surface; the same result holds true for a spherical surface. Indeed, the flux through any surface enclosing a single-point charge is independent of the shape or size of the surface.

Gauss’s Law states that the total electric flux through any closed surface enclosing a definite volume is proportional to the total (net) electric charge inside the surface. The previous example shows that the flux is independent of the sphere’s radius and depends on only the charge q enclosed by the sphere. At each point on the spherical surface, is perpendicular to the surface, and its magnitude is the same as in any other point.

where denotes an integral over a closed surface; the integral is always taken over a closed surface. This applies to closed surfaces of any shape or size, enclosing the charge The area elements and the corresponding unit vectors always point out of the volume enclosed by the surface. Then, the convention is that the electric flux is positive in areas where the electric field points out of the surface and negative where the field points inward. Also, recall that this agrees with our established convention that is positive at points where points out of the surface and negative at points where points into the surface.

For a closed surface enclosing no charge, Mathematically, this states that when a region contains no charge, any field lines caused by charges outside the region do not produce a net flux. Field lines that enter one side will always leave on the other.

Suppose the surface encloses not just one point charge but several charges At any point, the total electric field is the vector sum of the fields of the individual charges. Let be the total charge enclosed by the surface: then we can write an equation for each charge and its corresponding field and add the results. When we do, we obtain the general statement of Gauss’s Law:

A Gaussian surface is an arbitrary closed surface in three-dimensional space through which the flux is calculated. It is used in conjunction with Gauss's Law for the corresponding field by performing a surface integral in order to calculate the total amount of the source quantity enclosed, e.g., the amount of electric charge as the source of electric field; or vice versa: calculate the fields for the source distribution.

Gaussian surfaces are chosen to exploit symmetries of a configuration to simplify the calculation of the surface integral. The Gaussian surface is chosen such that for every point on the surface, the component of the electric field along the normal vector is constant, so the calculation will not require difficult integration as the constants that arise can be taken out of the integral.

We can take advantage of the symmetry of a spherical Gaussian surface when finding the electric field or the flux produced by

• a point charge
• a uniformly distributed spherical shell of charge
• any other charge distribution with spherical symmetry

The spherical Gaussian surface is chosen to be concentric with the charge distribution.

Example: Consider a charged spherical shell of negligible thickness, with radius uniformly distributed charge across the surface. We can use Gauss's Law to find the magnitude of the resultant electric field at an arbitrary distance r from the center of the charged shell.

Using spherical Gaussian surfaces with arbitrary radii, it is immediately apparent that for a spherical Gaussian surface inside the shell (radius r<R), the enclosed charge is zero as the charge is only distributed on the surface of the shell. Hence the magnitude of the electric field on the Gaussian surface (r<R) is zero and the net flux is also 0.

With the same example, we use a larger Gaussian surface to examine the outside of the shell. There is an enclosed charge, which is the total charge distribution and hence Gauss's Law will produce a non-zero electric field. The flux of the Gaussian surface is

The electric field magnitude E along the Gaussian surface is constant, so we can take it out of the integral. The integral of dA is simply which is the surface area of a sphere and so we have

By Gauss's Law, the flux is also

Equating both expressions for we can calculate the electric field.

This general result shows that any spherical charge distribution can be considered a point charge when observed from the outside of the charge distribution; this is a verification of Coulomb's Law.

A cylindrical Gaussian surface is used when finding the electric field or the flux produced by any of the following:

• an infinitely long line of uniform charge
• an infinite plane of uniform charge
• an infinitely long cylinder of uniform charge

Example: Electric field near an infinite line charge

Consider a point at a distance from an infinite line charge with charge per unit length (charge density) Imagine a closed surface in the form of a cylinder whose axis of rotation is the line charge. If is the length of the cylinder, then the charge enclosed in the cylindrical Gaussian surface is

Let a and c be the surfaces of the ends, and b be the surface of the wall of the cylindrical Gaussian surface. The total flux through the cylinder is given by the sum of the individual flux of each surface, or

Due to the infinite line charge, the electric field is always perpendicular to for surfaces a and c and is always parallel to for surface b; hence, the calculation is simplified by choosing a cylindrical surface.

The surface area of a cylinder is the circumference of an end multiplied by the length (or height): Therefore,

By Gauss’s Law, we can obtain an expression for the electric field:

Recall that the line charge density is the charge per unit length, or Hence, the electric field at point P at a distance r from the line charge is given by

Example: Non-uniform charge distribution

A solid non-conducting sphere of radius R carries a non-uniform charge distribution with charge density where is a constant and r is the distance from the center of the sphere. Find:

(a) the total charge on the sphere, and

(b) the electric field inside the sphere.

(a) Since we are asked to find the total charge, we need an expression for dQ and proceed with the integration.

We know that the total charge can be expressed as the product of the charge distribution and the total volume of the sphere, or

Recall that

And so,

To get the total charge Q, we integrate both sides, with the right-hand side integrated from 0 to R, which is the radius of the sphere.

(b) Let there be a concentric spherical Gaussian surface of radius r inside the given sphere such that r < R. To get the total charge inside the sphere, we can use the integral in (a) but only integrate from 0 to r.

By Gauss’s Law,

The normal vector of the spherical Gaussian surface is always parallel to and the surface area of this sphere is This simplifies the integral to And the electric field inside the sphere is

Example: A very long conducting cylinder (length L) carrying a total charge +q is surrounded by a conducting cylindrical shell (also length L) with a total charge −2q. Use Gauss’ Law to find:

(a) the electric field at points outside the conducting shell,

(b) the electric field in the region between the cylinders

(a) Let the whole thing be enclosed in a cylindrical Gaussian surface of length L and radius r, which is bigger than the radius of the cylindrical shell. The normal vector of the surface enclosing the length of the cylindrical shell is parallel to Note that the charge enclosed by the Gaussian surface is and the surface area of a cylinder is

Using Gauss’s Law,

(b) We enclose the conducting cylinder with a cylindrical Gaussian surface with length L and radius r greater than the radius of the conducting cylinder but smaller than the inner radius of the conducting cylindrical shell. The charge enclosed by this Gaussian surface is

Following the same steps as (a), then

In this tutorial, we learned about the concept of electric flux and used Gauss’s Law to simplify electrostatic calculations by exploiting the symmetry properties of systems; for example, a cylindrical body doesn’t look any different after being rotated around its axis, and a sphere looks just the same after being rotated about any axis through its center. We also understood how Gauss' Law relates the charge inside a surface to the electric field on the surface, and how this general law applies to other closed surfaces as well. Its applications mainly involve finding the electric field due to infinite symmetries and/or relating the electric field to the charge distribution. In the case of conductors, the charge enclosed by any surface inside the conductor is zero because the electric field inside a conductor is zero. Therefore any charge on the conductor must reside on its surface.