Why is gausss law useful

A uniform charge density in an infinite straight wire has a cylindrical symmetry, and so does an infinitely long cylinder with constant charge density An infinitely long cylinder that has different charge densities along its length, such as a charge density for and for , does not have a usable cylindrical symmetry for this course. Neither does a cylinder in which charge density varies with the direction, such as a charge density for and for. A system with concentric cylindrical shells, each with uniform charge densities, albeit different in different shells, as in Figure d , does have cylindrical symmetry if they are infinitely long.

The infinite length requirement is due to the charge density changing along the axis of a finite cylinder. Consequences of symmetry In all cylindrically symmetrical cases, the electric field at any point P must also display cylindrical symmetry. Cylindrical symmetry: ,. Gaussian surface and flux calculation To make use of the direction and functional dependence of the electric field, we choose a closed Gaussian surface in the shape of a cylinder with the same axis as the axis of the charge distribution.

The flux through this surface of radius s and height L is easy to compute if we divide our task into two parts: a a flux through the flat ends and b a flux through the curved surface Figure. The electric field is perpendicular to the cylindrical side and parallel to the planar end caps of the surface. The flux through the cylindrical part is. Thus, the flux is. Let us write it as charge per unit length times length L :.

The charge per unit length depends on whether the field point is inside or outside the cylinder of charge distribution, just as we have seen for the spherical distribution. Let R be the radius of the cylinder within which charges are distributed in a cylindrically symmetrical way. Let the field point P be at a distance s from the axis.

The side of the Gaussian surface includes the field point P. When that is, when P is outside the charge distribution , the Gaussian surface includes all the charge in the cylinder of radius R and length L. When P is located inside the charge distribution , then only the charge within a cylinder of radius s and length L is enclosed by the Gaussian surface:. Uniformly Charged Cylindrical Shell A very long non-conducting cylindrical shell of radius R has a uniform surface charge density Find the electric field a at a point outside the shell and b at a point inside the shell.

Hence, the electric field at a point P outside the shell at a distance r away from the axis is. The electric field at P points in the direction of given in Figure if and in the opposite direction to if. Electric field at a point inside the shell. For a point inside the cylindrical shell, the Gaussian surface is a cylinder whose radius r is less than R Figure.

This means no charges are included inside the Gaussian surface:. This gives the following equation for the magnitude of the electric field at a point whose r is less than R of the shell of charges. Significance Notice that the result inside the shell is exactly what we should expect: No enclosed charge means zero electric field.

Outside the shell, the result becomes identical to a wire with uniform charge. Check Your Understanding A thin straight wire has a uniform linear charge density Find the electric field at a distance d from the wire, where d is much less than the length of the wire.

A planar symmetry of charge density is obtained when charges are uniformly spread over a large flat surface. In planar symmetry, all points in a plane parallel to the plane of charge are identical with respect to the charges.

We take the plane of the charge distribution to be the xy -plane and we find the electric field at a space point P with coordinates x , y , z. Since the charge density is the same at all x , y -coordinates in the plane, by symmetry, the electric field at P cannot depend on the x — or y -coordinates of point P , as shown in Figure. Therefore, the electric field at P can only depend on the distance from the plane and has a direction either toward the plane or away from the plane.

That is, the electric field at P has only a nonzero z -component. Uniform charges in xy plane:. Note that in this system, although of course they point in opposite directions. Gaussian surface and flux calculation In the present case, a convenient Gaussian surface is a box, since the expected electric field points in one direction only. To keep the Gaussian box symmetrical about the plane of charges, we take it to straddle the plane of the charges, such that one face containing the field point P is taken parallel to the plane of the charges.

In Figure , sides I and II of the Gaussian surface the box that are parallel to the infinite plane have been shaded. They are the only surfaces that give rise to nonzero flux because the electric field and the area vectors of the other faces are perpendicular to each other.

Let A be the area of the shaded surface on each side of the plane and be the magnitude of the electric field at point P. Since sides I and II are at the same distance from the plane, the electric field has the same magnitude at points in these planes, although the directions of the electric field at these points in the two planes are opposite to each other. Magnitude at I or II:. If the charge on the plane is positive, then the direction of the electric field and the area vectors are as shown in Figure.

Therefore, we find for the flux of electric field through the box. Note that if the charge on the plane is negative, the directions of electric field and area vectors for planes I and II are opposite to each other, and we get a negative sign for the flux.

From Figure , we see that the charges inside the volume enclosed by the Gaussian box reside on an area A of the xy -plane. The direction of the field depends on the sign of the charge on the plane and the side of the plane where the field point P is located.

Note that above the plane, , while below the plane,. You may be surprised to note that the electric field does not actually depend on the distance from the plane; this is an effect of the assumption that the plane is infinite.

In practical terms, the result given above is still a useful approximation for finite planes near the center. Discuss the restrictions on the Gaussian surface used to discuss planar symmetry. For example, is its length important? Does the cross-section have to be square? Must the end faces be on opposite sides of the sheet?

Any shape of the Gaussian surface can be used. The only restriction is that the Gaussian integral must be calculable; therefore, a box or a cylinder are the most convenient geometrical shapes for the Gaussian surface. Recall that in the example of a uniform charged sphere, Rewrite the answers in terms of the total charge Q on the sphere. Suppose that the charge density of the spherical charge distribution shown in Figure is for and zero for Obtain expressions for the electric field both inside and outside the distribution.

A very long, thin wire has a uniform linear charge density of What is the electric field at a distance 2. A charge of is distributed uniformly throughout a spherical volume of radius Determine the electric field due to this charge at a distance of a 2.

Repeat your calculations for the preceding problem, given that the charge is distributed uniformly over the surface of a spherical conductor of radius A total charge Q is distributed uniformly throughout a spherical shell of inner and outer radii respectively.

Show that the electric field due to the charge is. When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface.

Use this information to determine the electric field of charge put on a 5. A large sheet of charge has a uniform charge density of What is the electric field due to this charge at a point just above the surface of the sheet?

Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. The radius of each rod is 1 cm, and we seek an electric field at a point that is 4 cm from the center of the rod.

Yes, the length of the rod is much greater than the distance to the point in question. With Gauss's law, we can even work with a curved surface, for the following reason: When a surface is curved, that curvature is only noticeable when a sufficient amount of that surface is taken into account e. In this gauss's law approach, we can make the cross-sectional area of the cylinder as arbitrarily small as we like, and the answer doesn't change.

As soon as we make the cross-sectional area "small enough" that the curved conducting surface is effectively flat i. This means that this answer applies at every conducting surface , if the density is evaluated at a specific position on the surface.

Yet another problem we have already solved! As we will see, this one is different from the previous two, in that the field will end up depending upon the dimensions of our gaussian surface. This gives us a field that is not uniform which it isn't! Once again, the trick is to define a gaussian surface where the field lines pass through parts of it at right angles, and other parts not at all. The obvious choice is therefore a cylinder. We know from symmetry arguments we have already made in the past that the field points radially outward from the line, which means that the field lines don't pass through the ends of the cylinder, contributing nothing to the total flux.

Though the curved surface of the cylinder, the electric field is perpendicular everywhere, and since the cylinder is centered at the line of charge, the field strength is the same everywhere. The total flux is therefore the electric field strength at the cylinder wall multiplied by its area:. The enclosed charge is the charge contained between the two ends of the cylinder, which is the linear charge density multiplied by the length of the segment, which is the length of the cylinder.

Applying Gauss's law therefore gives:. Again this is in agreement with the answer previously obtained Equation 1. The reader should not get the impression that electric fields only exist outside of charge distributions, though so far every example has been of this variety. Indeed Gauss's law is very useful for finding fields within charge distributions, and the process is really no different from what is outlined above. The latter calculation is as simple as those above — the field has spherical symmetry is radially outward , so we choose a spherical gaussian surface through which the field will pass orthogonally, and on which the field strength is constant , giving:.

Yes, the field looks exactly like that of a point charge! The density is constant, so the total charge is just the density multiplied by the volume of the charge. Note that this is not the volume of our gaussian surface , which resides outside the sphere, so:.

Okay, so what about within the charge distribution? So how does this change the answer? Well, there is less charge enclosed than in the previous case. Specifically, this time the entire gaussian surface is filled with charge. Plugging in this new, smaller volume gives:. Rather than getting weaker with an inverse-square dependence as it gets farther from the center, this field actually gets stronger linearly. Whenever one solves a problem that includes multiple regions like this one one region being inside the charge, and the other outside the charge , it is a good idea to check to make sure that the field is continuou s at the boundary.

We will see that this is also sometimes used as a condition that we impose to help us solve the problem. Let's take a moment here to demonstrate how problems where we are looking for fields within charge distributions can also be solved using the local form of Gauss's law. Using this method to solve for fields in empty space is fraught with mathematical nuance that we will avoid, but for regions containing charge it is quite workable, and perhaps even preferable in some cases.

Returning to the uniform sphere of charge, the spherical symmetry suggests that we write the divergence of the spherically-symmetric field in spherical coordinates. We now apply Gauss's law and integrate. Note that this is an indefinite integral, which requires the introduction of an unknown constant of integration.

To solve for this constant, we will need to know the boundary condition for the charge distribution. This is a universal feature of this method. Regarding the solution of the electric field strength outside the Gaussian surface, note that the Gaussian surface is a mathematical construct, and it doesn't actually exist in the real world. This means that once you solve an electric field problem using Gauss's law, you will have an equation that will allow you to calculate electric field strength at ANY distance from the associated charge, regardless of the distance from the Gaussian surface that you originally used.

Gauss' Law has basically three different, but very similar meanings in electro dynamics. It is in general equivalent to Coulomb's law, and so has as much applicability. Gauss' law in differential form allows for the integration of the electric flux on a surface to match it against the contained charge.

Sufficiently high symmetry like those scenarios David White mentions allows one to quickly determine the electric field. Thirdly, Already knowing the electric field, some relationships regarding flux take on a simplified form. The integration of the Maxwell Stress Tensor on a plane is closely related to symmetric Gauss's Law problems. Would recommend playing with those to get a feel for some more subtle applications of the law. Sign up to join this community.

The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Knowing when Gauss's Law is useful Ask Question. Asked 2 years, 9 months ago. Active 2 years, 9 months ago. Viewed times.

Improve this question. Add a comment.