3.1 Gauss’s Law from Office of Academic Technologies on Vimeo.

- Example 1: Electric field of a point charge
- Example 2: Electric field of a uniformly charged spherical shell
- Example 3: Electric field of a uniformly charged soild sphere
- Example 4: Electric field of an infinite, uniformly charged straight rod
- Example 5: Electric Field of an infinite sheet of charge
- Example 6: Electric field of a non-uniform charge distribution

**3.1 Gauss’s Law**

So far we have seen that the Coulomb’s law is the only mechanism to calculate the electric field generated by charged distributions. Now we’re going to introduce another law, which is a very important law in electromagnetic theory, and it is known as Gauss’s law.

Before we go into the details of Gauss’s law, let’s introduce the concept of flux by means of a mechanical example. Let’s denote flux by Phi, and consider a pipe with a rectangular cross section, and the water is flowing through this pipe with a laminar flow. In other words, there is no turbulence, and we can assume that the molecules of flowing water moving along some stream lines, such that we can assign a velocity vector to each water molecule, they will all have the same velocity in laminar flow, and flowing through this pipe from high pressure region to low pressure region.

We’re interested with the amount of water flowing through the cross sectional area of this pipe, and let’s represent that area by a cross sectional area. Let’s try the product of this area, with the velocity vector magnitude. When we do that and if we look at the resulting units, from the dimensional point of view, the area has the units of meters square, in SI units system, and the velocity has the units of meters per second, therefore this product is going to give us meter cube per second.

But meter cube is the unit of volume, and the second is the unit of time, therefore this product directly gives us the volume of water flowing through this cross sectional area per unit time, and this is what we were after, we wanted to figure out the amount of water flowing through this pipe per unit time, through the cross sectional area of this pipe per unit time. Therefore this quantity is nothing but what we can define as flux of water.

We can easily see that in order to define the flux, therefore, we need a surface that we can write down the flux with respect to that surface. As we choose that surface, let’s look at a more general case, I mean in this case we chose the cross sectional area of this pipe which is perpendicular to the velocity vectors, but what if we tilt the surface a little bit like this.

If we do that, naturally some of the water, which is flowing through this region, will not go through that surface, in this position what will count is the projection of the surface along a plane perpendicular to the velocity vectors. In other words, only this much area will be included in our flux definition, because outside of this region water will go through without passing through that surface.

So for a more general case, therefore, we have to include the orientation of the surface that we choose, and to do that first let’s talk about the area vector. In physics area is a vectoral quantity, which is always perpendicular to the surface, it is always perpendicular to the surface, and it has the magnitude which is equal to the surface of interest, or area of interest.

Therefore, if we’re talking about the area surface, area vector of this purple surface over here, it is going to be a vector which will be perpendicular to that surface, and it will have the magnitude of the surface area of this region, therefore this angle is 90 degrees. Similarly, if we consider the surface area of this cross sectional are, that is going to be a vector perpendicular to this surface, and that will have the magnitude of the area of that region, and let’s call that one as a prime, for example.

Now, since, depending upon the orientation of the surface that we choose, with respect to the vector field, in this case the velocity vector field, we’re going to generalize the concept of flux, and in order to do that let’s consider, let’s look at this diagram from the front view. If we do that, we have the cross sectional area oriented like this, and we have the tilted version of this, and the area vector of this surface is perpendicular to the surface in this form, the velocity vectors are pointing to the right, and as you recall what it counts is the projection of the surface, along the surface and that is perpendicular to the vector field. So in other words, we’re talking about this segment, because the velocity vectors below this region will be missing this tilted surface.

If we denote the angle that the area vector which is making with the velocity vector as Theta, and if we follow the directions of this angle, the directions of its sides, the sides of the angle, we see that this side is perpendicular to this line, and the velocity vector, which is the other side of this angle, is perpendicular to this line. Therefore this angle Theta, and this angle, they’re congruent angles. In other words, they have mutual perpendicular sides therefore they have to be equal to one another.

Now in this triangle what we’re after is this side, which is the adjacent side relative to that angle, the magnitude of this side is a, so we can write the area of this region as a times cosine of Theta. Now the flux expression in the arbitrary orientation, relative to the arbitrary orientation of the surface, can be expressed as, for the water flux, v times a cosine of Theta.

Well, v is the velocity, which is a vector, a is the magnitude of the area vector, and Theta, from our diagram, is the angle between these two vectors. Then this expression is nothing but vector dot product of velocity vector and the area vector, because when we take the dot product of two vectors we end up with a scalar quantity and that quantity is equal to product of the magnitudes of the vectors, and the cosine of the angle between those two vectors.

So if we’re interested with the flux of water relative to a surface a, then we simply take the dot product velocity, dot product of the water velocity with the cross sectional area, I’m sorry, the area of interest vector, and that eventually gives us the amount of water flowing through that specific surface per unit time.

Now we generalize this to any other case, any other vector field. In other words, if we just replace the velocity vector with the vector field that we are interested with, then that vector times dot product with the area vector of the surface of interest, will give us the flux of that vectoral quantity through that specific area. Since our vector field of interest is the electric field, then now we can go back and talk about the e field flux, therefore that flux is going to be equal to, we can say by directly by substituting electric field for the area vector, as e dot a.

If we consider a positive point charge, we know that that point charge generates an electric field which originates from the charge and goes to infinity, in radially outward direction, as we have seen earlier, filling the whole space surrounding the charge.

To be able to define the flux of this electric field, if we choose an open surface, a surface which does not occupy or surround a volume, let’s say something like this for example, what will happen is that we will determine the electric field flux passing through this surface, but we will never get the total flux associated with the electric field generated by this charge, because not all of the field lines will go through that surface.

Whereas in the case of water flux, which is flowing through a pipe, choosing an open surface like the cross sectional area of this pipe, will provide a case such that all the velocity vectors will go through that surface, therefore we’re going to be able to define the exact flux associated with that case, but in this case, an open surface will not do it.

So to be able to include, or enclose, all the field lines, we’re going to choose a closed surface. Closed surface s to define the total electric flux. In that case, all the field lines originating from the charge will go through this closed surface. Along the surface, if we choose an incremental surface element with surface area vector of d a, and the electric field passing through that surface a, making an angle of Alpha with d a, from the previous case we know that what it counts is the projection of the surface on a plane perpendicular to the electric field vector. In other words, the projection of this perpendicular to the electric field vector, we can express the incremental flux through the surface of interest as d Phi e, for the electric field, field vector e dotted with the incremental surface area vector of d a.

Once we calculate the flux through this incremental surface, then we can go ahead and do the same job for the next incremental surface along this hypothetical surface that we choose, and calculate the flux through that incremental surface and then go to the next incremental surface on this big closed surface s, and do the same thing for that, and once we get all these incremental fluxes through these incremental surfaces, we add all of them through the surface to be able to get the total flux.

And the addition process over here is, again, integration, integral of d Phi e, is going to give us the total flux Phi sub e, and since this integral is taken over a closed surface, a surface which encloses on itself, therefore it occupies, or surrounds a certain volume, and in order to difference this integration from an open surface integral, we will introduce a little circle on the integral side, so it is integrated over a closed surface s.

All right, now let’s look at the argument of this integral in detail, e dot d a, dot product of these two vectors, will be equal to e magnitude d a magnitude times cosine of the angle between them, and that angle is Alpha. Therefore we will have cosine of Alpha over here.

If you recall the electric field generated by this charge, q, since the distance between the point of interest and charges are from Coulomb’s law, we can express that as e is equal to q over 4 Pi Epsilon 0 r squared, therefore the expression becomes, q over 4 Pi Epsilon 0 r squared times d a cosine of Alpha. D a cosine of Alpha is the projection of this green surface which is on the surface s that we chose, on a plane that is perpendicular to the electric field. Electric field is in radial direction, surface which is perpendicular to that will fall on the surface of a sphere.

At this point let’s recall spherical coordinates. Spherical coordinates we have r Theta and Phi as our coordinates, such that r is the radial coordinate, the distance from the origin, and the angle that it makes with the vertical axis is Theta, and if we take the projection of this along xy axis, or xy plane, the angle that this line makes with the x axis is defined as Phi, therefore r Theta and Phi are the coordinates, the spherical coordinates system, and here we’re interested to express an incremental surface element, which is on the surface of this sphere, in terms of these coordinates.

To do that, first we’re going to just this place in Phi direction, a very, very small displacement, something like this, of course I’m going to draw over here in an exaggerated way, and this is going to be an arc length, and that arc length can be expressed as the radius times, that as we have seen earlier, times the angle that is sub tense. So since we displaced in this direction, tiny little bit of displacement, that angle is incremental angle of b Phi.

Now the radius over here is equal to this distance, and that distance, we can express that distance by using this right triangle, and it is an opposite side relative to angle Theta, therefore it is equal to hypotenuse r times sine Theta, therefore the radius over here is, this radius, is r sine Theta, the angle that the arc length sub tense is d Phi, so this distance is going to be radius, which is r sine Theta time d Phi. Now we’re going to along the surface in this direction, which is in the direction of Theta, as you see this is measuring the angular position relative to the z axis, and so we will displace in this direction. For that direction the radius is r, that distance is r, the angle displaced will be d Theta, the angle, again we’re measuring relative to the z axis so we’re just displacing from z, d Theta, then the arc length for this part will be equal to radius, which is r, times the angle that the arc length sub tense, d Theta.

We can complete this into a tiny small rectangle on the surface of the sphere, like this. That is on the surface of the sphere, and the area of that is going to be d a. So d a there, incremental surface element, surface element, on the surface of a sphere in spherical coordinates, let’s denote this with a prime so that we will not confuse with this d a, the a prime is going to be equal to the area of this tiny little rectangle, and the area of that is r sine Theta d Phi, times the other sides length, r d Theta.

So d a prime becomes equal to r square sine Theta d Theta d Phi. Now d a prime is on the surface of the spherical surface, and also we know that d a cosine Alpha 2 is the area of an incremental element on the surface of a sphere. So at this specific distance r, away from the center, these two areas should be equal to one another

Then we can replace the a cosine Alpha with it’s equivalent, d a prime over here, in that case our equation takes the form of e dot d a is equal to q over 4 Pi Epsilon 0 r squared times 4 d a cosine Alpha we will have r square since Theta d Theta d Phi. These r squares will cancel. If we integrate now both sides, e dot d a integrate it over the closed surface s, as you can see on the right hand-side of the integral we have two variables, Theta and Phi, therefore we’re going to have double integrals, q over 4 Pi Epsilon 0 sine Theta d Theta d Phi. And now we know that this whole expression is equal to electric field flux.

For the boundaries, d Phi, since we’d like to add all these incremental surfaces along the surface of the sphere, and in order to do that the corresponding Phi will vary, starting from zero, going all the way down to 2 Pi radians, and similarly angle Theta is going to start from Theta and it will go all the way down to here, to Pi radians. Then we can say that Theta is going to go from zero to Pi radians and d Phi integration will go from zero to Pi radians.

Here q is constant, 4 Pi Epsilon 0 is constant. Let’s take these outside of the integral. Phi sub e therefore will be equal to integral of e dot d a over a closed surface of s is equal to q over 4 Pi Epsilon 0, and first we will take the integral of sine Theta d Theta, which is going to give us minus cosine Theta, which will be evaluated at 0 m Pi, and integral of d Phi will just give us Phi, and which will be evaluated at 0 and 2 Pi.

Moving on, we will have q over 4 Pi Epsilon 0. If we substitute Pi for Theta in cosine, cosine Pi is minus 1, minus minus will make plus, and then we will have minus. Now we’re going to substitute 0 for Theta, for minus cosine Theta, and cosine of 0 is 1 so we’ll have another minus 1 from there.

For the second boundary, the boundary that is for the angle Phi, substituting 2 Pi will give us 2 Pi. If we substitute zero, it’s going to give us just zero. Minus and minus will make plus, and therefore we will have q over 4 Pi Epsilon 0. This bracket will give us 2, 2 times 2 Pi will give us 4 Pi, and the 4 Pi’s will cancel, so our final expression is going to be, electric field flux Phi sub e is equal to integral of e dot d a, integrated over a closed surface s, is equal to q over Epsilon 0.

Since the charge over here that we’re talking about is our source charge which generates the electric field, and we’re calculating the flux of that electric field, therefore any charge outside of the closed surface, over here for example like q prime or negative q prime there, will not have any interest of ours because they’re not going to contribute to the electric field generated by this charge, the electric field flux generated by this charge.

In order to differ this specific charge of charges from the other ones which are outside of that closed surface s that we chose, we’re going to introduce a subscript of enclosed, or simply enc to the right-hand side, for the charge and that is nothing but the net charge inside of the region surrounded by this closed surface.

And this expression is known as Gauss’s law, and it simply states that the integral of electric field dotted with an incremental surface area vector, integrated over a closed surface s is equal to net charge enclosed and the volume surrounded by this closed surface divided by the product of the total for this space Epsilon 0.

As you can see from this expression, now to be able to determine the electric field flux, that is our sole purpose, then we really don’t need to calculate this integral over here, we just look at the amount of charge inside of the region, or inside of the enclosed surface that we choose, and divide that by Epsilon 0, that’s going to give us the field flux.

Okay, when we look at the Gauss’s law, and if we write down the integral in explicit form, which is going to be e magnitude d a magnitude times cosine of the angle between these two vectors, integrated over a closed surface s, it’s a hypothetical surface that we choose, equal to, which is equal to q enclosed over Epsilon 0, and here where q enclosed is the net charge enclosed inside of the volume surrounded by the closed surface s.

But we can use this expression, this law, in order to calculate the electric field. If we can somehow take the electric field outside of the integral, now in other words, if we can take the electric field outside of the integral, then we can leave that equation, I mean that quantity, alone on one side of the equation, take the remaining part of the integral and calculate the electric field that way.

And obviously to be able to take the electric field outside of the integral, well the electric field magnitude should be constant along the surface all the time, and the other issue over here is the angle between e and d a, and that angle also should be always constant so that we can take that outside of the integral to, and then finally we can easily take the remaining part of the integral, d a’s integral over that surface, and solve for the electric field. By this way we can use the Gauss’s law to calculate the electric field.

Now we can have the feeling that in order to do these things the surface that we choose should have to satisfy some certain conditions. And those conditions are such that the magnitude of the electric field should be constant wherever we go along this closed surface that we choose to define the electric flux, and furthermore the angle between electric field vector and the incremental surface area vector should always remain constant.

If the surface s that we choose satisfies these conditions, then we can apply Gauss’s law in order to calculate the electric field. If we then make a note over here, we can say that Coulomb’s law is the most applicable law to evaluate the electric field. Its disadvantage is that the electric field, in electric field calculations depending upon the type of the distributions, we can end up with complex integrals.

Let’s say the disadvantage of Coulomb’s law is that depending on the type of the distribution one can end up with complicated integral which can be taken numerically only through computers. On the other hand, Gauss’s law, is easy to apply, but it has limited applications, therefore Coulomb’s law, which can be applied for all cases, but sometimes one can end up with difficult integrals, Gauss’s law, which we will see soon, is easy to apply, but it has limited applications.

In order to apply Gauss’s law, one has to find a closed surface, let’s say s, such that first magnitude of the electric field, e magnitude, should remain constant everywhere along the surface s. Second condition, the angle between electric field vector and incremental area vector d a, should remain constant along the surface s.

These two conditions, condition one and condition two, implies a certain, existence of a certain symmetry, and a charge or charge distribution that we’re dealing with. Therefore if the distribution has some certain symmetry which will allow us to be able to find a hypothetical surface as a closed surface, such that it will satisfy these two conditions. In other words, the magnitude of the electric field will be the same everywhere along that surface, and as well as the angle between an incremental area vector on the surface will remain constant all the time, so that this will allow us to take the electric field outside of the surface integral and therefore solve for it to get the magnitude of the electric field.