Example 2- Electric field of an infinite conducting sheet charge
Let’s now try to determine the electric field of a very wide, charged conducting sheet. Let’s say with charge density σ coulombs per meter squared. In this case, we’re dealing with a conducting sheet and let’s try to again draw its thickness in an exaggerated form. Something like this. For a conducting sheet like this, it’s charge is collected only along one of its surfaces. Let’s assume that it is charged positively and we can always visualize this huge, large sheet as a segment of a surface which eventually closes upon itself. As you remember, for conducting mediums, we cannot have any excess charge inside of the medium. Whatever the excess charge that we put inside of a conducting medium, it immediately moves to the surface.
If we can visualize this, again, as a closed surface which eventually closes upon itself, a conducting medium, and the whole charge is basically collected along this outer surface. There cannot be any charge enclosed inside of this conducting medium. To be able to calculate the electric field that it generates at a specific point in space, again, we will apply Gauss’s law and we will use pill box technique to calculate the electric field. The surface is going to be generating electric fields originating from the surface and going into the infinity and from the global point of view, the field lines are going to be originating from the distribution and going into the infinity.
Let’s say that our point of interest is somewhere over here so we’re going to choose a cylindrical pill box such that one of its surfaces pass through the point of interest and it goes through the conducting sheet, through the surface to the other side. The other side, the electric field is 0, here, E is 0 inside of the conducting medium. Therefore we will have electric field only along the right-hand side.
Gauss’s law states that integral of E dot dA over a closed surface is equal to q-enclosed over ε0. Let’s assume that the Gaussian surface, the pill box we choose, has the cross sectional area of A and it occupies this much of fraction of overall distribution. Again, since we are taking the integral over this cylindrical surface, we can divide this into different surfaces on an open surface which eventually makes the whole closed surface. Let’s number those surfaces as surface 1, surface 2 for the side surface, and surface 3 in the back, and surface 4.
Therefore the closed surface integral can be separate into the integral of the first surface of E dot dA, which is going to be E magnitude dA magnitude and for the first surface, electric field is to the right and the area vector, which is perpendicular to the surface, that too also pointing to the right, and the angle between these two vectors, therefore, which is 0 degrees, so we have cosine of 0 from the dot product, plus integral over the second surface. That is the side surface. There, dA is perpendicular to the surface pointing up, whereas the electric field vector is, again, pointing to the right, so the angle between these two vectors is 90 degrees.
Then in explicit form we have E dA cosine of 90 for the side surface plus integral over the third surface which is the side surface on the other side and in that side there’s no electric field so the integral is not going to contribute to the flux at all. And plus integral over the fourth surface, which is this one over here and, again, there’s not electric field over there. That too will not contribute to the flux. On the right-hand side we will have q-enclosed over ε0.
Since cosine of 90 is also 0, there will not be any contribution from the integral over the second surface. Once we add all these open surface integrals to one another, then we have the closed surface integral over this cylinder, this pill box. As long as we’re same distance away from the source, the electric field will have the same magnitude over that surface, so it is constant here, we can take it outside of the integral. E times integral over the first surface of dA will be equal to q-enclosed over ε0.
If we add all these dA‘s to one another over the first surface, which is the surface of circular surface of the cylinder, that is equal to the cross sectional area of the cylinder, and we call that area as A. We will end up with A from the integration. E times A will be equal to q-enclosed over ε0. q-enclosed is the net charge inside of the region surrounded by the Gaussian surface, in this case the cylinder. We can easily see that that cylinder occupies only this much of the charged sheet, therefore whatever the amount of charge along this surface is the charge that we call it as q-enclosed for this case. That will be equal to surface charge density, coulombs per meter squared, times the surface area of the region that we’re interested with and that is A.
Once we express q-enclosed in terms of the charge density which is given for this infinite conducting sheet of charge, we will have EA is equal to σA over ε0 for the right hand side of the Gauss’s law. Since A‘s are common in both sides, we can divide both sides to eliminate the cross sectional area and that also tells us that it doesn’t make any difference how big or how small we choose the cross sectional area of the Gaussian pill box. It eventually cancels leaving us the electric field from such a charge distribution, a conducting sheet of charge, is equal to σ over ε0.
As you can see, this is also a constant quantity and it is different than the electric field of an infinite insulating sheet of charge. If you recall that for an insulating infinite sheet of charge, we have found the electric field as σ over 2 ε0 because in the insulators, charge is distributed throughout the volume to the both sides of the surface, whereas in the case of conductors, the charge will be along one side of the surface only. Inside of the conducting medium, the electric field is always 0. That’s the difference between the conducting sheet and insulating sheet of charge.