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**Example 5- Electric field of an infinite sheet of charge**

Now let’s consider an example of infinite sheet of charge with surface charge density *σ* coulombs per meter squared. In this case, we have a charged plate and it is very large, going to plus infinity in both dimensions and minus infinity, let’s say, in these dimensions. Let’s assume that it is positively charged and the surface charge density is given as *σ* coulombs per meter squared.

We’d like to calculate the electric field that it generates at a certain distance away from this distribution. Let’s assume that our point of interest is somewhere over here. To be able to solve this problem, we’re going to apply Gauss’s law, which is **E** dot *d***A** integrated over a closed surface, *s*, is equal to net charge enclosed within the volume surrounded by this hypothetical closed surface *s* divided by *ε*0.

We’re going to chose, in this case, our closed surface as a cylindrical surface passing through the charge sheet. Something like this. Therefore, it goes through the sheet of charge through this region and extends to the other side of the sheet. Since this cylindrical surface looks like a pillbox, this method is also known as “pillbox method”.

The electric field generated by such a very wide sheet of charge is going to be originating from the sheet and extending to infinity on both sides. Therefore, on the right-hand side, they will be pointing to the right. On the left-hand side, they’re going to be pointing to the left, extending to the infinity.

When we look at our pillbox, which is a cylinder, it is made from the circular sides on both sides and also a rectangular surface wrapping around those circular regions. In that sense, if you consider our surface over here on the right-hand side of the plate relative to the circular side, electric field is going to be pointing to the right in this direction, and the surface area vector is going to be perpendicular to the surface. Therefore for this surface over here, it is going to be perpendicular to that and will be pointing in the same direction with the electric field vector.

For the side surface, though, which is this surface over here, the associated surface vector will be perpendicular to this surface. Here, for example, it will be pointing perpendicular to that point *d***A** and the electric field vector right at that point, again, will be pointing to the right direction relative to the surface. Therefore the angle between these two vectors for the side surface will be 90 degrees.

If you go back to the other side of the plate, now for the side surface electric field is, again, pointing to the left in this case. This surface area vector is perpendicular to the surface, to the side surface *d***A**. Therefore the angle between these two vectors will be just the 90 degrees. And if you consider this side of the cylindrical surface, electric field at that point is pointing to the left and corresponding area vector over here will also be in the same direction of this vector because it is perpendicular to the surface.

Now, by applying Gauss’s law and considering this cylindrical surface, pillbox, we can divide that pillbox into its surfaces which eventually makes the whole pillbox. For the right-hand side, let’s number these surfaces as surface 1, surface 2 and the other side, surface 3 and surface 4. So we can separate the whole closed surface integral of **E** dot *d***A** into the sum of open surface integrals which eventually makes the whole cylindrical surface. Therefore integral over the first surface, which is the circular surface, and for that surface, if we express **E** dot *d***A** in explicit form, we have *E* magnitude, *dA* magnitude, times cosine of the angle between them, which is cosine of 0.

Now, as we were choosing this pillbox, we choose it such that one of these circular sides passes through the point of interest, and as well as the whole pillbox, the whole cylindrical surface, passes through the plate. So this is the integral taken over one of these circular sides over here. Now, let’s take the integral over the second surface, which is the side surface of the cylinder on the right-hand side of the plate, and for that region we have *E* magnitude *dA* magnitude times cosine of the angle between these two vectors, and for that part it is 90 degrees.

Moving on, plus integral over the third surface, and third surface is the side surface of the cylinder on the other side of this plate. For that region, again, we have *E* magnitude *dA* magnitude times cosine of the angle between these two vectors and that is also 90 degrees. Plus integral of the fourth surface, and that is other circular surface which is on the left-hand side of this plate, and for that region we have *E* magnitude *dA* magnitude times, again, the angle between *E* and *dA* for that region is 0 degrees. That should be equal to *q*-enclosed over *ε*0.

Since cosine of 90 is 0, integrals over the second surface and the third surface will not contribute to our closed surface integration. Only contributions are going to come from the integrals over the first and the fourth, or over the circular surfaces of this pillbox of the cylinder. Cosine of 0 is 1, and when we look at our diagram, whenever we are on these circular surfaces, we will be same distance away from the charge, from the distribution. Therefore the magnitude of the electric field vector will be the same for these regions. Over these surfaces, electric field is constant; we can take it outside of the integral, and these are identical integrals. Therefore we end up with *E* integral over the first surface of *dA*, plus *E* times integral over the fourth surface of the cylinder should be equal to *q*-enclosed over *ε*0.

This is the hypothetical Gaussian surface that we choose. Therefore we can assign or we can choose any cross-sectional area that we’d like to. Let us say that *A* be the cross-sectional area of the pillbox that we choose. In that case, if we integrate *dA*, the incremental surface element, over this whole first surface, then eventually we’re going to end up with the area of that surface and that’s going to give us just *A*. Similarly, for the integral over the fourth surface, if we add all these *dA*‘s to one another throughout the circular surface 4, we will end up with the area of that total surface area of that surface.

So the left-hand side will eventually give us 2 times *E* times *A*, which is equal to* q*-enclosed over *ε*0. *q*-enclosed, again, by definition, is the net charge inside of the region surrounded by Gaussian surface. Gaussian surface is the co-surface of this pillbox which encloses only this shaded area of the charge distribution. Therefore, whatever the charge that we’re interested is, the amount of charge along this region, along this surface. We are given the surface charge density of the distribution. Therefore we can easily calculate the *q*-enclosed knowing this charge density. Since this area is equal to the cross sectional area of this cylinder and we said we chose this cylinder with a cross-sectional area of *A*, then we can express *q*-enclosed as the surface charge density *σ*, which is coulombs per meter squared, times the area of the region that we’re interested with, and that is *A*.

Once we express *q*-enclosed, in terms of the surface charge density, then our expression becomes 2*EA* is equal to *σA* over *ε*0. Dividing both sides by the cross-sectional area *A*, we can eliminate *A* on both sides and solving for the electric field, the magnitude of electric field generated by this infinite plate or sheet of charge, *E* becomes equal to *σ* over 2*ε*0.

One interesting in this result is that the *σ* is constant and 2*ε*0 is constant. Electric field from such a charge distribution is equal to a constant and it is equal to surface charge density divided by 2*ε*0.

Of course, infinite sheet of charge is a relative concept. Let’s recall the discharge distribution’s electric field that we did earlier by applying Coulomb’s law. Recall discharge distribution. In that case, we considered a disc charge which was uniformly charged throughout it’s surface with a radius *R*, and we calculate the electric field along its axis *z* distance from the center. We have found that the electric field was pointing along the axis in upward direction. The charge had the disc at positive charge of *Q* coulombs and we obtained a result for the electric field magnitude as *E* is equal to *Q* over 2*πε*0*R*2 times, in parentheses, 1 minus *z* over *z*2 plus *R*2 in square root.

Now, if we look at this distribution for a specific case such that we approach along the axis towards the center of the disc, in other words for points such that *z* is much, much smaller than the radius of the disc, then naturally *R*, since *z* is much, much smaller than *R*, *z* over *R* will be much, much smaller than 1. In order to take the advantage of this condition, if we rewrite the expression such that by taking the *R* outside of the square root bracket, then we will have *z* in the numerator, *R*2 is going to come out *R* outside, and inside of the square root we will have *z*2 over *R*2 and plus 1, closed parenthesis.

Since *z* over *R* is much, much smaller than 1, then *z*2 over *R*2 will be even smaller than 1, which can be neglected, therefore, in comparing to 1. Then the square root we will just end up with 1. Here also we have *z* over *R*, which is going to be also very, very small in comparing to this one. Then we can neglect that in comparing to that one also. So we can end up with an approximate expression for the electric field which will be equal to this quantity.

But if we look at over here, *π* times *R*2 will give us the area of this disc, and then we will have charge per unit area, and that is nothing but the surface charge density of this disc. Since *Q* over *πR*2 is equal to *σ*, then the expression reduces into a form of *σ* over 2*ε*0. You can easily see now this is the identical result that we obtained from the infinite sheet of charge distribution.

It indicates that, when we approached the disc so close such that our point of interest, interests distance along the axis is much, much smaller than the radius of the disc, then we will pursue this distribution as an infinite sheet of charge. For those points, the electric field that it generates will be a constant and it will be equal to surface charge density divided by 2 times the permittivity of free space.