Example 3- Electric field of a uniformly charged solid sphere
We will now calculate the electric field of a charged solid spherical distribution. Electric field of a uniformly charged, solid spherical charge distribution.
In this case, we have spherical solid object, like a solid plastic ball, for example, with radius R and it is charged positively throughout its volume to some Q coulumbs and we’re interested in the electric field first for points inside of the distribution. In order to do that, we’ll apply Gauss’s law.
Assume that our point of interest is located at this point, which is little r distance away from the center of the distribution. We will choose a spherical Gaussian surface, hypothetical closed surface. Let’s call this surface s1. The Gauss’s law is E dot dA integrated over this closed surface s1 is equal to q-enclosed over ε0.
Since this is a positive charge distribution, it is going to generate electric field radially outwards everywhere at the location of this hypothetical surface that we choose that is passing through the point of interest. The electric field, therefore, is going to be pointing radially out and so is everywhere else of this surface. And the incremental area vector will be perpendicular to the surface at this location; therefore, it is going to be pointing radially out at this location. Radially out, like this and here and there also.
Therefore, such a surface will satisfy the conditions to apply Gauss’s law because the electric field magnitude is constant. It is the same everywhere along the surface since, again, it will be same distance away from the source every point, at every point along the surface and the angle between E and dA is 0 wherever we go along this surface. The left-hand side of the expression is very similar to the previous examples that we did. For the spherical geometry, we will have E magnitude, dA magnitude times cosine of the angle between them, which is 0 degrees integrated over surface s1 is equal to q-enclosed over ε0.
Cosine of 0 is 1. Since electric field magnitude is constant along surface s, we can take outside of integral, leaving us E times integral of dA over the closed surface s1 is equal to q-enclosed over ε0. Again, integral of dA over the closed surface s1 means adding all these incremental surfaces to one another along the surface of this sphere s1, which will therefore eventually give us the total surface area of that sphere and that is 4π times its radius squared, which is little r2. And we have E times E and that will be equal to q-enclosed over ε0.
The left-hand side is done now; we’re going to look at the right hand side of this equation. Now, remember the charge is distributed along the volume of this solid, spherical object. So, what we mean by q-enclosed is the net charge inside of the region surrounded by this Gaussian sphere. Any charge outside of this region is of interest; therefore, we need to determine the q-enclosed, right over here, the total amount of charge in this shaded region.
To be able to express this amount of charge in this region and since this is a volume charge distribution, we’re first going to express the volume charge density and, as you recall, that was denoted as ρ. Volume charge density, which is going to give us total charge divided by total volume of the distribution or charge per unit volume. Total charge is Q and the total volume of the whole distribution is the volume of this big distribution sphere. That is 4 over 3 π big R3.
The q-enclosed is going to be ρ times the volume of the Gaussian sphere that we choose, which is sphere s1. Therefore, q-enclosed is going to be equal to Q over 4 over 3 πR3. This is charge per unit volume times the volume of the region that we’re interested with is, and that is 4 over 3 π times little r3 will give us q-enclosed. Here we can cancel 4 over 3’s, π‘s. Then we will end up q-enclosed as Q over big R3 times little r3.
Now I have to go back to the Gauss’s law. On the left-hand side, we have E times 4πr2 and, on the right-hand side, we have q-enclosed, which is Q over R3 times little r3 and we’ll divide this by ε0, q-enclosed over ε0. r2 and r3 will cancel and, solving for the electric field, we will have Q over 4πε0R3 times the little r. This expression will give us the electric field inside of this charge distribution. Again, in vector form, since it is in radial direction, you can multiply this by the unit vector pointing in radial direction.
Now, let us try to determine the electric field outside of this distribution, and that is electric field for little r is bigger than big R. And I will re-draw the diagram over here. This is our spherical charge distribution with radius R and it has its charge uniformly distributed throughout its volume. Now we’re interested with a point which is located outside of this distribution, r distance away from the center.
Again, using the symmetry of the distribution, we will choose a spherical Gaussian surface, a closed surface, passing through the point of interest. Let’s call this one s2. Again, exactly similar to the previous part, the electric field will be radially out everywhere wherever we go along this surface and the area vector will be also in radial direction. Therefore, the angle between the E and dA will be 0 and, since we’re same distance away from the source as long as we’re on the surface of the sphere, then the magnitude of the electric field will be the same everywhere along the sphere s2.
So, that surface satisfies the conditions to apply Gauss’s Law of E dot dA integrated over surface s2 in this case, which will be equal to q-enclosed over ε0. The left-hand side of this equation will be identical to the previous part will eventually give us E times the surface area of this surface s2 and that is 4πr2.
The right-hand side is the net charge, the q-enclosed, inside of the volume surrounded by this s2, which is this region and, as you can see, now, once we’re outside, the Gaussian surface encloses the whole charge on the distribution and that is Q. So q enclosed is equal to big q and the right-hand side will be equal to Q over ε0. And, solving for electric field, we will end up with Q over 4πε0r2. It is in the radial direction again, and we multiply with the unit vector in radial direction.
Again, as you can see, the result is identical with the point charge so whenever we are outside of this spherical distribution, the distribution is behaving like a point charge and we had a similar type of result for this spherical shell charge distribution. Then we can make an important note by saying that a spherical charge distribution, shell or solid, behaves like a point charge for all the exterior points as if its all charge concentrated at its center.
Alright, now if we look back to the results that we obtained from the inside and outside solution for the electric field of this charge distribution, for the electric field inside, E of r was equal to Q over 4πε0R3. Q over 4πε0R3 times little r for r is less than big R. For the outside, we obtained Q over 4πε0r2 such that the charge was behaving like a point charge.
If we plot the electric field as a function of the radial distance for these cases, let’s place our sphere over here as our distribution with radius r. Inside of the sphere, electric field increases linearly with r; therefore, inside it increases linearly and, as you can see, at r is equal to 0, the electric field is going to be 0 so it passes through the origin. Then once we leave this sphere, when we go outside of this sphere, then it decreases with 1 over r2 and it becomes 0 as r goes to infinity. So it is proportional to 1 over r2 outside of the sphere, and it is proportional to r inside of the sphere.
As you can see, the maximum value of the electric field occurs when little r becomes equal to the radius of the distribution and, at that point, the value of electric field is Q over 4πε0 big R2. And that’s, therefore, the electric field profile of such a charged solid sphere such that the charge is distributed throughout this volume uniformly. Inside, it is increasing linearly and outside, it decreases with one over r2 and goes to 0 when r approaches to infinity.