Example 6- Electric field of a non-uniform charge distribution
So far, we have studied the examples of distributions such that they had uniform charge distribution. In other words, charge density was constant throughout the distribution. Now, we’re going to consider an example such that the charge density is not constant. For that, let’s consider a solid, non-conducting sphere of radius R, which has a non-uniform charge distribution of volume charge density. ρ is equal to some constant ρs times little r over big R, let’s say where ρs is a constant and little r is the distance from the center of the sphere to the point of interest.
What we’d like to do is calculate the electric field of such a distribution at different regions. First, let’s try to figure out the total charge of the distribution as the question mark. We have a solid, spherical charge distribution — charge is not distributed uniformly throughout the volume of this object — such that it’s volume charge density varies with ρ is equal to ρs times r over R. So in other words, as we go away from the center of the distribution, which has a radius of big R, as we move radially out from the center of the distribution, the charge density increases. As a matter of fact, when little r becomes big R at the surface of the distribution, then it reaches to its maximum value, which is going to be equal to this constant, ρs coulombs per meter cubed.
Well, of course if the charge were distributed uniformly and therefore the charge per unit volume would have been the same at every point inside of this region, and to be able to get the total charge of the distribution, we would have directly taken the product of the volume charge density by the volume of the whole distribution, which would have given us the total charge. In this case, we cannot do that, we cannot take the product of charge density with the volume of whole distribution to be able to get the total charge because ρ is not the same at every point inside of this region. As little r changes, then the ρ will also change.
To be able to calculate the total charge density, we look at the density, we see that it varies with the radial distance R, so we’re going to assume that this whole distribution is made up from concentric spherical shells of incremental thickness. If you choose one of these shells at an arbitrary location inside of the distribution, something like this, that it has a very, very small thickness, a shell, and it has the radius of r and thickness of dr, and this dr is so small such that when we go from inner surface to the outer surface of this spherical shell, the change in density is negligible. In other words, this quantity change is so small, the little r, we can assume that throughout this thickness, ρ remains constant. If that is the case, then this will allow us to be able to calculate the amount of charge associated with this incremental shell.
Therefore we are interested with the amount of charge throughout the volume of this shell. Once we calculate that charge, which we will call that one dq, then we can go ahead and calculate the amount of incremental charge in the next incremental shell, and then the next incremental shell, and so on and so forth. Eventually we add all those incremental charges to one another throughout the volume of this whole distribution and get the total charge.
The incremental amount of charge that is distributed throughout this incremental spherical shell will be equal to ρ times the volume of incremental shell. Therefore it’s going to be equal to the charge density, which we assume that it remains constant throughout this very small thickness, ρs r over R. Now the volume of the incremental shell. That’s going to be the surface area of the sphere times its thickness. In other words, it’s going to be equal to 4πr2 times dr. So this expression gives us then ρs times 4π divided by big R and r times r2 will give us r3 dr. This is the amount of charge distributed through this incremental spherical shell. By using the same procedure, we calculate the amount of charge along the next shell, and then the next shell, and then we add all those charges to one another.
The addition process over here is the integration. Therefore we take the integral of both sides, so we will end up with the total charge on the left-hand side. If we integrate this quantity here, ρs 4π and big R, these are all constant, we can take it outside of the integral. Q therefore becomes equal to ρs 4π over R times integral of r3 dr. As we add all these incremental, spherical shells to one another, throughout the volume of the whole distribution, the associated radii of these shells will vary from 0, starting from the innermost one, and going out to the outermost one, from 0 to big R.
Q is going to be equal to ρs time 4π over R, integral of r3 is r4 over 4 which will be evaluated at 0 and big R. Here, we can cancel this 4 in numerator with the one in the denominator, leaving us Q is equal to ρs, and substituting big R for the little r we will have R4 over here, R in the denominator, and we also have π in the numerator. This R4 and that R will cancel, leaving us R3 in the numerator, so the net charge of this distribution is going to be equal to ρsπR3.
Now, let’s consider the same distribution and try to calculate the electric field inside of an arbitrary point in this distribution. Let’s redraw the distribution over here, our spherical distribution. Charge is distributed non-uniformly throughout the volume of the distribution, which has radius of big R, and the charge density was given as a constant ρs times little r over big R, and little r is the location of the point of interest. Let’s assume that our point of interest, P, is somewhere over here. Again, we are going to apply Gauss’s law and by using the spherical symmetry, we will choose a spherical Gaussian surface such that it is passing through the point of interest. Naturally, it will have the radius of little r.
Gauss’s law states that E dot dA integrated over this surface s is equal to net charge enclosed inside of the region surrounded by this Gaussian sphere divided by ε0. As we have seen in the case of previous examples for the spherical symmetry, the electric field, or the positive charge distribution, will be radially outward everywhere enhance along the surface of this Gaussian sphere, and the incremental surface area vector, which will also be in a radial direction as being perpendicular to the surface, that too will be pointing radially out everywhere. Therefore the angle between electric field vector and the surface area vector will be 0.
Then we can express the left-hand side in explicit for as E dA cosine of 0. Since we will be the same distance from the charge, as long as we are on the surface of this Gaussian sphere, the magnitude of the electric field will be constant over that surface, so we are able to take it outside of the integral and the right-hand side will be, again, q-enclosed over ε0.
So we will have E times integral of dA over the Gaussian surface s, since cosine of 0 is 1. Therefore E times the integral of dA over the closed surface s will be equal to q-enclosed over ε0. Adding all the incremental area vectors along this surface, we will eventually end up with the surface area of that sphere, which is going to be 4πr2. On the right-hand side, we will have q-enclosed over ε0. So, the left-hand side of the Gauss’s law is identical to the previous spherical symmetry problems.
Now we will look at the right-hand side. If we had uniform charge distribution throughout the volume of this distribution, as we had done in one of the earlier examples, we would have expressed the volume charge density of the distribution, and simply by taking the product of that density with the volume of that region that we’re interested in, which is the region inside of the Gaussian sphere, we would have ended up with the q-enclosed, the net charge inside of this region. Now, we cannot do that because the charge density is not constant. In other words, it is changing from point to point.
In order to get the q-enclosed throughout this region, which is the region surrounded by Gaussian sphere, as in the previous part of this problem, we are going to choose an incremental spherical shell at an arbitrary location, let’s say something like this. The thickness of this shell is so small that we can assume, as we go along that thickness, change in charge density can be taken as constant. So, we will calculate the amount of charge inside of the spherical shell and we will call that as dq, and then we are going to calculate the amount of charge in the next concentric spherical shell, and so on and so forth. Then we will add all of those dq‘s to one another throughout the region inside of this Gaussian sphere. In other words, we will apply the same procedure that we did in the previous part, except instead of integrating and adding the dq‘s from 0 to big R, now we are going to add them from 0 to little r. That’s the region of our interest.
In order to avoid confusion with the little r variable and the variable of the radius of these concentric shells, we’re going to call the radius of this shell as s, a new variable, and the thickness as ds. In terms of therefore, this, then we can express the density as ρs, we replace the little r with s, divided by now the radius of the whole distribution, which is big R.
Then, q-enclosed is going to be the sum of all of the dq‘s associated with these concentric spherical shells, which eventually makes the whole region occupied by the Gaussian sphere, and adding all those dq‘s, addition over here is, again, integration, where we’ll have the q-enclosed. Therefore in explicit form this is going to be equal to charge density, ρs, times s over R. This is charge per unit volume, times the volume of the region that we’re interested with. For dq, which is associated with the volume of this incremental, spherical shell, and the volume of that is surface area, 4πs2, times its thickness, ds. Here we will integrate this from 0 to little r. That’s our region of interest to be able to get the q-enclosed, the net charge inside of the volume surrounded by the Gaussian sphere.
Here again, 4π and R and ρs, these are all constant, so we can take it outside of the integral. Then q-enclosed becomes equal to ρs times 4π over big R integral of s times s2 is s3 ds integrated from 0 to r. Moving on, q-enclosed will be equal to ρs 4π over R, integral of s3 is s4 over 4, which we will evaluate at 0 and little r. q-enclosed will then be ρs, we can cancel this 4 and that 4, and we will have π over R and first we will substitute little r for the s, so we’re going to have r to the 4 and we will substitute 0, minus 0, which will give us 0. So q-enclosed is going to be equal to ρsπ over R times r to the 4th.
Now we will go back to our Gauss’s law expression and substitute this for q-enclosed. If we do that, we will have E times 4πr 2 is equal to ρs π over big R times little r4 divided by ε0 on the right-hand side of the Gauss’s law expression. Here, we can do some cancellations, dividing both sides by r2, we are going to end up with r2 on the right-hand side and we can cancel π‘s here. If we express then E, which will be equal to ρs divided by 4ε0R.
Now we can try to express this in terms of the total charge of the distribution. Remember that we have found in the previous part, that the Q was equal to ρsπR3. In order to have the same relationship, let’s multiply both the numerator and denominator of this expression by πR3. In doing that, we will have E is equal to ρsπR3, multiplying them by πR3, and also multiply the denominator by the same quantity in order to keep the ratio unchanged. So we will have 4πε0, R3 times R will give us R4. In this notation then, in the numerator, we will have the total charge of the distribution.
Let’s not forget the r2 term that we had left from this step, so we will have an r2 over here. Then the final expression for the electric field is going to be, in terms of the total charge of the distribution inside of the sphere, as Q over 4πε0R4 times r2. This is in radial direction, so we can multiply this by the unit vector pointing in radial direction in order to express the electric field in vector form. This result is for the case that the point of interest is inside of the distribution.
Now let us look at the electric field outside of this distribution for r is larger than R. The simplest way of handling such a problem is since we are dealing with a spherical charge distribution with radius big R and we are interested in the electric field at a point outside of the distribution, again applying Gauss’s law, we simply place a Gaussian sphere using the spherical symmetry passing through the point of interest.
In this case, E dot dA over this closed surface s will be equal to q-enclosed over ε0. The left-hand side will be identical to the previous part, which will eventually gives us electric field times the surface area of the sphere, which is 4πr2, and the q-enclosed in this case is the net charge inside of the region surrounded by this Gaussian sphere. When we look at that region, we see that it encloses all the charge distributed throughout the sphere. If you call that one Q, q-enclosed is going to be equal to big Q. Therefore we will have Q over ε0 on the right-hand side, and solving for electric field we will have Q over 4πε0r2. That is again a familiar result, which is identical to the point charge electric field.
Also, if you recall, we said that whenever we are dealing with spherical charge distributions, then for all exterior points, the system behaves as if all the charge is concentrated to its center, and it behaves like a point charge for all of the exterior points, therefore the problem reduces to a point charge problem, such that we are calculating the electric field that it generates at point P, which is r distance away from the charge, generating an electric field in a radially outward direction exactly like in this case. In other words, even before we apply these steps, we can say that the system will behave like a point charge and total electric field is going to be equal to this quantity.
In more explicit form, or in terms of the charge density, since Q, the total charge, was equal to ρsπR3, we can also express this as ρsπR3 over 4πε0r2. Both of them are correct answers for this case. Again, it is in radial direction, so we can express this in vector form like this. That is the electric field generated by this charge distribution at a point outside, r distance away from the center of the distribution.