Example 4: Potential of a disc charge distribution

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CORRECTION: pi at he numerator cancels with the pi at the denominator when the constant quantities taken out the integral at 8:30 min. Therefore pi square in the denominator of the final expression should be just pi.

Example 4- Potential of a disc charge distribution

As another example, let’s calculate the electric potential of a charged disc. Potential of a charged disc with radius R, and charge Q along its axis, z distance from its center.

In this case, we have a charged disc, with radius R and charge Q. Let us assume that the charge is distributed uniformly through the surface of this disc and we are interested with the potential that it generates, z distance away from the center along its axis.

If we recall the electric field of a similarly charged disc, we applied a procedure, such that we assumed the disc consists of concentric, incremental ring charges. If we choose one of them at an arbitrary radius, we can treat the amount of charge along this ring as, dq, so the amount of charge along this ring, of this incremental thickness, dr, as dq, and let’s denote this distance, instead of dr and r, and since we used the little r notation for the distance between the incremental charge or between the charge and the point of interest, and in order to differ from that, let’s call this one, s and ds.

The distance of this incremental charge ring to the point of interest is what we call, rd. Therefore this distance over here is s, the radius of the incremental ring, and we can calculate the potential generated by this ring charge at this location, which is going to be an incremental potential of dV, and that will be equal to dq over 4 πε0r.

Here, we can express dq in terms of the total charge of the distribution, and since this is a surface charge distribution, dq is going to be equal to surface charge density times the area of interest ,which is the area of this incremental ring, area of incremental ring charge. σ in exponential form will be the total charge of the distribution which is Q divided by total surface area of the distribution, and that is π times big R2 since radius is big R, in other words, this distance, πR2.

The area of the incremental ring charge can be obtained, by simply cutting open that incremental ring, we will end up with a rectangular strip such that its length is going to be equal to the circumference of this ring. That is 2π times s, the radius, and the thickness will be equal to ds. Therefore, the surface area of this rectangular strip, let’s call this one as, dA, will be equal to 2πs ds, and that is basically the area of this incremental ring.

Now, furthermore, we can express little r in terms of the distance z and radius of the incremental charge. Using this right triangle and applying Pythagorean Theorem, r is going to be equal to square root of z2 plus s2. Therefore the incremental potential generated by this incremental ring charge at the point of interest P is going to be equal to dq, that is σ times dA. If you write those quantities in explicit form you will have Q over πR2, and dA is 2πs ds divided by 4πε0 times R, and for R we will write down square root of z2 plus s2. This is going to be the incremental potential generated by this incremental ring at the point of interest.

Once we obtain this, then we go ahead and do the same calculation for the next incremental ring and then for the next incremental ring and then the next one, and we do this throughout the whole incremental rings, which eventually makes the whole distribution. Then since we are dealing with a scalar quantity we add them directly, and the addition process over here will be integration, which will give us the total potential of this disc charge.

If you look at our quantities in the integrand, Q, πR2, 4πε0, these are all constant. Let’s take these quantities outside of the integral. I will leave two inside of the integral for integration purposes and our variable is s. Of course we cannot take all those z2, where the z is constant, we cannot take it outside of the integral because it is in the square root with the variable s. s is the radius of these incremental rings and as we add them together, that radius will vary starting from the innermost ring with radius 0, to the outermost ring with radius big R, so the boundaries will go from 0 to R.

If you continue the potential will be equal to Q over πR2 times 4πε0, and coming from the denominator, and inside of the integral we have 2s ds divided by square root of z2 plus s2, integrated from 0 to R.

In order to take the integral we will make a change of variable transformation and we will say that let z2 plus s2 is equal to u. Then, if you take the derivable of both sides, the derivative z will give us 0 since it is a constant. Our variable is s2 derivative of that will give us 2s ds and that will that will be equal to du. As we can see, we already have this quantity in the numerator, that’s why I didn’t take 2 outside of the integral, if I had done that then you wouldn’t be able to 2s ds, I was going to multiply both numerator and denominator by 2 again. So instead of doing that twice I just left it over here, and so this term is going to be equal to du in terms of the new variable and the quantity in the denominator is going to be equal to square root of u.

Of course, since we changed the variable, the boundaries of the integral will change too, but we will not calculate those boundaries because we will go back to the original variable after we take the integral, so we will have integral of du over u to the power of one-half, or square root of u, integrated now from u1 to u2.

If we continue, this will be equal to Q over πR2 times 4πε0 and we can express this integral by moving u to the numerator, as u to the minus power, u to the power of minus one-half du, integrated from u1 to u2.

We can easily take this integral now, we will have, therefore, Q over πR2, times 4πε0. So if we add 1 to the power and divided by that quantity and if you add 1 we will have u to the plus one-half, and we will divide it by plus one-half. That will be result of the integration which will be evaluated at u1 and u2.

Moving on, the voltage or the potential is going to be equal to Q over, let’s go ahead and multiple these quantities, 4πε0R2, and this 2 will go to the numerator times u to the power of one-half evaluated at u1 and u2.

Now we can go back to our original variable. Then our expression will the Q over 4π2ε0R2, and as a matter of fact we can cancel this 2 and 4, and we are going to have just 2 in the denominator, 2πε0R2. And u was defined as z2 plus s2. So instead of u, we will have, on a square root of u, we will have, square root of z2 plus s2. Now we go back to our original boundaries, which will be evaluated at 0 and big R.

V is going to be equal to Q over 2πε0R2, open parentheses, first we will substitute R for s2. That is going to give us z2 plus R2 in square root. And then we will substitute 0 for s2, which will give us 0 here, and then we will end up with square root of z2 and that is going to come out as z.

That will be our final expression. So a uniformly charged disc with radius big R is going to generate a potential z distance along its axis from its center, which is going to be given by this quantity. Again, since we are dealing with potential it is a scalar quantity it does not have any directional properties.