Example 3- Potential of a ring charge distribution
All right. Now let’s try to calculate the potential of a charged ring. Let’s assume that we have a charged ring which has a radius of big R and we are interested with the potential that it generates z distance away from its center at this point P. Let’s assume also that the ring is uniformly charged along its circumference with a positive charge of Q coulombs.
Since we know the potential of a point charge that we calculated earlier was equal to q over 4πε0r, where q is the charge, and r is the distance between the charge and the point of interest, we’re going to approach this type of charge distribution problems to calculate the potential that they generate at a specific point in a similar way that we approached to the electric field problems. In those cases, as you recall, we choose an incremental charge element along the distribution at an arbitrary location and call the amount of charge associated with that segment as incremental charge of dq and treat this dq like a point charge to calculate its potential at the point of interest.
Once we do that then we go to the next incremental charge element, treat that like a point charge and calculate its potential at the point of interest and eventually do this throughout the whole distribution and finally add all those incremental potentials associated with those incremental charges throughout the distribution to be able to get the total potential. Since potential is a scalar quantity, then we don’t have to worry about any directional properties which are associated with the vectors. So the point charge potential was given in this form and if we go from here to an incremental charge, then the potential will be incremental potential dV and the charge will be incremental charge dq divided by 4πε0r.
Well, the distance between the charge of interest and the point of interest is what we call r, and this distance is, again, big R. Therefore we can express the incremental potential generated by this incremental charge at the point of location as dq over 4πε0 and the little r here, using this right triangle and applying Pythagorean theorem, since r2 is equal to big R2 plus z2, then the little r can be expressed as square root of R2 plus z2 in terms of the given quantities.
This will be the potential generated by this dq and then the potential of the next dq can also be calculated in a similar way and once it is done all throughout this ring charge and the total potential can be obtained by adding all these dV‘s. The addition process here is integration. Therefore integral of dq over 4πε0 square root of R2 plus z2 will give us the total potential of the system.
Here, of course, to be able to take this integral, we have to express dq in terms of the total charge of the distribution. In order to do that, we can easily see that dq has the arc length of ds and this arc length subtends an incremental angle of let’s say dΦ. Then we can express dq, the incremental charge, as the charge density. In this case we’re dealing with line charge density because charge is distributed along the length of this ring, and that is Q or the total charge divided by the total length of the distribution, which is the circumference of this ring, and that is equal to 2πR, times the length of the region that we’re interested in. Since length of dq is ds, then if we multiply this linear charge density λ by the length of the region that we’re interested with, then we will get the amount of charge along that length.
Okay. If we just take it one further step, ds, since it is arc length, can be expressed as the radius times the angle that it subtends which is R dΦ. Then dq becomes equal to Q over 2πR times R dΦ. Now we can substitute this into our integrand for dq. Then the potential expression will be equal to integral of, here we can cancel the R in the numerator with the one in the denominator, and we will have Q over 2π times dΦ, this is for dq, divided by 4πε0 times R2 plus z2 in square root.
In the integrand, Q is constant, 2π and as well as 4πε0 is constant, R is the radius of the ring, which is constant, and z is the distance from center to the point of interest, that is constant. Therefore all these quantities are constant and we can take them outside of the integral. Our variable is dΦ and as we integrate dq‘s, in other words, add all of them to one another along this ring, then the corresponding Φ is going to start from 0 and will go all the way around to 2π radians. Therefore the boundaries will go from 0 to 2π and these quantities are constant, we will take it outside of the integral. Then the potential becomes equal to Q over 2π times 4πε0 times square root of R2 plus z2 and integral of dΦ integrated from 0 to 2π.
Integral of dΦ is going to give us Φ and we will evaluate this at 0 and 2π. When we substitute 2π, we will have 2π, and when we substitute 0 for Φ, we will have just 0. V is going to be equal to Q over 2π times 4πε0 times squared root of R2 plus z2, and from the integration we will have just 2π. That 2π and this 2π will cancel then the potential of this ring charge, along its axis z distance from its center, will be equal to Q over 4πε0 times square root of R2 plus z2.