Example 3- Electric field of a charged disc along its axis
As another application of the Coulomb’s law, for the charged distributions, now let us consider a uniform charged disc. Let’s try to determine the electric field of such a distribution generates along its axis, electric field, of a uniformly charged disc, along its axis.
If we consider a disc, like your CD for example, with a radius of R, and let us assume it is charged and that the charge on the disc is positive and it is distributed uniformly through the surface of the disc. We’re interested with the electric field that it generates along its axis, let’s say some z-distance away from its center. As we can easily see the type of distribution that we’re dealing in this example is a surface charge distribution.
We’re going to calculate the electric field generated by this distribution by applying two different methods. The first one is, basically we will take the advantage of the solution of the previous example, which is the ring charge, and apply it to this case. We can assume that the disc charge is made from the sum of incremental ring charges, incremental concentric ring charges.
Since we have already calculated the electric field of a ring charge distribution along its axis in the previous example, we can adopt that solution for an incremental ring charge for this example. We can say that the incremental ring over here will have a charge of dq and the next ring will have a charge of dq and the next one will have a charge of dq. So we can look at each one of these incremental concentric ring charges, determine their electric field at the point of interest over here, and eventually the sum of these ring charges make the whole distribution by adding the electric fields that they generate vectorially to one another, we will get a total electric field of the distribution.
Now if we recall the ring charge solution from the previous example, we have found that the electric field that it generates is equal to the total charge of the ring times distance of the point of interest to the center of the ring divided by 4πε0 times z2 plus radius squared of the ring charge to the power 3 over 2. And the direction of this field was such that, it was an upward direction so if we choose that direction as our z-axis multiplying that by the unit vector k̂ pointing in z direction, we express the electric field generated by ring charge along this axis in this explicit form.
Since we’re going to talk advantage of this solution in order to calculate the electric field of a disc charge distribution, we have to adopt ring charge solution for the incremental ring charge. Therefore if we go from ring charge to incremental ring charge, the total charge associated with the ring charge will be incremental charge dq for the incremental ring charge. z distance, that is the distance between the center and the point of interest along the axis, will be same for both cases. The radius for the ring which was denoted capital R in the previous example will now be different depending upon the radius of the incremental ring that we’re dealing with. So if you call this distance as s, then the thickness of the ring can be denoted as incremental thickness of ds. Therefore the radius will go, to s.
Now, since the total electric field associated with the ring charge was E, for the incremental ring charge that will be incremental electric field generated by the charge dq associated with incremental ring. Therefore dE becomes equal to, we replace q with dq, z remains the same and in the denominator we will have 4πε0, z2 and instead of the capital R we will have s, so s2 over here to the power 3 over 2. Therefore every incremental ring will generate its own electric field along the axis in upward direction with this magnitude.
So if we add all these incremental electric fields then we will end up with the total electric field. Since all of these electric fields are going to be in the same direction then we don’t have to worry about the components and therefore the total electric field will be integral of dE, sum of all these incremental electric fields. Or in explicit form it’ll be z dq divided by 4πε0, z2 plus s2 to the power 3 over 2.
We have to put the integrand into integrable form and first we have to express dq in terms of the given quantities. Since we’re dealing with a surface charge distribution, in other words the electric field, I mean, the charge is distributed uniformly along the surface of this disc, dq, which is the amount of charge along this incremental ring is going to be surface charge density times the area of this ring. Surface charge density, as you’ll recall, is defined, denoted with σ and it is defined as the total charge of the distribution, which is Q, divided by the total surface area of the distribution, πR2.
If we multiply this quantity by the area of the region that we’re interested with, then we will get the amount of charge in that or along that region. So it means that we need to know the area of this ring. To be able to calculate that area, let’s cut this incremental ring open. If you do that, we’re going to end up with a rectangular strip, something like this. The length of that strip is going to be equal to the circumference of the ring, which is 2πs, and the thickness of the strip will be the thickness of the incremental ring, which is the s.
Now, knowing these quantities, one can easily calculate the area of the strip, which is the area of the ring, and that will be equal to, if you call that area as dA, 2πs times ds. So once we determine σ, the incremental charge dq along the incremental ring will be surface charge density σ times the area of the incremental ring. In explicit form dq will be equal to Q over πR2, total charge divided by total area of the distribution, times the area of the incremental ring and that is 2πs ds.
Okay, now we can go back to our integral equation. Electric field will be equal to integral of z times, for dq we will write down Q over πR2 times 2πs ds, and that is the incremental charge along the incremental ring, divided by 4πε0, z2 plus s2. Our variable is s, which is the radii of the incremental rings, and as we add all these ring charges to make the whole distribution of this disc, then we will start from the innermost ring which will have the radius of 0 and we will go all the way along the surface of the disc up to here. Therefore the s will vary from 0 to the outermost ring which will have the radius of big R. The boundary of the integral will, therefore, go from 0 to big R.
Again, when we look at the integrand, we see that Q is constant, z is constant, πR2, 4ε0, these are all constant. Therefore we can take these quantities outside of the integral. And therefore we will take them, we just take this π here — as a matter of fact, that π and this π will cancel and Qz over R2, 4πε0, these are all constant and we will take them outside of the integral. Of course we will leave terms involving s inside of the integral because our variable over here is s, which is the radius of the incremental ring charge.
This is constant. Let me take it outside of the integral, leaving us electric field which is equal to Qz over 4πε0R2 and inside of the integral we will have 2s ds divided by z2 plus s2 to the power 3 over 2. Let’s not forget this power. And the boundaries are from 0 to R.
Now the task becomes taking this integral. To be able to do that we will use a simple transformation. We’re going to say let z2 plus s2 is equal to u. If that is the case, then taking the derivative of both sides since z is constant, that’s just going to go as 0 and from the derivative of s2 we will have 2s ds, that will be equal to du. When we look at these terms we see that we have 2s ds in the numerator and relative to our new transformation, this is du and this is the reason that I left 2 inside of the integral instead of taking it out or simplifying with the 4 in the denominator because to be able to have du in the integrand. If I had done that earlier then I should, again, I should have multiplied both numerator and denominator by 2 to be able to get the du term. For the denominator we have z2 plus s2 which we define that as u that is in the power bracket of u to the power of 3 over 2.
So the integral reduces into this simplified form with respect to this new variable and it becomes Qz over 4πε0R2 times integral of du over u to the power of 3 over 2. Since we changed the variable, naturally the boundaries of the integral will also change as u1 and u2. I’m not going to calculate the u1 and u2 because after taking the integral we will go back to the original variable, which is s.
Alright, well, we can easily take this integral. If we look at over here we can express this as u to the power minus 3 over 2 du and that will be equal to, we will add 1 to the power and then divide it by that quantity. If we add 1 to the power we will have minus 3 over 2 plus 1 will give us u to the power minus one-half and we will divide this by minus one-half, leaving us -2u to the power one-half or 2 over square root of u. So the integral is going to be equal to Qz over 4πε0R2 and we will have -2 divided by square root of u. We defined u as z2 plus s2 and we will evaluate this integral at 0 and R.
At this point we can cancel 2 with this 4, leaving us 2 in the denominator. Let’s go ahead now substitute the boundaries. Electric field is going to be equal to Qz over 2πε0R2, open parentheses minus, first we will substitute R for s, that’s going to give us 1 over square root of z2 plus big R2 and then minus, now we’re going to substitute 0 for our expression, for s, and another minus sign is going to come out from here. Just let me write it down in explicit form, 1 over z2 plus s2 is basically 0 here in this case, close parentheses, close parentheses.
Electric field is going to be about equal to Qz over 2πε0R2 open parentheses, minus minus here will make plus z2 in the square root will come out as z. Therefore that term is going to give us plus 1 over z and the other term will remain as it is, -1 over square root of z2 plus R2.
We can take this one more step further, and if we put this z into this bracket then z over z will give us just 1, so we’ll have Q over 2πε0R2, open parentheses, 1 minus and when we multiply we’ll have z at the numerator here divided by z2 plus R2, square root. And though we can leave it in this form, since all the electric fields associated with the incremental charges are pointing in upward direction, all the these dE‘s, then the total electric field will also be pointing in upward direction. In vector form therefore we can represent or write down this equation as electric field vector is equal to its magnitude, this term, times the unit vector pointing in z direction.
We can look at the similar type of approximations for the case that z is much, much greater than R as we did in the previous example. If z is much, much greater than R, then R over z will be much much smaller than 1. So it means that in the most crude approximation we’re going to able to neglect R over z ratio in comparing to 1. To be able to have that ratio, let us express, our equation in the form by taking the z2 outside of the square root. In doing that, since z2 is going to come out from square root as z, inside of the square root we will have 1 plus R2 over z2, since there’s no multiplayer of R with z, and close parentheses.
This z and z will cancel and we’re going to be able to neglect R2 over z2 in comparing to 1, a most crude approximation. Then we will have just 1 in the denominator and we have 1 in the numerator, 1 over 1 will give us 1 and 1 minus 1 will give us 0. When we look at this result, therefore, we can say that the most crude approximation is going to lead us to the result what we would expect at infinity. In other words if you go really infinite distance away from the disc charge, we will indeed end up with 0 electric field because the electric field decreases, we know that, with the square of the distance.
But we’d like to obtain an expression at an intermediate region such that we will end up with that non-zero result but also take the advantage of the approximation. And in order to do that, as we have seen earlier in this case so far, dipole example, we’re going to apply binomial expansion. Recall binomial expansion. It was such that 1 plus x to the power n, factorial under the condition that x is much, much smaller than 1 can be expanded as 1 plus nx over 1 factorial, plus n, n minus 1, x2 over 2 factorial plus the higher order terms.
So we’re going to look at the quantity in the denominator over here before we neglect R2 over z2 in comparing to 1 directly. We will try to apply binomial expansion for this term. If we move the square root term to the numerator, then it’s going to be 1 plus R2 over z2, to the power minus one-half. Now, R2 over z2 is indeed much, much smaller than 1, and in the formula of binomial expansion will be equal to -1 over 2.
So applying the formula for this case since the ratio satisfies much more than 1 condition we can express this, parentheses, 1 plus, and n is minus one-half, and the x is in this case R2 over z2 and plus the higher order terms. But as a first step we will neglect second and higher order terms. Remember the idea was, after this process, at the end of the non-zero result, that our approximation is successful. If we end up with 0 again, then we go ahead and expand or include the second order of terms and apply the same procedure. Again if it turns out to be 0 we go ahead and do the third order, include the third order terms and so on and so forth.
So we here we continue, we will have 1 minus R2 over 2z2 as an approximate value for this bracket. So we’re going to just go ahead and replace 1 plus R2 over z2 term in our equation with the approximate value over here. That’s going to give us electric field is equal to Q over 2πε0R2. And inside of the bracket we will have 1 minus, and for the 1 over square root part we will replace that with its approximate value which is 1 minus R2 over 2z2.
Moving on, that’s going to give us Q over 2πε0R2, 1 minus 1 plus R2 over 2z2, close parentheses. Plus 1 and minus 1 will cancel and we’re going to end up with R2 in the numerator, R2 in the denominator will cancel and the final expression is going to be E is equal to Q over 2 times 2 is 4, 4πε0, z2. Again, we ended up with the familiar expression that we have seen before. In this case the disc charge is behaving like a point charge. Disc charge behaves like a point charge for all the points along the axis for z is much, much greater than the radius of the disc. In other words, for further points such that their location relative to the center of the disc along the axis, much, much greater than the radius of the disc and if we’re that much away from the distribution, then we will perceive that distribution like a point charge.
Alright. Now, let us solve the same problem by applying another method. Here we have our disc distribution with radius R and we’re interested in the electric field that it generates some z distance away along its axis from its center. We assume that it is positively and uniformly charged to a Q coulombs. By looking at the type of the distribution we can easily see that it’s a surface charge distribution and, like in the case of the all other distribution charge problems, we choose an incremental element within the distribution at an arbitrary location and treat the amount of charge associated with that incremental like a point charge.
So to do that, let’s choose an incremental surface element here, a very, very small, tiny surface with a surface area of dA and the amount of charge associated with it as dq and treat this element like a point charge so it will look like therefore that we have positive point charge with a charge of dq sitting here and it will generate an electric field at the point of interest such that it will be pointing in a radially outward direction and the incremental charge will generate an incremental electric field of dE.
By looking at the symmetry of the distribution since it is symmetric about the axis, we can always find another incremental surface element with an incremental charge of dq right across from it, which will in turn generate its own electric field at the point of interest, and that will be pointing along the line joining these two points and it will have the magnitude of dE. Since these dq‘s, the incremental charges, have the same magnitude of charge and they’re going to be same distance away from the point of interest, then they will have the same magnitude.
Using the symmetry, we can see that for every dq that we will choose, we’re going to have a symmetrical one across from it along this distribution. Like in the case of ring charge, their electric fields, therefore, are going to align along the surface of a cone. To get the total electric field from the vector sum of all these vectors simply by considering these 2, for simplicity, we introduce a coordinate system with horizontal and vertical axes in this form and resolve the vectors into their components by taking their projections along these axes.
So when we do this we will see their horizontal components and call them as dEh with a line along the horizontal axis but they’re going to be pointing in opposite directions with the same magnitude, therefore, when we add them, they will cancel one another. On the other hand, vertical components will be pointing in the same direction along the vertical direction and they will add and eventually they will determine the electric field of the disc charge along the axis of the disc.
So we can make a note here — let’s say this is our method two — by saying that dE horizontals cancel, again, due to the symmetry and the total electric field becomes the sum of all of the vertical components and the summation over here is integration.
Now we can see that the total electric field of this disc is going to be pointing along the axis in an upward direction. Now we have these dEverticals in their explicit form and to be able to do that we’re going to introduce an angle either the incremental electric field vector making with the horizontal axis or with the vertical axis. Let’s choose this one and take the advantage of the right triangles which forming as we draw the component diagrams of the electric field vector with respect to the coordinate system that we chose.
By looking at this triangle here, and it’s an electric field triangle, we can easily see that the vertical component is the adjacent side relative to this angle. Trigonometry function associated with the adjacent side is the cosine. Therefore dEvertical will be equal to hypotenuse of the right triangle, which is dE, times cosine of θ.
Again, we could have easily chosen the other angle, the angle that the electric field vector makes with the horizontal axis and called it Φ, for example. In that case the vertical side will be the opposite side and the trigonometry function associated with the opposite side is sine. Therefore we would’ve ended up with dE times sine of that angle.
Okay. So here is the dE in explicit form or dE vertical. Now we can easily see that we can easily determine the magnitude of the electric field by applying Coulomb’s law. So this will be determined from Coulomb’s law. If we do that, dE is going to be equal to 1 over 4πε0 times the magnitude of the charge, which is the dq, divided by the square of the distance between the charge and the point of interest and we call that distance or denote that distance with little r.
Let’s show the position of the arbitrary charge, the location of the arbitrary charge that we choose. Or I shall say that the location of the incremental charge we choose within the distribution relative to the center as s, this distance. Well, if this angle is θ here, then this angle will also be θ. Now, by looking at this big triangle, which is a triangle forming from the distances, we can express cosine θ and as well we can express r in terms of z and distance s. So in that triangle, by applying Pythagorean Theorem, r2 is going to be equal to square of s plus square of z, from the Pythagorean Theorem, or r is going to be equal to square root of s2 plus z2.
Now using the same triangle, cosine of θ will be equal to ratio of adjacent sides, which is z, to the hypotenuse, and that is r. Therefore z over r is equal to z over square root of s2 plus z2 will give us the cosine of angle θ. Therefore our total electric field expression will be integral of dEverticals, and that is dE times cosine of θ, which will be equal to integral of dq over 4πε0r2, and r2 is s2 plus z2, times cosine of θ, and that is z over square root of s2 plus z2.
One thing left in the integrand is the expression of dq, in terms of the total charge of the distribution. To do that, let’s look at our incremental element first in a little bit exaggerated diagram. Here’s our disc, and I’m going to plot the incremental element in a little way, a bit exaggerated picture so we can easily see what’s going on. Something like this.
Now, remember, this element is very very small, so small such that we can choose the amount of charge associated with it or treat it like a point charge. The radius of this or the distance of this element to the center is s and therefore this thickness will be just a small increment from that distance. I should draw this like this. s and this distance over here is the incremental distance, ds, just a little bit increment to the radial distance. Let us assume that the angle that the arc length subtends is an incremental angle of dΦ. If that is the case, then, the arc length, which is this distance, is going to be equal to s dΦ, the radius times the angle that is subtends.
The area of this incremental surface element, dA is going to be equal to, if we treat this tiny little element like a rectangle with length of s dΦ and the thickness of ds, then we will get the area of this incremental surface element.
The amount of charge associated with this or distributed along this surface is what we called incremental charge of dq. So we can express that dq as surface charge density σ, charge per unit area, times times the area of the region that we’re interested with. The area of the region is the area of this incremental element. Therefore s ds dΦ is going to give us what dq is. Here, σ is the total charge of the distribution divided by the total surface area of the distribution, which is πR2.
If we go back to our integral then we will have integral, for dq we have Q over πR2 times s ds dΦ divided by 4πε0. If we look over here we also have term z at the numerator, let’s add that over here also, z, and in the denominator we have s2 plus z2 is multiplied by the square root of s2 plus z2, which is going to give us to the power of 3 over 2 over here.
Now, when we look at the integrand we see we have two variables as we integrate or as we add all these incremental dq‘s to one another. When we do that, the radial part will vary starting from the innermost element, 0, up to the outermost, which will have the radius of R, and that is the radius of the whole disc. And as we do that also, this angular part is going to start from 0 and we will go all the way around of this disc, therefore it will go to 2π radius. We have two variables over here: s and Φ. Therefore we’re taking a surface integral and we will represent that with double the integration like this.
Now, taking double integral is not any different than taking a single integral. We simply take the integrals relative to each variable and the Φ is going to go from 0 to 2π radians and s is going to go from 0 to the radius of the disc. When we look at our expression also we see that Qz and πR, ε0, these are all constant, we can take it outside of the integral. And if we rearrange our integral, then it’s going to be something like this, Qz over πR2 coming from the numerator down to the denominator, times 4πε0, and the first integral is s ds over s2 plus z2 with respect to the variable s to the power 3 over 2, which will be taken from 0 to R. And the second one is going to be taken from 0 to 2π of dΦ.
Integral of dΦ will give us just Φ, which will evaluated at 0 and 2π. Therefore this integral is going to give us nothing but 2π. To be able to take the s integral, again, we’re going to do the transformation of the variable by saying let s2 plus z2 is equal to u, then, our variable is s, 2s ds — the derivative of z is 0 because it is a constant — will be equal to du.
Now, as you can see, to be able to have du term over here we have to multiply both numerator and denominator by 2. So if we multiplied by 2 and divide it by 2 nothing will change, but in this notation with respect to this new variable we’re going to have du at the numerator and use the power 3 over 2 at the denominator. Therefore E is going to be equal to Qz over πR2 times 4πε0. Integral of dΦ will eventually just give us 2π — let’s put that over here — and we’re only left with the s integration and after the transformation we end up with du over u to the power of 3 over 2. Since we changed the variable, boundaries will also change and they will become u1 and u2. Again, we’re going to go back to the original variable, therefore we don’t need to calculate u1 and u2.
Here, this π and that π will cancel and let us not forget the 2 over here also in the denominator. Okay. This 2 and that 2 will cancel. And we will have Qz divided by 4πε0z2, the integral of du over u to the power of 3 over 2 will give us -2 over square root of u evaluated at u1 and u2. Going back to the original variables we will have Qz over 4πε0R2 — this quantity over here is R2 — and -2 over square root of u is square root of s2 plus R2 evaluated at 0 and R. Actually, this is s2 plus z2.
Alright. Now substituting the boundaries, electric field is going to be equal to, let’s cancel this 2 with this 4, which will give us 2 in the denominator. Qz divided by 2πε0R2, substituting R for s2 we will have minus 1 over R2 plus z2 to the square root. And substituting 0 for s will give us z2 in the denominator inside of the square root so it’s going to come out as z, another minus from the integration boundaries will make this sign positive. Therefore we will end up with 1 over z for the second term.
Again, if you just put z inside the brackets multiplying each term by z, electric field is going to be equal Q over 2πε0R2 times 1 over z minus z over square root of z2 plus R2, close parentheses. And after we multiply z the second term actually will just give us 1. And that will be the final expression.
If we compare this expression with the result that we obtained from the previous method and before we do that, let’s represent this in vector form with its direction, therefore multiplying with the unit vector along positive z direction, since it is pointing in upward direction, now we can compare it with the result that we obtained from the method 1. We see that when we do that it is exactly the same result, showing us that these two methods, whichever the one that we like, will eventually lead us to the right answer.