Integrals in Electrostatic Problems

Transcription

1 PHYS 119 Integrals in Electrostatic Problems Josh McKenney University of North Carolina at Chapel Hill (Dated: January 6, 2016) 1

2 FIG. 1. Three positive charges positioned at equal distances around an observation point x. I. SUPERPOSITION PRINCIPLE FOR DISCRETE CHARGES As you re all familiar with by now, the electric field and potential due to a few point charges can be found pretty easily using the superposition principle and Coulomb s law. With V = 1 q 4πε 0 r (1) E = 1 q ˆr, 4πε 0 r2 (2) the potentials and fields coming from each individual charge can simply be added together, component-wise for E, and directly for V. Starting from an arrangement of individual point charges, we re going to build the framework for how integrals arise out of the problems we ve been doing in studio. Let s first consider a simple configuration involving three positive charges, all of charge +Q (Fig. 1). By the superposition principle, we can add up these three charges as E = 1 ( Q 4πε 0 r1 2 ˆr 1 + Q r ˆr 2 + Q r 2 3 ˆr 3 ), (3)

3 FIG. 2. Seven positive charges positioned at equal distances around an observation point x. or, to allow for an arbitrary number N of charges, E = N i 1 Q 4πε 0 ri 2 ˆr i. (4) In this form, we see that each individual charge, indexed by i, has its own associated vector r i. Additionally, we can see that if we keep allowing each point charge to have charge +Q, then as our number of charges N increases, this summation will increase without limit. In order to keep the overall charge constant, let s take a single amount of charge Q and divide it among the N charges, so that each has a charge of Q/N. Let s do this for the case N = 7 (Fig. 2). To go any further, we re going to need to figure out what the r i terms are. This becomes very easy if we choose our point x to be the origin, because then, all of the r i s are just the negative of the position vectors of the charges themselves. In general, r i will point from the source charge to the observation point. For constructing the vector, this means final initial, where final corresponds to the observation point, and initial corresponds to the location of the charge contributing to the field. Mathematically, r i = r obs r q, (5) and here, r obs = 0. Let s call the distance from each charge to the origin R. Then the position vector of each charge is R cos(θ i ) ˆx + R sin(θ i ) ŷ, (6) 3

4 FIG positive charges positioned at equal distances around an observation point x. where θ i corresponds to the angle of the i th charge measured counterclockwise from the horizontal. This is just the position in polar coordinates. We chose these coordinates because they happen to suit the way the charges are arranged that s all. Thus, the vector that points from the charge to the origin (and goes into Coulomb s law) is just the opposite, r i = R cos θ i ˆx R sin θ i ŷ. (7) We can use symmetry arguments to reason that any ˆx components of E should cancel out with charges on the opposite side of the semicircle, so we can ignore those parts and from now on focus only on the ŷ part of E. Let s call the chunk of the total charge on each point charge Q. For our N = 7 case, Q = Q/7. Compared to the first example, where Q would be Q/3, we see that the amount of charge on each individual point charge decreases as we increase the total number of point charges while keeping the overall charge constant. Next, we re going to see what happens when we make N very large. II. SUPERPOSITION PRINCIPLE FOR A CONTINUOUS DISTRIBUTION OF CHARGES Imagine that we have a set of 101 charges that we have to add up by the superposition principle, one by one, to find the net field (Fig. 3). You don t want to do this, and I don t 4

5 FIG. 4. An infinite number of positive charges positioned at equal distances around an observation point x. want to do it either. The good news is, counter-intuitively, that if we make the number N way bigger, not just 101, then the problem actually becomes much easier. If we make the number of charges N bigger, we ve already seen that Q, the amount of charge on each individual piece, gets smaller. If we make N incredibly large far bigger than we can actually count then Q gets very close to zero. However, if we do this, it becomes difficult to speak of individual charges that we can add up by the superposition principle. We no longer have one piece here, another piece there, and so on. Now, we have a solid, continuous piece of charge (Fig. 4). Since we no longer have individual charges to count and add up, we need a way of describing how much charge is at each location as we did before. Really, we re still breaking it up into sections like we did with Q/3, Q/7, etc., but we re making the sections have almost zero width. So, we still need to know how much charge is in each tiny section, which means we need to know how much charge is on a given section of the semicircle. In this case, if the charge is evenly smeared over the half circle, then the fraction of the total charge Q that is on a given arc of the circle ought to be equal to the fraction of the total semi-circumference that is covered by that same arc. So, for a given sweep of angle θ, the fraction of the total angle is θ/π. Thus, the amount Q on an arc described by θ is Q θ/π. 5

6 Going back to our summation for E, we now have E = N i 1 4πε 0 Q R 2 sin θ i ŷ. (8) If we take the limit N, then Q 0, and lim Q 0,N N i 1 4πε 0 Q R 2 sin θ i ŷ, (9) which is exactly the definition of the integral, 1 dq sin θ ŷ. (10) 4πε 0 R2 (The integral symbol is actually just a fancy old way of writing the letter s, standing for sum just like sigma.) Notice that we ve dropped the index i since our angle has now become continuous. How do we know the bounds of our integral? We have to follow the charge. Wherever it goes is where we go, adding up the contribution at each point where we find a small piece of charge. Though we may be used to thinking of an integral as the area under the curve for some function, this idea limits what we can actually do with integrals. When we say we integrate over something, we just mean that we follow that particular quantity around as our guiding parameter and do the summation that the integrand tells us to do. dq means follow the charge, and in this case, θ happens to be a perfectly convenient parameter for telling us where the charge is (If we were looking at a linear wire of charge, some parameter like x would be used instead). We ve already established that Q = Q θ/π, so it s not much of a stretch to state that dq = Qdθ/π. If we move from the first charge (on the right) to the last charge (on the left), then our angle θ goes from 0 to π. Taking all of this into account, our integral becomes leading to a final answer of π 0 1 sin θ Q 4πε 0 R 2 π dθ ŷ, (11) E = 1 Q ŷ. (12) 2π 2 ε 0 R2 It should be noted that we didn t directly look at the charge density λ on the semicircle. We dealt with it indirectly via the fraction of the total angle, but if we wanted an actual charge density in terms of length, we would need to incorporate the total arc length as λ = Q/L, where L = πr. Then, instead of integrating over just the angle θ, we would 6

7 need to integrate over the arc length, R dθ. The factors of R end up canceling anyway, but it may be helpful to see it carried out so that the same reasoning can be applied to other arrangements of charge. Also, Coulomb s law only applies to point charges! We re allowed to use it in this integral because we decided that we would break up the charge into infinitesimally small points along the arc point charges. Spheres of charge have the same form of electric field as point charges, but you should only consider that a coincidence due to spherical symmetry and nothing more. In general, you need to integrate Coulomb s law for all the tiny point charges that make up a distribution, and the final result you get for the field will usually not look like 1/r 2. 7