Separation of Variabes and a Spherica She with Surface Charge In cass we worked out the eectrostatic potentia due to a spherica she of radius R with a surface charge density σθ = σ cos θ. This cacuation invoved severa steps, so et s go back through it and make sure that everything we did is cear. First, what are we trying to do? We want to find the potentia due to the charge distribution described above. More specificay, we want to determine the potentia inside the sphere, r < R, outside the sphere, r > R, and of course we woud aso ike to know what happens at r = R. So the region V where we want to determine the potentia is everywhere, or a space. To pose a probem with a unique soution we need to describe any charge inside this region, and aso specify the potentia on the boundary of the region. The charge inside the region has aready been described: there is a surface charge density σθ = σ cos θ at r = R. What about the boundary of the region? In probems where we want to know the potentia everywhere, the boundary is basicay r. Why is that? You can think of everywhere as the inside of a giant ba. The boundary of the ba is just the sphere that surrounds it. You make the ba bigger and bigger, so that it encoses more and more space, by etting its radius become arbitrariy arge. This may sound a bit hand-wavy, but if we took the time we coud make it more precise. The bottom ine is that we need to say what the potentia does as r if we want to find a unique expression for the potentia everywhere. Now how do we approach this probem? As is usuay the case, it s good to start by asking what sorts of simpifications we can make. Wi the potentia depend on a three coordinates as we move from point to point, or is it simper than that? In this probem you get the feeing that it shoud be simper. Why is that? This is an exampe of an azimuthay symmetric system. A system has azimutha symmetry if a important aspects of the system ike the distribution of charge, or the potentia on various surfaces are symmetric about the same axis. That is the case here: a the charge is the same distance from the center of a sphere, and the amount and sign of the charge at a given point on the sphere ony depends on the poar ange θ. So the charged spherica she ooks exacty the same from ϕ =, ϕ = π/, ϕ = π/7, or any other vaue of the azimutha ange ϕ. How does azimutha symmetry hep us? As we saw in cass, we have access to some powerfu techniques for determining the potentia in probems with azimutha symmetry. At most of the points where we want to find the potentia there is no charge, right? A the charge is on the she at r = R, so when r < R or r > R the potentia satisfies Lapace s equation V =. Our strategy in a probem ike this to find the appropriate soutions of Lapace s equation in regions where there is no charge ike r < R and r > R, and then patch them together where those regions overap r = R in this case, where the charge is ocated. We re motivated to approach the probem this way because Lapace s equation for azimuthay symmetric probems can be soved quite generay Remember, the boundary is a surface that encoses the region you are interested in. You may actuay need more than one surface to encose the region. For instance, to describe the potentia between the was of a cyindrica she you woud need to specify the potentia on both the inner and outer was of the cyinder. In fact, we aso have access to powerfu techniques that hep with probems where there is much ess symmetry. Separation of variabes can be used to buid up soutions to probems that depend on r, θ, and ϕ.
using separation of variabes. In any region where there is no charge, the potentia can be written as 3 V r, θ = A r + B r + P cos θ. The A and B are constants, and the P cos θ are Legendre poynomias. 4 The coefficients A and B te us how much of each separabe soution r P cos θ and /r + P cos θ to incude to get the actua potentia we want. Think of them ike the components of a vector, which te us the contributions needed in the ˆx, ŷ, and ẑ directions to get a specific vector. Now, we ve seen in previous probems invoving spheres, cyinders, etc that the potentia may behave differenty inside and outside the object. For exampe, outside a spherica she with a constant surface charge density the potentia fas off ike /r, but inside that sphere it is constant. So we expect that in a probem ike this the potentia might ook different inside and outside the sphere. That means there are two different regions where we need to consider how the potentia behaves. We write the potentia in the two regions as V in r, θ = V out r, θ = r + B in r + P cos θ 3 A out r + B out r + P cos θ. 4 Notice that the potentia has the same genera form in both regions powers of r time Legendre poynomias, but we assume that the detais the specific coefficients are different. That is, the coefficients and A out probaby aren t the same, and ikewise for B in and B out. The boundary conditions and other information for this probem wi be used to identify these coefficients, eading us to the specific soution we want. How do we figure out what the coefficients are? We, what do we know about the probem? As we discussed in cass, the potentia shoud approach zero as we go very, very far away from the she. After a, it s just a bob of charge, so as we get far away we expect the potentia to drop off approach zero at east as fast as the potentia for a point charge. That tes us something about the potentia outside the sphere; what about inside? Anything might be hepfu, so what can we say? We, at the very east we can say that since there is no charge inside the sphere, we don t expect the potentia to become too arge anywhere. Does that make sense? Think about the potentia due to a point charge: it becomes very arge as you get cose to the point charge, and in fact it is infinite right on top of the charge because /r diverges at r =, but it doesn t do that anywhere ese. Since there are no charges inside the sphere, the potentia shoud at east be finite i.e., not divergent everywhere inside the sphere. This sounds ike I m stating the obvious, but as we see in a moment it turns out to be a usefu observation. What about where inside and outside meet? We know the potentia might behave differenty inside and outside the sphere, but it must be continuous at the surface of the sphere where the two regions overap. So our expressions for the potentia inside and outside the sphere shoud agree at the surface of the sphere: V in R, θ = V out R, θ. Finay, there is one other thing we know that wi be very important. The potentia is aways continuous, but not the eectric 3 This refers to a particuar choice of origin and orientation of our coordinates! In this probem the origin is at the center of the sphere, and the poar ange was defined as soon as the charge density was described as σ cos θ. 4 If you don t remember very much about Legendre poynomias, you shoud definitey read the reevant sections in Griffiths or your Math Methods book. Aso, it s reay important to remember that the Legendre poynomias that appear in this genera soution of Lapace s equation are functions of cos θ, and not θ.
fied. When a surface has a charge density on it, this wi cause a discontinuity in the component of the eectric fied norma to the surface: E out Ein = σ 5 ε surface In this case the surface is a sphere, so the direction norma to the surface is ˆr. The r-component of the eectric fied is V/, which means that the potentia has to satisfy 5 V out V in = σ cos θ. 6 r=r r=r ε So, et s summarize what we know about the potentia. Here are the four conditions we expect it to satisfy V out r, θ as r V in r, θ is finite inside the sphere 3 V in R, θ = V out R, θ 4 Vin Vout r = R = σ cos θ ε Let s go through these conditions one at a time and see how they take us from the genera soution 3 and 4 to the unique soution of this probem. The first condition tes us that V out shoud go to zero as r. If we are very far away from the sphere it shoud just ook ike a itte bit of charge. Since a point charge has a potentia proportiona to /r, the potentia for this charged spherica she shoud approach zero at east as fast as /r if we move very far away from it. If we ook at the genera form 4 for the potentia in the region r > R, we see that it contains terms proportiona to both positive and negative powers of r, as we as a constant term the = term. But if the non-negative powers of r were present then we woudn t have a potentia that goes to zero as r! Therefore, the ony way that 4 can be consistent with the requirement V out as r is if A out = for a vaues of. But be carefu! Since this is a statement about the potentia as r, it is ony teing us something about V out. That is, we have earned that A out =, but we have not earned anything about the coefficients and Bin that appear in V in. Now et s move on to the second condition, which tes us that the potentia is finite inside the sphere. The inside region corresponds to r < R, so we need to examine the genera soution of Lapace s equation to see if anything coud go wrong in this region. There are two types of terms in 3: terms proportiona to non-negative powers of r the r terms, and terms that invove negative powers of r the /r + terms. Do either of these do anything screwy in the region r < R? Yes: any negative powers of r woud diverge at r =. Since the potentia is finite inside the sphere, it must be that the specific V in we re after has no terms with negative powers of r. In other words, the coefficients of negative powers of r in 3 must be zero: B in =. At this point we ve used the first two conditions to narrow down a genera soution of Lapace s equation we re on our way to finding the unique soution we want. Soutions consistent with these 5 Remember that E and V are reated by E = V. The minus sign is important in 6! 3
first two conditions have the form V in r, θ = V out r, θ = r P cos θ 7 B out r + P cos θ. 8 To narrow things down further, et s ook at the third condition. The potentia must be continuous, so our expressions for the potentia inside and outside the sphere must be the same at r = R, where inside and outside overap. This means that 7 and 8 have to be equa at r = R R P cos θ = B out R + P cos θ. 9 This makes it seem ike the coefficients and Bout must be reated somehow. To get the exact reation, we wi use the fact that the Legendre poynomias P cos θ are orthogona on the interva θ π. Since we re trying to determine the potentia everywhere r <, θ π, ϕ < π, the equaity 9 has to be true over these same vaues of θ. If we move both series to one side of the equation we get R B out R + P cos θ =. Orthogonaity of the Legendre poynomias means that this can be true for a θ π if and ony if every term in the series is zero. 6 Since each term in has to be zero on its own, we see that R B out = R+ B out This is true for a vaues of, so for a the terms in the series 8. = R +. Important: Starting from 9, use the orthogonaity reation for Legendre poynomias to prove that is true. First mutipy both sides of 9 by P k cos θ sin θ where k is arbitrary and then use dθ sin θ P cos θ P k cos θ = + δ,k. 3 So now, after appying the first three conditions on the potentia, we have V in r, θ = V out r, θ = r P cos θ 4 R + r + P cos θ. 5 6 It s exacty the same as saying that A xˆx + A yŷ + A zẑ = can ony be true if the components A x, A y, and A z are a zero. Since ˆx, ŷ, and ẑ are orthogona, you can t have non-zero x, y, and z components somehow adding up to zero. 4
We re amost done! We started with two infinite series, with each series having two unspecified sets of coefficients in it. After appying our first three conditions we ve narrowed that down to one set of unknown coefficients. We ve gone from 4 unknowns to unknowns! To determine the remaining unknowns in 4 and 5 we wi use the fourth and fina condition on the potentia. We know that the eectric fied shoud be discontinuous at r = R since there is a surface charge there. Specificay, we have Vin V out = σ cos θ. 6 r=r ε This shoudn t be too hard to impement! First, et s take the r-derivatives of 4 and 5. V in r, θ V out r, θ = = r P cos θ 7 + Evauating these expressions at r = R and putting them in 6, we get R + r + P cos θ. 8 + R P cos θ = σ cos θ ε. 9 Now a that s eft is to use the orthogonaity reations for the Legendre poynomias to figure out what the unknown coefficients are. First we mutipy both sides of 9 by a factor of sin θ and an arbitrary Legendre poynomia P k cos θ + R P cos θ P k cos θ sin θ = σ ε P k cos θ cos θ sin θ. Next, integrate both sides of this equation with respect to θ, from to π + R dθ P cos θ P k cos θ sin θ = σ dθ P k cos θ cos θ sin θ. ε The integra on the eft-hand side of this equation is zero uness = k, in which case it is / k + So for the eft-hand side of we get + R dθ P cos θ P k cos θ sin θ = dθ P cos θ P k cos θ sin θ = k + δ,k. + R k + δ,k 3 = k Rk. 4 What about the right-hand side of? We ve seen this integra before, and we evauated it by noting that cos θ is the same thing as P cos θ. Then σ ε dθ P k cos θ cos θ sin θ = σ ε dθ P k cos θ P cos θ sin θ 5 = σ ε k + δ k,. 6 5
So comparing our resuts for the two sides of, we get k Rk = σ ε k + δ k,. 7 If k =, or k =, or k is equa to any integer other than, then the right-hand side is zero. This means that k is zero for a those vaues of k! But if k = then the right-hand side is non-zero, so we get a non-zero vaue for. We find = σ 3 ε 8 k = k. 9 That s it! We used our fourth condition on the potentia to determine the remaining unknowns in our expressions 4 and 5 for the potentia. When we re-write those expressions using our resuts for k, we get Putting everything together, our fina answer is V in r, θ = σ 3 ε r P cos θ 3 V out r, θ = σ 3 ε R 3 r P cos θ 3 σ r cos θ r R 3 ε V r, θ = σ R 3 3 ε r cos θ r R 3 There are a few fina points to make. First, notice that the expressions for the potentia inside and outside the sphere agree at r = R. We made a point of enforcing that with our third condition, but it s aways good to check we coud have made an agebra mistake somewhere aong the way. Now, in both those regions we were soving Lapace s equation. But reay we ve done more than that, because there is charge at r = R. So by joining our two soutions of Lapace s equation together in the right way at the right pace, we ve actuay found the correct soution of Poisson s equation for a spherica she with given surface charge density. Second, we didn t actuay use the specific form of the surface charge density unti we got to 9. If we had a spherica she with a different azimuthay symmetric surface charge density σθ we coud just jump right to the expressions 4 and 5 for the potentia, and determine the unknown coefficients as above, starting from the condition + R P cos θ = σθ ε. 33 Third, et s consider what V ooks ike far away from the sphere. We see from 3 that the potentia is proportiona to /r outside the sphere. Can that be right? This is just a bob of charge, so if we re far away it ooks ike a point. Shoudn t we get a /r potentia? Not so fast! Look at the tota charge on the sphere: π q tot = da σ = dϕ dθ R sin θ σ cos θ = π R dθ sin θ cos θ =. 34 6
To a first approximation the potentia outside the sphere does ook just ike a point charge, but it s a point charge with with zero charge! This just tes us that there is no /r term in the potentia. Finay and most importanty, we found a unique soution to our probem, as promised. How do we know? Because we started with 4 sets of unknowns, Bin, Aout, and B out and we finished with zero unknowns. Everything has been naied down, and there are no other possibiities. If we got to the end of the probem and sti had unknowns eft, it woud mean that we forgot some important condition on the potentia. 7