Scott Hughes 8 February 2005 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2005 Lecture 3: Electric field energy. Potential; Gradient. Gauss's law revisited; divergence. 3.1 Energy in the ¯eld Suppose we take our spherical shell of radius r, with charge per unit area σ = q=4¼r2, and squeeze it. How much work does it take to do this squeezing? To answer this, we need to know how much pressure is being exerted by the shell's electric ¯eld on itself. Pressure is force/area; σ is charge/area; force is charge times electric ¯eld. Hence, our intuitive guess is that the pressure should be Pguess = (electric ¯eld)(charge/area) = 4¼σ2 : (We have used the fact that the electric ¯eld just at the surface of the shell is 4¼σ.) This is almost right, but contains a serious flaw. To understand the flaw, note that the electric ¯eld we used here, E = 4¼σ, includes contributions from all of the charge on the sphere | including the charge that is being acted upon by the ¯eld. We must therefore be overcounting in some sense | a chunk of charge on the sphere cannot act upon itself. It's not too hard to ¯x this up. Imagine pulling a little circular disk out of the sphere: The correct formula for the pressure will be given by P = (electric ¯eld in hole)(charge/area on disk) = (electric ¯eld in hole)σ : We just need to ¯gure out the electric ¯eld in the hole. Fortunately, there's a clever argument that spares us the pain of a hideous integral: we know that the ¯eld of the hole plus the ¯eld of the disk gives us the ¯eld of the sphere: Ehole + Edisk = Esphere = 4¼σ Just outside the sphere = 0 Just inside the sphere. 23 What's Edisk? Fortunately, we only need to know this very close to the disk itself: from the calculations we did in Lecture 2, we have Edisk = +2¼σ Outer side of disk = ¡2¼σ Inner side of disk. We're now done: Ehole = 2¼σ (Pointing radially outward) which means that the pressure is P = 2¼σ2 : (Purcell motivates the factor of 1=2 di®erence by pointing out that one gets the pressure by averaging the ¯eld at r over its discontinuity | the ¯eld is 0 at radius r ¡ ², it is 4¼σ at r + ². This argument is totally equivalent, and is in fact based upon Purcell problem 1.29.) Now that we know the pressure it is easy to compute the work it takes to squeeze: dW = F dr = P 4¼r2 dr = 2¼σ2 dV : On the last line, we've used the fact that when we compress the shell by dr, we have squeezed out a volume dV = 4¼r2 dr. Now, using E = 4¼σ, we can write this formula in terms of the electric ¯eld: E2 dW = dV : 8¼ The quantity E2=8¼ is the energy density of the electric ¯eld itself. (In SI units, we would 2 have found ²0E =2.) It shows up in this work calculation because when we squeeze the shell we are creating new electric ¯eld: the shell from r ¡ dr to r suddenly contains ¯eld, whereas it did not before. We had to put dW = (energy density) dV amount of work into the system to make that ¯eld. 3.2 Electric potential di®erence I move a charge q around in an electric ¯eld E~ . The work done by me as I move this charge from position ~a to ~b is given by ~b Uba = F~me ¢ d~s Z~a ~b = ¡ F~q ¢ d~s Z~a ~b = ¡q E~ ¢ d~s : Z~a F~me is the force that I exert. By Newton's 3rd, this is equal and opposite to the force Fq that the electric ¯eld exerts on the charge. This is why we flip the sign going from line 1 to line 24 2. Make sure you understand this logic | that sign flip is an important part of stu® we're about to de¯ne! On the last line, we've expressed this in terms of the electric ¯eld. Recall that one of the things we learned about electric ¯elds is that they generate conservative forces. This means that this integral must be independent of the path we take from ~a to ~b. We'll return to this point later. Since the work we do is proportional to the charge, we might as well divide it out. Doing so, we de¯ne the electric potential di®erence between ~a and ~b: ~b Ába = ¡ E~ ¢ d~s : Z~a This function has a simple physical interpretation: it is the work that I must do, per unit charge, to move a charge from ~a to ~b. Its units are energy/charge: erg/esu in cgs, Joule/Coulomb in SI. Both of these combinations are given special names: the cgs unit of potential is the statvolt, the SI unit is the Volt. 3.3 Electric potential Let's say we ¯x the position ~a and consider the electric potential di®erence between ~a and a variety of di®erent points ~r. Since the point ~a is ¯xed, we can speak of the electric potential Á(~r) relative to the reference point ~a. The language we use here is very important | \potential" (as opposed to \potential di®erence") only makes sense provided we know what is the reference point. All potentials are then de¯ned relative to that reference point. In a large number of cases, it is very convenient to take the reference point ~a to be at in¯nity. (We'll discuss momentarily when it is not convenient to do so.) In this case, the electric potential at ~r is ~r Á(~r) = ¡ E~ ¢ d~s : Z1 Suppose E~ arises from a point charge q at the origin: ~r Á(~r) = ¡ E~ ¢ d~s 1 Z r q = ¡ dr 1 r2 q Z = : r Note that using this formula the potential di®erence has a very simple form: ~b Ába = ¡ E~ ¢ d~s : ~a q Z q = ¡ b a = Á(~b) ¡ Á(~a) : As the name suggests, the potential di®erence is indeed just the di®erence in the potential between the two points. 25 By invoking superposition, we can generalize these formulas very simply: Suppose we have N point charges. Using in¯nity as our reference point, the potential at some point P is given by N q Á(P ) = i ; ri Xi=1 where ri is the distance from P to the ith charge. We can further generalize this to continuous distributions: for a charge density ½ distributed through a volume V , ½ dV Á(P ) = ; ZV r where r is the distance from a volume element dV to the ¯eld point; for surface charge density σ distributed over an area A, σ dA Á(P ) = ; ZA r for a line charge density ¸ distributed over a length L, ¸ dl Á(P ) = : ZL r 3.4 An important caveat There is one massive caveat that must be borne in mind when using these formulas: they ONLY work if it is OK to set the reference point to in¯nity. That, in turn only works if the charge distribution itself is of ¯nite size. As a counterexample, consider an in¯nite line charge, with charge per unit length ¸. Using Gauss's law, it is simple to show that the electric ¯eld of this charge distribution is given by 2¸ E~ = r^ ; r where r^ points out from the line charge, and r is the distance to the line. Let's ¯rst work out the potential di®erence between two points r1 and r2 away from the line: r2 Á21 = ¡ E~ ¢ d~s Zr1 r2 dr = ¡2¸ Zr1 r = 2¸ ln (r1=r2) : Suppose we make r1 our reference point. Now, look what happens as we let this reference point approach in¯nity | the potential diverges! For a charge distribution like this, the ¯eld does not fall o® quickly enough as a function of distance for in¯nity to be a useful reference point. In cases like this, we need to pick some other reference point. In such a situation, it is crucial that you make clear what you have picked. 26 3.5 Energy of a charge distribution If we can use in¯nity as a reference point, then we can write down a formula for the energy of a charge distribution that is sometimes quite handy. From an earlier lecture, we worked out the formula 1 q q U = i j : 2 rij Xi Xj6=i A useful way to rearrange this formula is as follows: 1 qi U = qi 2 rij Xi Xj6=i 1 = q Á(r ) : 2 i i Xi On the second line, we have used the fact that the quantity under the second sum is just the electric potential of all the charges except qi, evaluated at qi's location. Now, we take a continuum limit: we put qi ! ½ dV and let the sum become an integral. We ¯nd 1 U = ½Á dV : 2 Z This formula is entirely equivalent to other formulas we worked out for the energy of a charge distribution. Bear in mind, though, that we have implicitly assumed that the potential goes to zero in¯nitely far from the charge distribution! 3.6 Field from the potential Consider the potential di®erence between the point ~r and ~r + ¢~r: ~r+¢~r ¢Á = ¡ E~ ¢ d~s Z~r ' ¡E~ (~r) ¢ ¢~r : Now, expand the vectors in Cartesian coordinates: ¢Á ' ¡(Exx^ + Eyy^ + Ezz^) ¢ (¢xx^ + ¢yy^ + ¢zz^) ' ¡Ex ¢x ¡ Ey ¢y ¡ Ez ¢z : Dividing, taking the limit ¢ ! 0, and invoking the de¯nition of the partial derivative, we ¯nd @Á @Á @Á E = ¡ ; E = ¡ ; E = ¡ ; x @x y @y z @z or @Á @Á @Á E~ = ¡x^ ¡ y^ ¡ z^ @x @y @z ´ ¡r~ Á : 27 On the last line we have de¯ned the gradient operator, @ @ @ r~ = x^ + y^ + z^ : @x @y @z The electric ¯eld is thus given by the gradient of the potential.