1 Review of Electrostatics

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1 1 Review of Electrostatics The first 11 sections you can omit 1.1 and 1.13) of this chapter in Jackson should only be revision; little is done there that you shouldn t have seen before presumably...). Therefore, I will assign it for you to read carefully, and will only go over the salient points here. 1.1 Electrostatic Field sections J1.1 J1.4) Coulomb s law gives the force exerted on a point-charge q by a point-charge q 1 : F = k e q 1 q x x 1 x x ) where x x 1 is the distance vector from the source charge q 1, located at position x 1, to q, located at x, with respect to an arbitrarily chosen origin. In I units, k e = 1/ǫ = N m /C ; in CG units, it is simply equal to 1. In the first eight chapters of these notes, whenever I deviate from writing the general k e, I will adopt I units as is done in the first ten chapters of Jackson. We also define the Coulomb electric field whose source isq as: Ex) = k e q x x x x 3 1.) Again, x x is the distance vector from the source at x to the observation or field) point x, where there may or may not exist a point-charge see fig. J1.1). o F = qe. The electric field obeys the principle of superposition. o, when the source is a collection of N point-charges q i sitting at their respective positions x i, the electric field at observation point x is: Ex) = k e N q i x x i x x i 3 This is easily extended to continuous source charge distributions. We only give the expression for a distribution extending over a : Ex) = k e ρx ) x x x x 3 d3 x 1.3) where ρ is the charge density in the distribution. Note that only Cartesian components can be used in the integral, because only then are the unit vectors inx x constant with respect to integration, so that they can be pulled out of the integral. The remaining three integrals can then be evaluated in any coordinate basis you wish. Note also that you can always write a discrete distribution of N point-charges at positions x i as: ρx) = q i δx x i ) The Coulomb field, as well as any electric field, satisfies Gauss law in differential form: E = k e ρ = ρ/ǫ in I units) 1.4) or, via the divergence theorem, in the equivalent integral form: E da = k e q in 1.5) where q in = ρx )d 3 x is the total net charge enclosed by the. The field of any static charge distribution can be written as a superposition of Coulomb fields. We call such fields electrostatic. 13

2 1. calar Potential sections J1.5 J1.7) Because it is a central field, the Coulomb field has vanishing curl. Therefore, by the superposition principle, all electrostatic fields satisfy E = everywhere. This is equivalent to saying that E dl = for any closed integration path. Because E = for any electrostatic field E, there exists associated with this vector field a scalar field Φx) such that E = Φ. From the fundamental theorem for gradients section.3): b a E dl = b a Φ) dl = Φa) Φb) 1.6) Therefore, the electrostatic potential at any pointp isφp) = P E dl, whereo is an arbitrary reference point, usually O chosen where we want the potential to vanish. The Coulomb potential of a point-charge q isφx) = k e q/ x x. The potentialφ, like the fielde, obeys the superposition principle. Then the potential of a localised charge distributionρ is: ρx ) Φx) = k e x x d3 x 1.7) where the reference point is taken at infinity, hence the restriction to localised distributions. Combining E = Φ with Gauss law yields the Poisson equation: In vacuum, this becomes the Laplace equation, Φ =. Φ = k e ρ 1.8) 1.3 Matching Conditions on Electrostatic E and Φ Let ˆn be a unit vector normal to the boundary of some, directed from the in region to the out region. The electrostatic-field component tangent to any is continuous across the : E out E in ) ˆn = 1.9) For conductors, on the other hand, the normal component is discontinuous when there is a local charge densityσ: E out E in ) ˆn = = k e σ 1.1) and the local field due toσ has magnitude πk e σ, so that the total field has to jump by k e σ across the. The potential is continuous across the boundary, but its gradient in the direction normal to the inherits the discontinuity ine: n Φ out n Φ in = k e σ 1.11) where the normal derivative, n Φ = Φ) ˆn, is evaluated at the. For non-conductors, we will have to be a bit careful see chapter 4 in Jackson) with the normal components. 1.4 Boundary-alue Problem with Green Functions; Uniqueness of a olution sections J1.9 J1.1) In realistic electrostatic problems, we rarely know the whole charge distribution that produces the potential observed at a point. What we most often do know and often can control) is the potential over specific s such as conductors. These s can enclose regions with or without charge inside. The proper approach then involves solving the Poisson or Laplace equation, subject to boundary conditions B.C.). There are two very important questions that must be answered: What kind of information do we need to supply in order to obtain a solution to the Poisson equation? If we find a solution, is it unique? 14

3 1.4.1 Uniqueness of the solution It is convenient to address the first question before the second: is a solution of the Poisson equation that satisfies boundary conditions B.C.) on a closed unique? uppose there exist two solutions,φ 1 andφ, of Poisson s equation that satisfy the same B.C., on a closed. Define Φ 3 Φ Φ 1. From Green s first identity eq..6) withf = g = Φ 3, we have: [Φ 3 Φ 3 + Φ 3 ) ]d 3 x Φ 3 n Φ 3 da The integral is zero because either Φ 3 = because we have specified the potential) or n Φ 3 = on the because we have specified the normal derivative). Also, Φ 3 = inside the. Therefore, Φ 3 ) d 3 x =, which can only be true if Φ 3 = inside the, so that Φ 3 is a constant there. Then, if Φ 3 = on the, it vanishes everywhere inside, whereas Φ 3 can be a non-zero constant inside if n Φ is specified on the. We conclude thatφ 1 = Φ inside up to a possible irrelevant additive constant), and that the electrostatic field is uniquely determined. The importance of this result cannot be exaggerated. It means that any function that satisfies Poisson s equation and the B.C. is the solution, no matter how it was found! It also means that if Φ is specified on the, its normal derivative is also determined and cannot be specified arbitrarily; if it is n Φ that is specified, then we lose almost all control over Φ which is determined up to a constant olution of boundary-balue problems with Green functions J1.8 1) To answer the first question, we introduce functions Gx,x ), called Green functions. A Green functiongfor a linear differential operatorlis defined as a solution of the equationlgx,x ) = δx x ), where x and x are two points in the manifold on which the functions are defined. Thus, in 3-dimensional space, the Green function for the Laplacian satisfies: Gx,x ) = δx x ) 1.1) Now takef = Φ Φ is the potential) and g = G in Green s second identity eq..7)): Φ G G Φ ) d 3 x Φ n G G n Φ ) da After using the Poisson equation in the second integral we obtain: [ Φx )δx x ) + k e ρx )Gx,x ) ] d 3 x = Φ n G G n Φ ) da If x lies outside the arbitrary, the first integral is zero. If x lies inside the, we can extend the first integral to cover all space without changing anything, and integrate. Rearranging then yields: Φx) = k e ρx )Gx,x )d 3 x + 1 G n Φ Φ n G ) da 1.13) where ρ is the charge density and is the boundary of. The normal derivatives are to be evaluated on before integrating. You may think that we have answered the first question: specifying ρ over a and Φ and its normal derivative as B.C. over the enclosing the gives the potential everywhere via the above equation. There is a subtle and important point here, however: as we have seen above, Φ and n Φ are not independent on the. o we are not free to specify them both arbitrarily as such values will in general be inconsistent. We may only specify one or the other! In that sense, our expression for Φ should not be considered a solution yet, but just an integral equation. After all, the integrals come from an identity. pecifyingφon the gives Dirichlet B.C., whereas specifying n Φ gives Neumann B.C. In the case of conducting s, the latter are equivalent to specifying the charge density σ = ǫ n Φ. How do we get a solution for Φ then? In principle, this is simple. To a particular solution of the defining equation 1.1), we can add any function Fx,x ) that satisfies the Laplace equation: Fx,x ) =, to obtain the most general Green function for the Laplacian. All we have to do is find af that eliminates one of the two integrals. 15

4 For instance, uppose we wish to specifyφfreely on the Dirichlet problem). Then we should haveg D x,x ) = x. The solution for Φ would then be: Φx) = k e ρx )G D x,x )d 3 x 1 Φx ) n G D x,x )da 1.14) This means that inside a charge-free region, the potential is determined once its values on the boundary of the region are specified. Note also that if the is all space, the integral at infinity vanishes ifφx) at least as fast as1/ x x since da R ). In that case we can take F = and, comparing eq. 1.1) and.18), G D x,x ) = 1/ x x because this already satisfiesg D = at infinity, and we end up with our earlier Coulomb expression: Φx) = k e d 3 x ρx )/ x x. imilar considerations apply with Neumann B.C., ie. when n Φ is known on the boundary. Then, if the is bounded by two s, one closed and finite and the other at infinity see Jackson p. 39), we have: Φx) = k e ρx )G N x,x )d 3 x + 1 G N x,x ) n Φx )da 1.15) o our first question has now been answered: supply the charge density inside a, together with Dirichlet or Neumann B.C. on the boundary of the ; then Φ inside the can be found. Incidentally, a useful property of Dirichlet Green functions for the Laplacian or any other differential operator) follows directly by taking f = Gx,x ) and g = Gx,x ) in Green s second identity, eq..7): [ Gx,x ) Gx,x ) Gx,x ) Gx,x ) ] d 3 x [ = Gx,x ) n Gx,x ) Gx,x ) n Gx,x ) ] da The right-hand side vanishes because G = for Dirichlet B.C. on the boundary. ince the is arbitrary, implementing the defining equation 1.1) for Green functions yields, : G D x,x) = G D x,x ) 1.16) This symmetry of Dirichlet Green functions in their two arguments corresponds to the interchangeability of source point x and observation point x. It provides a good check on candidate Green functions for a Dirichlet problem. 1.5 Electrostatic Energy section J1.11) Potential interaction) energy If a point-charge q is sitting at position x where there exists an electrostatic potential Φx), work had to be done to bring it there from infinity. This work is W = qφx), and is done by an external agent against the force associated with the potential acting on the charge q. is: We know that the potential produced at a pointx j where there is no charge) by a collection ofj 1 discrete point-charges j 1 q i Φx j ) = k e x i x j Then, to bring one extra charge, q j, to x j from infinity requires an energy q j Φx j ). By virtue of being located at x j in the potential of the other j 1 charges, q j now has a potential energy equal to the work done by the external agent to bring it there, namely q j Φx j ). o the total work needed to build a configuration of N charges, each at point x j, is: W = = 1 = 1 i<j i j q i k e q j x j x i i j k e q j q i x i x j q j Φx j ) 1.17) 16

5 where Φx j ) is the potential at x j from all charges other than the charge sitting there. By the same token, W is the potential energy of the distribution in other words, the energy required to assemble it from charges initially located at infinity. For a continuous distribution, this generalises to: W = 1 ρx)ρx ) x x d 3 xd 3 x = 1 ρx)φx)d 3 x 1.18) except that there is a subtle but very important difference with the potential energy of a discrete distribution. In eq. 1.17), the potential at the points where the charges are sitting does not contribute, whereas in eq. 1.18) the integral receives contributions from the potential at all points in the. The latter, but not the former, includes the self-energy of the charge distribution Energy density in an electric field In terms of the field, we have, in I units: W = 1 ρx)φx)d 3 x 1.19) = ǫ E)Φx)d 3 x = ǫ ) EΦ)d 3 x E Φd 3 x = ǫ EΦ) da + E d 3 x ǫ = E d 3 x 1.) Note that even if if the initial integral in effect includes only the region whereρ, starting on the third line, it must extend over all space so as to capture all the energy! In this form,w can be interpreted as the energy stored in the field, with ǫ E / the local energy density. Note that this is always positive, and that, therefore, the energy given by eq. 1.18) must also be positive, something which is not obvious there. As mentioned above, the apparent discrepancy with the energy of a discrete distribution, which can be negative, stems from the fact that the latter cannot include the self-energy of the charges. ee also the example involving two point-charges on p. 4 in Jackson. Because the electrostatic energy density is quadratic in the field, it does not obey the superposition principle; only the field does. To find the total energy density from more than one field contribution, you must first add the fields before using ǫ E /. Finally, we define the capacitance of a conductor as the amount of charge per unit potential that sits on it, when all other conductors are maintained at zero potential. If this latter condition is not met, the relationship between charge and potential, while still linear, is more complicated: Q i = C ij j where j is the potential on the j th conductor, and the C ij i j) are called induction coefficients. 17