Introduction to Electrostatics

Chapter 1 Introduction to Electrostatics Problem Set #1: 1.5, 1.7, 1.12 (Due Monday Feb. 11th) 1.1 Electric field Coulomb showed experimentally that for two point charges the force is -proportionaltoeachofthecharges, -actsinthedirectionalongthelinejoiningchargesand -fallsofasinversesquareofthedistance,i.e. In a fixed coordinate system (and in SI units), F = k q 1q 2 ˆr. (1.1) r2 F = 1 q 1 q 2 x 1 x 2 2 (x 1 x 2 ) x 1 x 2 = 1 q 1 q 2 x 1 x 2 3 (x 1 x 2 ). (1.2) It convenient to define an electric field of the second point charge as such that E 1 q 2 x 1 x 2 3 (x 1 x 2 ) (1.3) F = q 1 E. (1.4) Since E is a vector field it must have a certain direction and a magnitude in every point in space and time. One can always think of electric field as afunctionsofspatialx (in the first part of the course) and also temporal t (in the second part of the course) coordinates even if the dependence is often suppressed for brevity of notations. 4

CHAPTER 1. INTRODUCTION TO ELECTROSTATICS 5 If the system consists of N charges q 1,q 2...q N at positions x 1, x 1...x N, then from the additivity of forces we get additivity of electric fields E(x) = 1 N q i x x i 3 (x x i) (1.5) or in continuum limit E(x) = 1 ρ(x ) x x x x 3 d3 x, (1.6) where ρ(x) is charge density. For example, ρ(x) = q 0 δ (3) (x x 0 ) for a point charge N ρ(x) = q i δ (3) (x x i ) for a collection of point charges ρ(x) = λδ (2) (x x 0 ) for a line of charges ρ(x) = σδ (1) (x x 0 ) for a plane of charges Remind Yourself: Probability distributions, Generalized functions, Delta [not a] function... 1.2 Gauss s Law Take a point charge q and enclose arbitrary smooth surface around it. The electric field at the surface is E(x) = q 1 ˆr (1.7) r2 where the origin is chosen to be at the location the charge. If ˆn be a unit vector normal to the enclosing surface, then E ˆnda = q 1 ˆr ˆnda = q 1 cos(θ)da. (1.8) r2 r2 For the spherical coordinates r 2 dω=cos(θ) da where Ω is a solid angle and we obtain the Gauss s law for a point charge in integral form E ˆnda = q 1 ˆr ˆnda = q dω = q. (1.9) r2 ɛ 0

CHAPTER 1. INTRODUCTION TO ELECTROSTATICS 6 For a distribution of charges the Gauss s law is given by E ˆnda = 1 ρ(x)d 3 x. (1.10) ɛ 0 From the divergence theorem, i.e. A ˆnda = Ad 3 x (1.11) we get Ed 3 x = 1 ɛ 0 ρ(x)d 3 x, (1.12) which must be true for an arbitrary enclosing surface. Therefore the integrants must be equal, i.e. E = ρ(x) ɛ 0. (1.13) This the Gauss s law in differential form which is one of four Maxwell equations. For example, we can apply the Gauss law to a surface x y charge density at z =0,then top E 2 ẑ dx dy E 1 ẑ dx dy = 1 ɛ 0 or [(E 2 E 1 ) ẑ = σɛ0 ] σδ(z)dz (1.14). (1.15) z=0 Remind Yourself: Spherical coordinates, Change of variables, Jacobian matrix... 1.3 Scalar Potential It looks as if the electric field E(x) is a vector quantity, however in electrostatics (first part of the course) it is actually a gradient of ascalarquantity. This was shown by an experimental observation that the electric field has a vanishing curl, i.e. E =0. (1.16) But because ( Φ) = 0 (1.17)

CHAPTER 1. INTRODUCTION TO ELECTROSTATICS 7 it is tempting to define a scalar potential Φ such that E Φ. (1.18) Evidently the exact values of Φ cannot be observed, but the differences in Φ are relate to E which has a physical meaning. For example, the work done to move charge from point A to point Bin electric field is B B B W = F dl = q E dl = q Φ dl = q(φ B Φ A ) (1.19) A A A regardless of the path of integration. Thus, one can think of qφ as a potential energy of a charge in electric field. It should also be clear that the total work over a closed path is always zero. Then one can apply the Stoke s theorem, A dl = ( A) ˆnda (1.20) to confirm that E =0. If we apply the new definition of the electric potential to Coulomb s law, then or Φ(x) = 1 ρ(x) x x x x 3 d3 x = 1 Φ(x) = 1 where we also used the identity ρ(x) 1 x x d3 x (1.21) 1 ρ(x) x x d3 x + C (1.22) x x x x 3 = 1 x x. (1.23) If we apply the Gauss s Law (in differential form) then we arrive at the Poisson Equation 2 Φ= ρ ɛ 0 (1.24) or in the absence of charges at a Laplace equation 2 Φ=0. (1.25) Since 2 1 r = 1 1 r 2 r r2 =0 for r 0 (1.26) r r

CHAPTER 1. INTRODUCTION TO ELECTROSTATICS 8 and formally, 2 1 r d3 x = ( 1 ) ( ) 1 ˆn da = r 2 dω= 4π (1.27) r r r 2 1 x x = 4πδ(x x ) (1.28) is the Poisson equation of a unit point charge at x. Remind Yourself: Divergence theorem, Stokes theorem... 1.4 Green functions Using the divergence theorem we can derive very useful Green identities. Since (φ ψ) =φ 2 ψ + φ ψ (1.29) and φ ψ ˆn = φ ψ (1.30) (where is the normal derivative at the surface S directed outwards) then from the divergence theorem we obtain (φ 2 ψ + φ ψ)d 3 x = φ ψ ˆn da = φ ψ da. (1.31) By interchanging φ andψ andsubtractingthetwo equations fromoneanother we get ( (φ 2 ψ ψ 2 φ)d 3 x = φ ψ ψ φ ) da. (1.32) Let G(x, x )= 1 + F (x, x x x ) be a solution of the Poisson equation for unit charge where F (x, x ) is a solution of the Laplace equation. Then for an observation point at x, wecanchooseψ = G and φ =Φ,then ( 4πΦ(x )δ(x x )+ 1 ɛ 0 Gρ(x ) and thus, 1 ) d 3 x = ρ(x )Gd 3 x + 1 ( G 4π Φ Φ ) G da = ( Φ G G Φ { Φ(x) ) da. (1.33) for x V = S 0 for x / V = S. (1.34)

CHAPTER 1. INTRODUCTION TO ELECTROSTATICS 9 If the electric field falls off faster than 1/R and G 1/R, thenthe second integral vanishes and we are back to a familiar expression. If there is no charge inside surface S then the first integral vanishes and we get a solution for the potential in terms of boundary conditions at the surface. Looks as if this is a Cauchy boundary value problem, but it is actually not. 1.5 Dirichlet and Neumann boundary conditions To solve for the potential one must choose an appropriate boundary conditions. In particular there are two types of boundary conditions which eliminate one of the boundary integrals. For Dirichlet boundary conditions (i.e. when Φ(x) is known at the surface) we choose F such that G D =0at the surface and Φ(x) = 1 ρ(x )G D d 3 x 1 Φ 4π G Dda. (1.35) For Neumann boundary conditions (i.e. when Φ(x) is known at the surface) we choose F such that G N = 4π at the surface and S Φ(x) = Φ S + 1 ρ(x )Gd 3 x 1 G N 4π Φda, (1.36) where Φ S is an average value of the potential at the surface and can usually made to vanish with surface at infinity. One can also prove (by contradiction) that specification of a given boundary conditions (i.e. Neumann or Dirichlet) leads to a unique solution. Let us assume that there are two solutions Φ 1 and Φ 2 of the Poisson equation. Then U =Φ 1 Φ 2 must satisfy a Laplace equation 2 U =0 (1.37) and U =0or U =0of the surface for Dirichlet or Neumann boundary conditions respectively. From Green s first identity (U 2 U + U U)d 3 x = U U da (1.38)

CHAPTER 1. INTRODUCTION TO ELECTROSTATICS 10 or U 2 d 3 x =0. (1.39) Therefore U = const and a given Dirichlet or Neumann boundary condition uniquely specify a solution up to an additive constant. This also mean that acauchyboundaryvalueproblemwouldbeillposedingeneral. It should now be clear that in order to find a solution for a given problem in electrostatics all that we have to do it so find an appropriate Green s function. 1.6 Energy density We have already argued that the total potential energy should be W = qφ. If we are to bring an additional charge to a collection of n 1 charges, then the work that needs to be done is W = q n Φ= q n n 1 q i x i x j. (1.40) If we are to assemble from scratch a charge distribution made out of multiple charges the total potential energy would be W = q i q j x i x j = 1 2 j<i q i q j x i x j (1.41) or continuum limit W = 1 8πɛ 0 ρ(x)ρ(x ) x x d3 x d 2 x = ɛ 0 2 Φ(x) 2 Φ(x)d 3 x. (1.42) After integration by parts (and neglecting the boundary term) we get W = ɛ 0 Φ 2 dx = ɛ 0 E 2 dx. (1.43) 2 2 Then we can identify the integrant with energy density w = ɛ 0 2 E2 (1.44) Note that the energy density is always positive which is due to self-energy contribution.

CHAPTER 1. INTRODUCTION TO ELECTROSTATICS 11 Consider a collection of perfect conductors. The potential of i th conductor can be written as V i = p ij Q j (1.45) where p ij depends only on the geometry of the conductors. We can invert the equations to obtain Q i = C ij V j (1.46) where C ii are called capacitances and C ij (for i j) arecalledcoefficientsof induction. Then the potential energy of the system of conductors is or W = 1 2 W = 1 2 Q i V i = 1 2 Q i V i = 1 2 C ij V i V j (1.47) p ij Q i Q j (1.48) Note that variations of W with respect to V i s or Q i s can be used to obtain C ij or p ij.