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ELECTRODYNAMICS—lecture notes second semester 2004 Ora Entin-Wohlman References: 1. J. D. Jackson, “Classical Electrodynamics”, Wiley. 2. G. B. Arfken, "Mathematical methods for Physicists", Academic Press. 1 Summary of vector analysis r r ∂F ∂F ∂F ∇F = xˆ + yˆ + zˆ 1. Gradient, ∇ , ∂z ∂x ∂y r r a. ∇F (r ) is perpendicular to the surface of constant F that r the point r . r r r dF b. The gradient of F (r ) : ∇F = r dr includes r r ∂Vx ∂V y ∂Vz r ∇ ⋅V = + + ∇ ⋅ , 2. Divergence, ∂x ∂y ∂z r r a. ∇ ⋅ V is the flux per unit volume through an infinitesimal closed surface, r r lim 1 r ∇ ⋅V = V ⋅ nˆ da v → 0 v ∫S xˆ r r ∂ r ∇ × V = 3. Curl (rotor), ∇ × , ∂x Vx yˆ ∂ ∂y Vy zˆ ∂ ∂z Vz a. The n̂ component of the rotor is related to its circulation by r r lim 1 r r nˆ ⋅ ∇ × V = V ⋅ dl S → 0 S C∫ where n̂ is normal to the plane where C lies. r r Exercise: Given the vector V = r F (r ) , find F (r ) so that r r r r ∇ ⋅V = ∇ ×V = 0 . ∂2F ∂2F ∂2F + + 2 4. Div.grad=Laplacian, ∇ , ∇ F = ∂x 2 ∂y 2 ∂z 2 2 2 r r 5. Rot × grad =0, ∇ × ∇F = 0 r r r 6. Div.rot=0, ∇ ⋅ ∇ × V = 0 r r r r r r r 2 7. Rot × rot, ∇ × (∇ × V ) = ∇(∇ ⋅ V ) − ∇ V 8. Gauss’s theorem r rr r ∫ V ⋅ nˆda = ∫ dr ∇ ⋅ V S V Important consequence: r rr r 1 r r r ⎧− 4π ∫ dr ∇ ⋅ ∇ r = − ∫ dr ∇ ⋅ r 3 = ⎨⎩ 0 r r r r Exercise: show that when V = A × P, where A is a constant vector, Gauss’s r rr r ˆ n × P da = d r ∇ × P . Hint: use theorem gives ∫ ∫ S V r r r r r r r r r r r r r r r nˆ ⋅ A × P = −nˆ ⋅ P × A = −nˆ × P ⋅ A and ∇ ⋅ A × P = P ⋅ ∇ × A − A ⋅ ∇ × P . 9. Stoke’s theorem r r r r V ⋅ d l = ∇ ∫ ∫ × V ⋅ nˆda S r r r Exercise: show that when V = AF , where A is a constant vector, r r ˆ n × ∇ Fda = Fd l . Hint: use Stoke’s theorem gives ∫ ∫ S r r r r r r ∇ × ( AF ) = F∇ × A + (∇F ) × A . 3 Electrostatics 1. Coulomb’s law r r r The electric field, E (r ) , at point r , due to a collection of point-charges, r qi , located at ri , is r r n r r r − ri E (r ) = ∑ qi r r 3 | r − ri | i =1 When those charges can be described as a charge-density (of dimension r charge per unit volume), ρ (r ' ) , then r r r r r r r − r' E (r ) = ∫ dr ' ρ (r ' ) r r 3 | r − r '| The two expressions coincide for n r r r ρ (r ) = ∑ qi δ (r − ri ) i =1 The delta-function Dirac delta-function obeys the following in one-dimension: δ ( x − a) = 0, x ≠ a a+ ∫ dxδ ( x − a) = 1 a− and has the properties 4 ∫ dxf ( x)δ ( x − a) = f (a) δ ( f ( x) ) = ∑ i 1 df dx ∫ dxf ( x)δ ' ( x) = − f ' (a) δ ( x − xi ) xi ∞ Exercise: calculate ∫ dxf ( x )δ ( ax + b) = −∞ 1 b f (− ) . |a| a In more than one dimension, the delta-function becomes a product. In Cartesian coordinates r r δ (r − R) = δ ( x − X )δ ( y − Y )δ ( z − Z ) In spherical coordinates r r δ (r − r ' ) = Exercise: calculate r r 1 δ (r − r ' )δ (cosθ − cosθ ' )δ (ϕ − ϕ ' ) r2 r r r dxdyF ( r ) δ ( A r + B ), where A is a two by two ∫ matrix, B and r are two-dimensional vectors. Written explicitly, we have r dxdyF ( r )δ ( A11 x + A12 y + Bx )δ ( A21 x + A22 y + B y ) . ∫ r r r r −1 Hence, Ar = r ' and A r ' = r . It remains to note that ∂x ∂x ' dxdy = ∂y ∂x ' ∂x dx ' dy ' r ∂y ' 1 −1 dx ' dy ' = F (− A B ) . ∂y | det A | , to obtain | det A | ∂y ' 5 Examples: 1. Gaussian representation of a one-dimensional delta-function, the limit α → 0 of ⎡ ⎛ x ⎞2 ⎤ δ ( x) = exp ⎢− ⎜ ⎟ ⎥ α π ⎣⎢ ⎝ α ⎠ ⎦⎥ 1 2. Gaussian representation of a three-dimensional delta-function, the limit α → 0 of ⎛ 1 ⎞ δ (r ) = ⎜ ⎟ ⎝ 2π ⎠ 3/ 2 ( ) ⎡ 1 ⎤ exp x2 + y2 + z2 ⎥ − 3 2 ⎢ α ⎣ 2α ⎦ 1 3. Other one-dimensional delta-functions: (a) Lorentzian--the limit α → 0 1 1 ; of δ ( x) = απ 1 + ( x / α ) 2 (b) of sin( x / α ) 1 δ ( x) = = πx 2π 1/ α ixt dte ∫ −1 / α (c) and of δ ( x) = 1 α e −( x / α ) , x > 0 4. From Gauss’s theorem, δ (r ) = − 1 2⎛ 1 ⎞ ∇ ⎜ ⎟ 4π ⎝r⎠ 6 2. Gauss’s law n r r r ˆ E ⋅ n da = 4 π q = 4 π d r ∑ i ∫ ∫ ρ (r ) i =1 S S V n V q E r r r r r r r r r r − r' ∇ ⋅ E ( r ) = ∫ dr ' ρ (r ' )∇ ⋅ r r 3 = 4πρ ( r ) | r − r '| r r r ∇ × E (r ) = 0 explanation: r rr r ∇ ⋅ 3 = 4πδ ( r ) r and r rr ∇× 3 = 0 r Example: Calculate the electric field of an infinite cylinder, of radius a, charged with a constant charge density λ per unit length. We use cylindrical coordinates, ρ , ϕ , z . By symmetry, the field is only along the ρ̂ -direction. Consider a volume of length L around the cylinder. By Gauss’s law, ⎛Q⎞ ⎛λ⎞ EL 2πρ = 4π ⎜ ⎟ Lπρ 2 = 4π ⎜ ⎟ Lπρ 2 , ρ < a , ⎝ SL ⎠ ⎝S⎠ ⎛Q⎞ ⎛λ⎞ EL 2πρ = 4π ⎜ ⎟ Lπa 2 = 4π ⎜ ⎟ Lπa 2 , ρ > a , ⎝ SL ⎠ ⎝S⎠ giving r 2λ E ( ρ ) = ρˆ 2 ρ , ρ < a , a 7 r 2λ E ( ρ ) = ρˆ , ρ > a. ρ 3. scalar potential r r r r ρ (rr ' ) r r E (r ) = −∇ ∫ dr ' r r ≡ −∇Φ (r ), | r − r '| r r r ρ (r ' ) Φ (r ) = ∫ dr ' r r | r − r '| r r1 r r1 r rr r r 2 1 = −∇ ⋅ 3 = −4πδ (r ) since ∇ = − 3 ⇒ ∇ ⋅ ∇ ≡ ∇ r r r r r r r ∫ E ⋅ dl = −(Φ B − Φ A ), B Thus: A (work per charge) r r E ∫ ⋅ dl = 0 Exercise: Find the potential of a uniformly charged straight wire of length 2l . The charge per unit length is λ . Hint: ∫ dx 1 + x2 = ln( x + 1 + x 2 ) . Let the wire be along the z-axis, such that the x-y plane cuts it at its middle point. Then for points on the x-y plane we have l Φ( ρ ) = λ ∫ −l dz ' ρ + z' 2 2 = λ ln l + l2 + ρ 2 − l + l2 + ρ 2 For points in the x-z plane 8 l Φ ( x,0, z ) = λ ∫ −l dz ' ρ + ( z − z' ) 2 2 = λ ln z + l + ( z + l) 2 + x 2 z − l + ( z − l) 2 + x 2 By symmetry, the potential at any point is Φ ( x, y, z ) = λ ln z + l + ( z + l) 2 + x 2 + y 2 z − l + ( z − l) 2 + x 2 + y 2 . The equi-potential surfaces are ellipsoids. Consider points on the x-z plane. An ellipse there is given by x2 z 2 + =1 a2 c2 The sum of the distances from the points (0, l) and (0,−l) should be constant for points on the ellipse. Namely, z2 z2 2 2 a (1 − 2 ) + ( z − l) + a (1 − 2 ) + ( z + l) 2 = const . c c 2 For z=0 this gives 2 a + l and for z=c (namely, x=0) it gives 2c. 2 2 2 2 2 Therefore, c = a + l . Inserting into the result for Φ ( x,0, z ) , we find that the equi-potential curves are Φ 0 = λ ln c+l . c−l 4. surface charge distributions n E2 r σ (r ) E1 9 ( ) r r E 2 − E1 ⋅ nˆ = 4πσ discontinuity in the normal component of the electric field. The tangential components are continuous. Exercise: The electric field of a uniformly charged (infinite) plane, of charge σ per unit area. By symmetry, (for a plane perpendicular to the z-axis), r r ⎧ zˆE ( z ), E (r ) = ⎨ ⎩− zˆE (| z |), z > 0, z < 0. Then, applying Gauss’s law for a cylinder along the z-direction, the electric field on each side of the plane is discontinuity as given above. E ( z ) = 2πσ ,with the 5. Poisson and Laplace equations When there is charge density, the potential satisfies the Poisson equation: ∇ 2 Φ = −4πρ . Otherwise, it satisfies the Laplace equation: ∇ 2Φ = 0 . Example: Let us find the charge distribution giving rise to the scalar potential e −α r ⎛ α r ⎞ Φ (r ) = q ⎜1 + ⎟ r ⎝ 2 ⎠ Let us the Laplacian of this potential (in Cartesian coordinates) 10 ∂r ∂Φ ∂ x = q F (r ), Φ(r ) = ∂x ∂r ∂x r α α2 ⎞ −αr ⎛ 1 ⎟⎟ F (r ) = −e ⎜⎜ 2 + + r r 2 ⎝ ⎠ ⎛ F (r ) ⎛ x ⎞ 2 ∂ ⎞ ∂2 ∂ x ⎜ ⎟ Φ = = + ( r ) q F ( r ) q F ( r ) ⎜ ⎟ ⎜ r ⎟ ∂x 2 ∂x r ∂ r r ⎝ ⎠ ⎝ ⎠ Therefore, ∇ 2 Φ(r ) = q −αr 3 e α and 2 ρ ( r ) = − 2πqe −α r α 3 6. Electrostatic potential energy The potential energy of a collection of point charges is: qi q j 1 W = ∑∑ r r 2 i j | ri − r j | The potential energy of continuous charge distribution is: r v r r ρ (r ' ) ρ (r ) 1 r r r r r r 1 1 W = ∫ ∫ dr ' dr r r = ∫ dr ρ (r )Φ(r ) = − ∫ dr Φ(r )∇ 2 Φ(r ) 2 | r − r '| 2 8π W= r r r 2 1 r 2 1 d r r d r E ∇ Φ = | ( ) | 8π ∫ 8π ∫ Exercise: Find the electrostatic energy of a uniformly charged sphere of radius a. Let us first find the electric field, using Gauss’s law. By r r symmetry, E ( r ) = rˆE (r ). Then, 11 ⎧ 4πa 3 r>a ⎪⎪ 4π 3 ρ = 4πQ, 2 E ( r ) 4πr = ⎨ 3 3 ⎪4π 4πr ρ = 4πQ r , r < a ⎪⎩ 3 a3 The energy is then ∞ a 2 ⎤ 3 Q2 4π 2 ⎡ 2 r 2 1 W= Q ⎢ ∫ drr 6 + ∫ drr 4 ⎥ = 8π a r ⎦ 5 a a ⎣0 7. Conductors a. b. c. d. The The The The electric field is zero inside a conductor (at equilibrium). potential inside a conductor is constant. charge density inside a conductor is zero. electric field on the surface of a conductor satisfies: where r Et = 0 En = 4πσ , σ is the charge density on the surface. Example: the electric field of a conducting sphere, of radius a, carrying r r E (r ) = rˆE (r ) . The Laplace equation is then a charge q. By symmetry, (everywhere, except on the surface) ∇ 2Φ = 1 d ⎛ 2 dΦ ⎞ ⎜r ⎟ = 0, r 2 dr ⎝ dr ⎠ with Φ ( a ) = Φ 0 , Φ ( ∞ ) = 0. 12 The solution of this Laplace equation is in general Φ(r ) = A +B r The boundary conditions require A = aΦ 0 , B = 0. Thus the potential is a a Φ (r ) = Φ 0 , and the electric field is E (r ) = 2 Φ 0 , from which we find r r q q Φ 0 = 4πσa = 4π a = . 4πa 2 a Example: a grounded sphere, of radius a, in the presence of an external r r field, E (r ) = E0 zˆ . The boundary conditions on the potential are then Φ (a, θ ) = 0, Φ (r → ∞, θ ) = − E0 z = − E0 r cos θ . The Laplace equation is then 1 ∂2 1 ∂ ⎛ ∂Φ ⎞ ( ) sin r θ ∇ Φ (r , θ ) = Φ + ⎟ =0. ⎜ r ∂r 2 r 2 sin θ ∂θ ⎝ ∂θ ⎠ 2 1 ∂2 1 ∂ ⎛ 2 ∂Φ ⎞ (rΦ) = 2 [Note: ⎟ .] The solution which includes just terms ⎜r 2 r ∂r r ∂r ⎝ ∂r ⎠ linear in cosθ is Φ (r ,θ ) = A C cos θ +B+ + Dr cos θ . r r2 [To check this, it is convenient to put µ = cosθ .] B is just a constant, so we take it to be zero. Also, D = − E0 . In addition, A=0 because the electric field for cosθ = 0 [the plane z=0] has opposite signs above and below the plane z=0. Finally, 13 Φ ( a, θ ) = leading to C cos θ − E0 a cos θ = 0, a2 C = E0 a 3 . The solution for the potential is therefore a 3 cos θ Φ ( r , θ ) = E0 − E0 r cos θ , r2 which yields for the electric field ⎛ 2a 3 ⎞ Er (r , θ ) = E0 ⎜⎜1 + 3 ⎟⎟ cos θ r ⎠ ⎝ ⎛ a3 ⎞ Eθ (r , θ ) = − E0 ⎜⎜1 − 3 ⎟⎟ sin θ ⎝ r ⎠ The surface charge density on the sphere is σ (θ ) = E0 and the total surface charge is 3 cos θ , 4π π ∫ dθ sin θσ (θ ) = 0 . 0 This leads to a dipole-like behavior: the second terms in the result for the electric field can be put in the form with r r r 3rˆ( p ⋅ rˆ) − p E (r ,θ ) = , r3 r p = E0 a 3 rˆ cos θ − θˆ sin θ = E0 a 3 zˆ , since ( ) rˆ = xˆ sin θ cos ϕ + yˆ sin θ sin ϕ + zˆ cos θ , θˆ = xˆ cos θ cos ϕ + yˆ cos θ sin ϕ − zˆ sin θ , ϕˆ = − xˆ sin θ sin ϕ + yˆ sin θ cos ϕ . 14