3. Calculating Electrostatic Potential

Transcription

1 3. Calculating Electrostatic Potential 3. Laplace s Equation 3. The Method of Images 3.3 Separation of Variables 3.4 Multipole Expansion

2 3.. Introduction The primary task of electrostatics is to study the interaction (force) of a given stationary charges. since F = q test E ˆ E = ( R ) d R ρ τ this integrals can be difficult (unless there is symmetry) E = E xˆ + E yˆ + E zˆ + x y z we usually calculate V = ( ) ρ dτ R E = V This integral is often too tough to handle analytically.

3 3.. In differential form V ρ = ε (Poisson s equation) to solve a differential eq. we need boundary conditions. In case of ρ =, Poisson s eq. reduces to Laplace s equation Or in general V = V V V + + = x y z The solutions of Laplace s eq are called harmonic function.

4 3.. Laplace s Equation in One Dimension d V = V = mx + b d x m, b are to be determined by B.C.s e.q., V ( x = ) = 4 m = V = x + 5 V ( x = 5) = b = 5 V=4 V= 5 x. V(x) is the average of V(x + R) and V(x -R), for any R: [ ] V ( x) = V ( x + R) + V ( x R) Laplace s equation tolerates no local maxima or minima.

5 3..3 Laplace s Equation in Two Dimensions A partial differential eq. : x There is no general solution. The solution will be given in 3.3. We discuss certain general properties for now.. The value of V at a point (x, y) is the average of those around the point. V ( x, y) = Vdl π R circle. V has no local maxima or minima; all extreme occur at the boundaries. V V + = y

6 3..4 Laplace s Equation in Three Dimensions The value of V at point P is the average value of V over a spherical surface of radius R centered at P: V ( p) 4π R sphere As a consequence, V can have no local maxima or minima, the extreme values of V must occur at the boundaries. Example: q V =, = r + R rrcosθ r r q Vave = [ r R rr cos θ ] R sinθdθ dϕ 4π R + = q r R rrcosθ π rr + The same for a collection of q by the superposition principle. = Vda q q = rr + = = r ( r R) ( r R) Vat the center of the sphere

7 3..5 Boundary Conditions and Uniqueness Theorems First uniqueness theorem : the solution to Laplace s equation in some region is uniquely determined, if the value of V is specified on all their surfaces; the outer boundary could be at infinity, where V is ordinarily taken to be zero. V a V b in D, V = mx + b a one end b other end V is uniquely determined by its value at the boundary.

8 Proof: Suppose V, V are solutions V = V = V = V? V3 = V V V3 V V = = at boundary V = V V3 = V 3 = everywhere at boundary. hence V = V everywhere

9 3..5 The first uniqueness theorem aslo applies to regions with charge. Proof. at boundary. ρ V = ε V3 = V V V = ε V3 = ρ ρ V V = ε + ε = V = V V = 3 V3 =, i. e., V = V Corollary : The potential in some region is uniquely determined if (a) the charge density throughout the region, and (b) the value of V on all boundaries, are specified. ρ

10 3..6 Conductors and the Second Uniqueness Theorem Second uniqueness theorem: In a region containing conductors and filled with a specified charge densityρ, the electric field is uniquely determined if the total charge on each conductor is given. (The region as a whole can be bounded by another conductor, or else unbounded.) Proof: Suppose both E and E are satisfied. E = ρ E ε = ρ ε and E da = Qi, E da = Q ε ε ith conducting surface outer bourndary ith conducting surface E da Q E da Q = tot, ε = ε outer bourndary i tot

11 3..6 define E3 = E E E3 = ( E ρ ρ E) = E E = ε ε = E da = E da E da = 3 for V 3 is a constant over each conducting surface ( V E ) = V ( E ) + E ( V ) = E volume E3 dτ = ( V3E3 ) dτ = V surface 3E3 da = V3 E 3 da = surface E = i. e., E = E 3

12 3. The Method of Images 3.. The Classical Image Problem 3.. The Induced Surface Charge 3..3 Force and Energy 3..4 Other Image Problems

13 3. The Method of Images A charge q head d above an infinite grounded plane: Boundary conditions: What is V (z>)?. V = when z =, since the plane is grounded. V far from the charge, ie. x + y + z >> d The first uniqueness theorem guarantees that there is only one solution. If we can get one by any means, that is the only answer.

14 3.. Trick : Forget the plane, consider another charge q at (,,-d), for this configuration. q q V ( x, y, z) = + x + y + z d x + y + z + d ( ) ( ) E= E zˆ, E = E = ( E = ) at z =, V = for z = z x y and for Has same boundary conditions as the original problem, so by the uniqueness theorem this is the solution for original problem for z> in the original problem but we only care z >, z < is not a concern

15 3.. The Induced Surface Charge ( x + y + ( z d ) ) ( r + d ) 3/ 3/ σ = ε q q σ = 4π z x + y + ( z d ) x + y + ( z + d ) σ ( x, y) = π σ ( r, z = ) = π qd qd V z z= z= total induced charge π qd Q = rdrdφ = q 3/ π ( r + d )

16 3..3 Force and Energy The charge q is attracted toward the plane. The force of attraction is q( q) q F = zˆ = zˆ d ( d ) ( d ) With point charges and no conducting plane, the energy is W = qivi p i= ( ) i ( d ) Eq.(.36) q q = ( q) ( q) d d + + ( d d ) q =

17 3..3 () For point charge q and the conducting plane at z = the energy is half of the energy given at above, because the field exist only at z,and is zero at z < ; that is W = q 4 ( d ) or d d q W = F dl = dz πε ( z) 4 q d q z πε = = ( d ) ( dl = dlzˆ )

18 In general for any stationary distribution of charge q q q q q 3 q 3 q d d q Find the force on charge q.

19 3..4 () Conducting sphere of radius R (, θ ) V r q = + q r r q q = + rrˆ asˆ rrˆ bsˆ q q = + a r r rˆ sˆ b rˆ sˆ r b Image charge r r ŝ at q = b = R a R q a = r + a racosθ = r + b rbcosθ

20 q q V ( R, θ ) = + = a R R rˆ sˆ b rˆ sˆ R b a R q R R = b q = b R b = a b R q = q = q R a Force F qq = = 4 ( a b) πε ( a R ) q Ra [Note : how about conducting circular cylinder?]

21 3.3 Separation of Variables 3.3. Fourier series and Fourier transform 3.3. Cartesian Coordinate 3.3. Spherical Coordinate

22 3.3. Fourier series and Fourier transform Basic set of unit vectors in a certain coordinate can express any vector uniquely in the space represented by the coordinate. e.g. V N = i= V, x V iˆ i = V xˆ + V Completeness: a set of function f n ( x) if f ( x ) = C f ( x ) x y n = yˆ + V n z n zˆ in 3D. Cartesian Coordinate. xˆ, yˆ, zˆ, V are unique because y Vz are orthogonal. iˆ ˆj = i i is complete. for any function f ( x) = j j Orthogonal: a set of functions is orthogonal b if f ( x ) f ( x ) dx = a n m =const for for n n = m m

23 3.3. () A complete and orthogonal set of functions forms a basic set of functions. e.g. π sin kx sin nx dx = π π cos kx cos nx dx = π π sin kx cos nx dx = π { if k n π if k = n { if k n π if k = n k, n N sin cos ( kx) = sin( kx) odd ( kx ) = cos( kx ) even sin(nx) is a basic set of functions for any odd function. cos(nx) is a basic set of functions for any even function. sin(nx) and cos(nx) are a basic set of functions for any functions.

24 3.3. (3) for any ( x) f, f ( ) ( x) f ( x) g x = odd ; h( x) ( x) = g( x) h( x) f + odd even [sinkx] [coskx] = f ( x) + f ( x) even Fourier transform and Fourier series ( ) = ( ) n + n (*) f x A sin nx B cosnx n= π An = f ( x ) s in ( n x ) d x π π π B n = f ( x ) c o s ( n x ) d x π π π B = f A π π ( x ) dx = } n =,,

25 3.3. (4) Proof π π (*) sin( ) π ( ) kx dx f x sin kxdx π π π n n= π = n π n = ( sin cos ) sin( ) = A nx + B nx kx dx n ( ) sin ( ) A sin nx kx dx = A π if k k π Ak = f ( x) sin( kx) dx π

26 3.3. (5) π π (*) cos( ) π ( ) cos( ) π π π n n= π = n π n= kx dx f x kx dx ( sin cos ) cos ( ) = A nx + B nx kx dx B cosnx coskx dx π Bn = f ( x) cos( nx) dx π π π B = f ( x) dx π π n ={ B π for k = B k π for k =,,

27 3.3. Cartesian Coordinate Use the method of separation of variables to solve the Laplace s eq. Example 3 V ( y V ( y V ( x V ( x = ) = = π ) = = ) = V ) ( y) Find the potential inside this slot? Laplace s eq. x V V + = y set V ( x, y) = X ( x) Y ( y) Y x X Y + X = y

28 3.3. () X X x f ( x) depedent x only + Y Y y g ( y) depedent y only = f and g are constant d X f ( x) = = C X dx g( y) d Y = = Y dy C set so C = = C k d X dx = k X C + C = d Y dy = k Y C, C are constant f ( x) + g( y) = kx kx X ( x) = Ae + Be, Y ( y) = C sin ky + Dcosky kx kx V ( x, y) = ( Ae + Be )( C sin ky + Dcos ky)

29 3.3. (3) B.C. (iv) V ( x ) A =, k > V ( x, y) B.C. (i) V ( y = ) = D = kx V ( x, y) = Ce sin ky = kx e ( C sin ky + Dcos ky) B.C. (ii) V ( y = π ) = sin kπ = k =,,3 The principle of superposition V ( x, y) = B.C. (iii) V ( x = ) = V ( y) V ( y ) = Cke k= kx sin ky π Ck = V ( y)sin ky dy π k= Ce kx sin ky A fourier series for odd function

30 3.3. (4) For V ( y V = C ) k = constant V π = sin ky dy π if k = even V = ( cos kπ ) = 4V kπ if k = odd kπ 4V k V sin y V ( x, y) = e sin ky = tan ( ) π k π sinh x k=,3,5,

31 3.4 Multipole Expansion 3.4. Approximate Potentials at Large distances 3.4. The Monopole and Dipole Terms Origin of Coordinates in Multipole Expansions The Electric Field of a Dipole

32 3.4. Approximate Potentials at Large distances Example 3. A dipole (Fig 3.7). Find the approximate potential at points far from the dipole. q q V ( p) = ( ) r+ r r d ± = r + ( ) rd cosθ ( d cos d = r ± θ + r ) r>> d 4r d r ( cos θ ) r d d θ ± r± r r r r d ( ) cosθ r r r + - qd cosθ V ( p) r ( cos ) ( cos θ )

33 3.4. Example 3. For an arbitrary localized charge distribution. Find a systematic expansion of the potential? V ( p) r = r r ρ dτ = r + r rr cosθ r r r r = r [ + ( ) ( )cos θ ] r r r r = r + = ( )( cos θ )

34 3.4.() r r r 8 6 = ( + ) = ( + + ) r r 3 r r 5 r 3 r 3 θ θ θ r r r 8 r r 6 r r = [ ( )( cos ) + ( ) ( cos ) ( ) ( cos ) + ] r r 3 r r r r r = [ + ( )cos θ + ( ) ( cos θ ) + ( ) ( cos θ cos θ ) + ] r n= ( r ) n = P n (cos θ ) ρ dτ r n ( ) n θ ρ τ n+ n= r 3 3 r r r V ( r) = ( r ) P (cos ) d = [ ρ dτ + r cos θ ρ dτ + ( r ) ( cos θ ) ρ dτ + ] Multipole expansion Monopole term Dipole term Quadrupole term

35 3.4. The Monopole and Dipole Terms monopol e dipole V mon ( p) = dominates if r >> A physical dipole is consist of a pair of equal and opposite charge, ±q p = qr qr = q( r r ) = qd Q r r r V ( p) = r cosθ ρ dτ V dip dip ( p) = = rˆ r ρ dτ p rˆ r p r ρ dτ q r n = = i= i i P + + r cosθ = rˆ r dipole moment (vector)

36 3.4.3 Origin of Coordinates in Multipole Expansions p = r ρ dτ = ( r d) ρ dτ = r ρ dτ d ρ dτ = p dq if Q = p = p

37 3.4.4 The Electric Field of a Dipole A pure dipole V dip P rˆ Pcosθ r r ( r, θ ) = = E E E V Pcosθ r = = r 3 r V Psinθ θ = = r θ 3 r V ϕ rsinθ ϕ P 3 r E ( r, θ ) = (cosθ rˆ + sin θ ˆ θ ) dip = =