3. Charge elements (1): rod, ring, disk Line Ring , radius R Disk , radius R [C/m] O z dz dQ = .dz d R [C/m] dQ = .R.d P Symmetry: If homogeneous: dQ = .2 a.da [C/m 2 ] d a r P dE dQ = .dA = (a.d ) da a r ! 0 < a < R da
4. Charge elements (2): thin plate dQ = dA = dx.dy if plate large : dE // e z P Thin plate , [C/m 2 ] x y z (1) z P dE r e r dE = dE x e x + dE y e y + dE z e z (2) (2) if = f (x) only: x Use result for long wire: z dE r e r P z P with d = . dx dE in XZ-plane = dA = dx.dy , at (x,y) (1)
5. Charge elements (3): tube and rod Avoid using r for radius !! Tube , radius R [C/m 2 ] dz d dA = R. d . dz dQ = R. d . dz Cross section: thin ring, radius R dz d Cross section: ring, radius a, width da Solid rod , radius R [C/m 3 ] dV = (r.d ).dr.dz dQ = (r.d ).dr.dz
6. Arc angle and Solid angle r 1 radian [rad] angle with arc length = radius r. 2 radians on circumference. 1 steradian [sr] solid angle with surface area = r 2 . There are 4 sr on the surface. d = (sin .d ) d r sin Solid angle:
7. Charge elements (4): on/in a sphere dA = u.v = (R.sin .d ).(Rd ) 0 < < 2 ; 0 < < dV = u.v.w = (r.sin .d ).(rd ).dr 0 < < 2 ; 0 < < 0 < r < R R d d R.sin u v Spherical surface element r d d r.sin u v Spherical volume element w R
8. Flux : Definition = 0 = A.S. cos = ( A.e n ) S e n e n e n A S S S 1. Homogeneous vector field A ( e n , A ) = 0 = 90 o Def.: = c.A.S Choice: c 1 2. General vector field A For small surface elements dS: A and e n are constant
9. Gauss’ Law (1): derivation Result is independent of the shape of the surface e n ’ dA’ Flux E ’ through surface A’: A’ R dA e n A Flux E through sphere A: O q Charge q in O
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11. Gauss-boxes (1): symmetry 1: symmetry present: everywhere 2+3: no symmetry: averaged E over surface can be calculated only: 1 2 3 A A A
12. Gauss boxes (2) : solid sphere 1. Homogeneously distributed over volume (non-conducting) 2. Conducting: homogeneously distributed over surface A Solid sphere: radius R, charge q 0 R r 0 E ~ 1/ r 2 1 2 1+2
13. Gauss boxes (3) : hollow spheres E R 1 R 2 r O O 2 3 1. Non-conducting; homog. charge Q 2. Conducting; charge Q 3. Idem, with extra q in O Radii: R 1 < R 2 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + 1 + + + + + + + + + + + + + - - - - - - - -
14. Gauss-boxes (4): at plates Enclosed charge: S Gauss: 2. E S = S / 0 E = ½ / 0 Convenient choices for surface A: E = 0 or E // dA or E dA large plate, homog. charge [C/m 2 ] x y z Symmetry: E // + e y resp. -e y S Take cylinder for Gauss box Non-zero contributions to from top and bottom only
15. Gauss-boxes (5): around rods Convenient choices for surface A: E = 0 or E // dA or E dA Enclosed charge: L Gauss: E. 2 r.L = L / 0 long bar, homog. charge [C/m] z Non-zero contributions to from side wall only Take cylinder for Gauss box L r Symmetry: E radial direction
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19. Electrostatic Crane (2) From Coulomb: No stable equilibrium possible. Question : Does “Gauss” provide a “ simple” proof? From “Gauss”: this is possible only when at P negative charge present !! Conclusion : no negative charge at P no stable equilibrium Approach : Gauss sphere around P P q,m Q Q Q Q Z For stable equilibrium: all forces on P should point inward sphere
20. Gradient in 2 dimensions P Arrows represent vectorial function: grad f Top view: Steepest slopes: P x y f (x,y) Scalar function: f (x,y)
21. Electric Potential (1): Definition Field force F E Work needed for transport of q, from A to B, “opposing field force” Work per Coulomb = potential difference Work is independent of path choice Conservative field E ~ e r / r 2 dr e r r dl A B Q
22. Electric Potential (2): Reference Circulation-free field If A (= reference point) : “ Potential in B with reference in : In general: Q A B dr e r r dl Q A B
23. E = - grad V A B A’ B’ y x V(x,y) dx dV dy x x+dx df f(x)
27. Particle transport and Flux Particles : density n per m 3 ; velocity v m/s (position dependent) Flux : nr. of particles passing per sec . All particles within v meter from A will pass within 1 sec. from now: n.v = flux density [part. m -2 s -1 ] Suppose all particles carry charge q : Charge flux = current: j = current density Flow tube : contains all particles going through dA dA A v
28. Point Charge opposing Conductor Opposing charges: dipolar field Field lines and equipotential surfaces Charge opposing conductor: image charge inside conductor + +
29. Point charge opposing flat conductor Question: determine (r) [C/m 2 ] Applying image charge and Coulomb: + d Q Cross section E Gauss box: ring-shaped box with radii: r and r+dr r 0
30. Electric field at Interface Surface charge: [C/m 2 ] Thin Gauss box: E - E = / 0 E e n Rectangular loop: E // = E //
31. Energy of/in Charge Distribution Energy for creation = energy for destruction Total Energy released upon destruction : E tot + .............................................. + 2 Q 2 Q 3 Q 4 r 12 r 14 Q 1 r ij =r ji
32. Energy to Charge a Capacitor Question : Energy to charge capacitor C up to end charge Q E Suppose: “Now” already charge Q present potential V=Q / C Adding charge q : does this cost extra energy q.V ? No, because during addition V will change! V = f (Q) Approach : add dQ ; then V remains constant This will cost energy: dE = V.dQ = (Q / C) . dQ In total, charge from 0 to Q E :
33. Change Spacing in a flat Capacitor Important variables: V (=const.) ; C, Q, E all f (s) Energy balance: net supplied energy = growth of field energy Supplied: mechanical and electrical (from battery) ~ A ; ~ s -2 ; ~ V 2 Question: Energy and force to change spacing s to s+ds F Suppose : capacitor connected to battery, potential V s V
34. Conductor: Field at Boundary Everywhere on A: E dA Suppose charge density [C/m 2 ] is f (position) Gauss: E . dA = dA / 0 E = / 0 Self-field of box dA: E s = / 2 0 Field from other charges outside box dA: E oc = / 2 0 A + + + + + + + + + + + + + + + + + + + + E =0 dA E E s E s E oc E oc A dA E
35. Electrical Dipole (1): Far-field Dipole moment: p = q a Potential field (with respect to ) : Far-field approximation: P - q + q a Definition: - q + q a r - r + P - q + q r - r + r a. cos a e r p
36. Electrical Dipole (2): components E = E r e r + E e + E e E = - grad V r P d d e e e r ds r ds ds
37. Electrical Dipole (3): field lines E = E r e r + E e E r E r e r e Field lines perpendicular to equipotential planes
38. Electrical Dipole (4): potential energy Potential energy of a dipole in an homogeneous external field E ext Dipole: p = q a Energetically most favourable orientation: = 0 Minimum energy level E (0). Work needed to rotate dipole from = 0 to : Set zero level: E (0) = - pE ext Potential energy: E ( )= - pE ext .cos E ( )= - p . E ext E ext a. cos E ( ) - E (0) =
39. Electrical Dipole (5): torque Dipole: p = q a Torque at angle : = p E ext | | = 2 F E . ½ a. sin = p E ext sin Direction given by right-hand rule: Torque of a dipole in an homogeneous external field E ext a. cos E ext Electrical force F E on + and - charge
40. Polarization of a Dielectric V = s . dA dQ= n Q s .dA = n p .dA = P . dA = P . dA Internal polarization shows itself as bound surface charge dA Unpolarized molecule ( n mol / m 3 ) E ext = 0 E ext 0 s Separation of charges Dipole: p = q s s = s + - s - s + s - dA Charge transport through internal surface element dA : dQ = N + Q - N - (-Q) = (N + + N - )Q = n.VQ , with V = vol s dA
41. Volume polarization Surface Charge dz Approach : replace dz- integration with dr -integration dp = np.dv = np.dS .dz = P.dS .dr / cos P.dS = P.dS Polarization = surface charges !! z x y P = n p Suppose : n dipoles/m 3 ; each dipole moment p Assume : Z-axis // p A Question : calculate V in A dz A dr r dr = dz. cos r r top r bottom z’ dS dv=dz.dS d d
42. Dielectric Displacement Consider field as representing imaginairy flow D 0 of charges: D 0 = 0 E 0 = Q 0 / S [C/m 2 ] Total charge delivered by battery: Q = Q 0 + Q V 0 +Q 0 -Q 0 1. Upon closing switch: Plates will be charged: Q 0 E 0 Between plates: no flow of charges, but E -field present 2. Insert dielectricum, with V 0 = const. - Q + Q E remains const. extra charge Q needed Now in dielectricum real flow of (bound polarization) charges Q = P.S P V 0 E E Field Free Bound Charges
43. E -field at interface Given : E 1 ; 1 ; 2 Question : Calculate E 2 Needed: “Interface-crossing relations”: Relation D and E: D = 0 r E E 1 E 2 1 2 1 2 Gauss box (empty!): D 1 .A cos 1 - D 2 .A cos 2 =0 D 1 = D 2 Circuit: E 1 .L sin 1 - E 2 .L sin 2 = 0 E 1 // = E 2 //
44. Electret: D and E not always parallel . E D Electret P - - - +++ E = ( D-P ) / 0 D = 0 E + P P D 0 E
45. Capacitors: series and parallel C 1 and C 2 : same V C 1 and C 2 : same Q Q 1 + Q 2 = Q 0 V 1 + V 2 = V 0 V 1 = V 2 = V 0 Q 1 = Q 2 = Q 0 C 0 = C 1 + C 2 C 0 ± Q 0 C 0 ± Q 0 C 1 C 2 ± Q 2 ± Q 1 Series C 2 C 1 ± Q 2 ± Q 1 Parallel