FdM 1 Electromagnetism University of Twente Department Applied Physics First-year course on Part I: Electricity: Slides © F.F.M. de Mul

FdM 2 Analysis:  long wire: [C/m] cylindrical symmetry E-field of a long wire dE y Symmetry  y-component only !! e r. e y = cos  r, e r and  are f (z) : Conclusion: E radial symmetry r dz erer dE Approach: z O Charge element: dQ=.dz   y z y x P yPyP Problem: E P in point P(0,y P,0)

FdM 3 Charge elements (1): rod, ring, disk LineRing, radius R Disk, radius R [C/m] O z dz dQ =.dz  dd R [C/m] dQ =.R.d  P Symmetry: If  homogeneous: dQ = .2  a.da  [C/m 2 ] dd  a r P dE dQ = .dA =  (a.d  ) da a  r ! 0 < a < R da

FdM 4 Charge elements (2): thin plate P Thin plate,  [C/m 2 ] x y z (1) zPzP dQ =  dA =  dx.dy dE r erer dE = dE x e x + dE y e y + dE z e z if plate  large : dE // e z (2) (2) if  = f (x) only: x Use result for  long wire: z dE r erer P zPzP with d = . dx dE in XZ- plane = dA = dx.dy, at (x,y) (1)

FdM 5 Charge elements (3): tube and rod Avoid using r for radius !! Tube, radius R  [C/m 2 ] dz dd dA = R. d . dz dQ =  R. d . dz Cross section: thin ring, radius R dz d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

FdM 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:

FdM 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  dd R.sin  u v Spherical surface element r  dd  dd r.sin  u v Spherical volume element w R

FdM 8 Flux  : Definition enen enen enen A S S S  1. Homogeneous vector field A    (e n, A) = 0  = 90 o Def.:  = c.A.S Choice: c  1  = 0  = A.S. cos   = (A.e n ) S 2. General vector field A For small surface elements dS: A and e n are constant

FdM 9 en’en’ dA’ Flux  E ’ through surface A’:  A’ R dA enen A Flux  E through sphere A: Gauss’ Law (1): derivation O q Charge q in O Result is independent of the shape of the surface

FdM 10 q needs not to be in O charge outside: no net flux more charges in A: Gauss’ Law (2): consequences R O q dA enen en’en’ dA’ A  A’ Result is independent of the shape of the surface Flux: Consequences:

FdM 11 Gauss-boxes (1): symmetry 1: symmetry present: everywhere 2+3: no symmetry: averaged E  over surface can be calculated only: A A A

FdM 12 Gauss boxes (2) : solid sphere A Solid sphere: radius R, charge q 1. Homogeneously distributed over volume (non-conducting) 2. Conducting: homogeneously distributed over surface 0 R r 0 E ~ 1/ r

FdM 13 Gauss boxes (3) : hollow spheres E R1R1 R2R2 r O O Non-conducting; homog. charge Q 2. Conducting; charge Q 3. Idem, with extra q in O Radii: R 1 < R

FdM 14 Gauss-boxes (4): at plates 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 Enclosed charge:  S Gauss: 2. E S =  S /  0 E = ½  /  0

FdM 15  long bar, homog. charge [C/m] z Gauss-boxes (5): around rods Convenient choices for surface A: E = 0 or E // dA or E  dA Non-zero contributions to from side wall only Enclosed charge: L Gauss: E. 2  r.L = L /  0 Take cylinder for Gauss box L r Symmetry: E radial direction

FdM 16 Divergence Theorem B A V x z y B dy dx dz P Outgoing flux: Right side: Left side: + total in y-direction: Total through the block : Integrate over volume V :

FdM 17 “Gauss” in differential form Divergence = “micro”-flux per unit of volume (m 3 ) : V dV Assume in V:  (x,y,z) [C/m 3 ] “Gauss” in differential form: Application:  (x,y,z) to be calculated from E (x,y,z)

FdM 18 Approach: Displace m to P’ from equilibrium by (dx,dy,dz); Calculate direction of Coulomb-force dF e ; From everywhere, dF e should point to P Result: for at least several (if not all ?) displacements: direction of dF e is outward from P. Conclusion: no stable equilibrium possible Question: 1.Can m be hold in equilibrium? 2.Can z P be changed by changing q (or the Q’s)? FeFe FgFg P q,m Q Q Q Q Z Mass m with charge q Electrostatic Crane (1)

FdM 19 Approach: Gauss sphere around P Electrostatic Crane (2) P q,m Q Q Q Q Z From Coulomb: No stable equilibrium possible. Question: Does “Gauss” provide a “simple” proof? For stable equilibrium: all forces on P should point inward sphere From “Gauss”: this is possible only when at P negative charge present !! Conclusion: no negative charge at P  no stable equilibrium

FdM 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)

FdM 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 erer r dl A B Q

FdM 22 Electric Potential (2): Reference Circulation-free field Q A B If A (= reference point)   : “Potential in B with reference in  : In general: dr erer r dl Q A B

FdM 23 E = - grad V A B A’ B’ y x V(x,y) dx dV dy xx+dx df f(x)

FdM 24 Vector fields: rotation-free A ~ r 0 = c A ~ r - 1 A ~ r - 2

FdM 25 Vector fields: non-zero rotation 2a v0v0 River flow If linear profile: X Y  r v  v J B

FdM 26 Who suffers greatest risk ? Earth surface Equipotential plane Head: r, q Earth: R, Q Sphere: Equipotential   Smallest head, greatest risk.

FdM 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

FdM 28 Charge opposing conductor: image charge inside conductor + Point Charge opposing Conductor + Opposing charges: dipolar field Field lines and equipotential surfaces

FdM 29 + d Q Cross section Point charge opposing flat conductor Question: determine  (r) [C/m 2 ] Applying image charge and Coulomb: E Gauss box: ring-shaped box with radii: r and r+dr r 0

FdM 30 Electric field at Interface E enen Surface charge:  [C/m 2 ] Thin Gauss box: E   - E   =  /  0 Rectangular loop: E //  = E // 

FdM 31 Energy of/in Charge Distribution Q2Q2 Q3Q3 Q4Q4 r 12 r 14 Q1Q1 Energy for creation = energy for destruction Total Energy released upon destruction : E tot r ij =r ji 2

FdM 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 :

FdM 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

FdM 34 Conductor: Field at Boundary Everywhere on A: E  dA Suppose charge density  [C/m 2 ] is f (position) A E=0 dA E EsEs EsEs E oc 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 dA E

FdM 35 Electrical Dipole (1): Far-field -q-q +q+q a Definition: Dipole moment: p = q a -q-q +q+q a r-r- r+r+ P Potential field (with respect to  ) : -q-q +q+q r-r- r+r+ r  a.cos  a Far-field approximation: P   erer p

FdM 36 Electrical Dipole (2): components E = E r e r + E  e  + E  e  E = - grad V r   P dd dd ee ee erer ds r ds  ds 

FdM 37 Electrical Dipole (3): field lines E = E r e r + E  e  ErEr EE  r erer ee Field lines perpendicular to equipotential planes

FdM 38  E ext 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 a.cos  E (  ) - E (0) =

FdM 39 Electrical Dipole (5): torque Dipole: p = q a Torque of a dipole in an homogeneous external field E ext a.cos   E ext Torque at angle  :  = p  E ext Electrical force F E on + and - charge |  | = 2 F E. ½ a. sin  = p E ext sin  Direction given by right-hand rule:

FdM 40 Polarization of a Dielectric dA Unpolarized molecule ( n mol / m 3 ) E ext = 0 V = s. dAdQ= n Qs.dA = n p.dA = P. dA = P .dA Internal polarization shows itself as bound surface charge E ext  0 s Separation of charges Dipole: p = q s s = 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

FdM 41 Volume polarization  Surface Charge z x y P = np 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  Approach: replace dz-integration with dr-integration dp = np.dv = np.dS .dz = P.dS .dr / cos  P.dS  = P.dS  Polarization = surface charges !! r r top r bottom z’ dS  dv=dz.dS  d d

FdM 42 Dielectric Displacement V0V0 +Q 0 -Q 0 1. Upon closing switch: Plates will be charged:  Q 0 E0E0 Between plates: no flow of charges, but E-field present Consider field as representing imaginairy flow D 0 of charges: D 0 =  0 E 0 = Q 0 / S [C/m 2 ] 2. Insert dielectricum, with V 0 = const. -Q-Q +Q+Q  E remains const.  extra charge  Q needed Now in dielectricum real flow of (bound polarization) charges  Q = P.S P V0V0 E E Total charge delivered by battery: Q = Q 0 +  Q Field Free Bound Charges

FdM 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 E1E1 E2E2 11 22 1 2 Gauss box (empty!): D 1.Acos  1 - D 2.Acos  2 =0  D 1  = D 2  Circuit: E 1.Lsin  1 - E 2.Lsin  2 =0  E 1// = E 2//

FdM 44 Electret: D and E not always parallel E D Electret P E = (D-P) /  0 D =  0 E + P. P D 0E0E

FdM 45 Capacitors: series and parallel C 1 and C 2 : same V C 1 and C 2 : same Q C0C0 ±Q0±Q0 C0C0 ±Q0±Q0 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 C1C1 C2C2 ±Q2±Q2 ±Q1±Q1 Series C2C2 C1C1 ±Q2±Q2 ±Q1±Q1 Parallel

FdM 46 Capacitors: spheres and cylinders Parallel Series Parallel Series