UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede LECTURE NOTES 9 AC Electromagnetic Fields Associated with a Parallel-Plate Capacitor Let’s investigate the nature of AC electromagnetic fields associated with a parallel-plate capacitor, e.g. with circular plates of radius a separated by a small distance d a as shown in the figure below – we will neglect edge effects here: 3-D View: V V z d V z 0 Vo At DC (f = 0 Hz), we know the static solution to this problem, namely that the {free} charge Qfree on the capacitor is related to the potential difference V across the capacitor’s plates by: Q free C V where the capacitance of the capacitor is: C o A d (Farads) for d a ; the area of one plate of the parallel plate capacitor is A a 2 . Since there is no free electric charge between the plates of the parallel plate capacitor, then for d a , the solution to Laplace’s Equation 2V r 0 {derived from Gauss’ Law E r free r o 0 , with E r V r } yields: V V z d V z 0 z d z 0 E r d But: E r Eo zˆ between the plates of the parallel plate capacitor for d a V V z d V z 0 Vo 0 Vo Eo d E r Vo d zˆ free o nˆ1 Side View: where: free Q free A free z d Q free free z 0 Q free And: A A V C V o AVo o o o Eo d A Ad AV V C V o o o o o Eo A Ad d © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 1 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede We ask: What happens when we slowly raise the frequency from f = 0 Hz (static E-field) to f > 0? e.g. Apply a sinusoidally time-varying potential difference across the plates of the capacitor of the form: V t Vo eit Vo cos t i sin t single frequency, f 2 e.g. using a sine-wave function generator , as shown in the figure below: Function Generator V t Vo eit ˆ E r t z zˆ Eo eit zˆ with: Eo Vo d , For d a : d d The potential difference V t and electric field E t vs. time t: E t E t Eo eit Maxwell’s Equations must be obeyed in the gap-region between the parallel plates of capacitor, where: free r , t bound r , t 0 and: J free r , t J bound r , t 0 : E r , t 0 2) No magnetic charges / monopoles : B r , t 0 Maxwell’s Displacement Current: E r , t B r , t t 3) Faraday’s Law: 4) Ampere’s Law: B r , t o o E r , t t o J D r , t where: J D r , t o E r , t t 1) Coulomb’s Law: 2 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede Ampere’s Law (with Maxwell’s Displacement Current) in integral form tells us that: B r , t da S B r , t d o S C J r , t da n.b. not a closed surface! S o D 1 B r t d 2 , C c t J D r , t da o o E r , t t da E r, t t da S S Using Stokes’ Theorem where: c 2 1 o o Let us consider a contour path of integration C1 enclosing the surface S1 as shown in the figure below: da danˆ dazˆ da d d (in cylindrical coordinates) E r , t Eo eit zˆ and: 1 B r , t d c 2 C1 t E r, t t da S1 Note that: B r , t B , t ˆ due to the circular/azimuthal symmetry associated with this problem. d d ˆ dˆ , and da danˆ dazˆ by the right-hand rule, da d d zˆ , da d d in cylindrical coordinates, thus B d , and E da , and: 1 1 2 E t E t B , t 2 2 E t 2 B , t 2 ˆ 2 ˆ c t c 2 t 2c t E t i Eo eit i E t But: E t Eo eit t i i B , t 2 E t ˆ 2 Eo eitˆ Bo eitˆ 2c 2c Bo i B , t Bo eitˆ 2 Eo eitˆ i 2 Eo eitˆ 2c 2c n.b. B , t also oscillates sinusoidally like E t but is 90o out-of-phase with E t . © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 3 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede Note also that B , t is linearly proportional to ρ (the radial distance from the axis of capacitor) and that B 0 0 at the center of the capacitor: NOTE: B a 0 because E a 0 for d a in our model of this capacitor. Thus, we see that for ω > 0, (i.e. there exists) an azimuthal, ρ-dependent and time-varying i Eo it e ˆ in the gap region of the parallel-plate capacitor, magnetic field B , t 2 2c for d a . Note also that the azimuthal magnetic field is also linearly proportional to 2 f , thus as the frequency increases, this magnetic field also increases in strength. Note that for ω = 0, B 0 as we obtained for the static limit case! Furthermore, because the capacitor now has a non-zero magnetic field associated with it, for ω > 0, the complex, frequency-dependent impedance Z R i (Ohms) {where R = AC resistance and = AC reactance} of the parallel-plate capacitor is no longer just: Z i i 1 C (Ohms) where 1 C = the AC capacitive C C C reactance of the capacitor (Ohms), with (complex) AC Ohm’s Law: V I Z Because of the existence of the magnetic field in gap-region of -plate capacitor, EM energy can also be/is stored in the magnetic field of -plate capacitor due to the inductance, LC (Henrys) associated with the parallelplate capacitor and hence it has an inductive reactance of L L and hence has an inductive complex impedance associated with it, of Z L i L i LC (Ohms). Since the inductance associated with this capacitor is in series with its capacitance, we add the two impedances: 1 ZCTOT ZC Z L i C i L i i LC C 1 LC ZCTOT i C The {complex} form Ohm’s Law {here} is thus: V I ZCTOT 4 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede Note that at low frequencies 0 for the parallel-plate capacitor with d a , the capacitive reactance 1 C L and thus Z TOT 0 Z 0 . However, at very C L C C high frequencies (ω → ∞), C L Z TOT c C Z L , i.e. in the very high frequency limit, this capacitor instead behaves like a pure inductor!!! Note also that the electric, magnetic and total EM energy densities in the gap-region of the parallel plate capacitor, respectively are: 2 1 1 2 EM u E r , t o E r , t , uM r , t B r , t and uTOT r , t uE r , t uM r , t 2 2o i Now because the capacitor has a non-zero time-varying magnetic field: B , t 2 Eo eitˆ 2c Faraday’s Law E r , t B r , t t tells us that there will be an additional {induced} electric field, because B r , t is also varying in time!!! Faraday’s Law in integral form is: S E r , t da t t m B r t da , S t t B r, t da is the magnetic flux (Webers = Tesla-m2) enclosed by the surface S where m S t m E r t d B r t da , , C S t t where the contour C around a closed path of integration encloses the surface S through which t B r, t da passes. magnetic flux m at time t. Applying Stokes’ Theorem, we have: S Now B Bˆ (i.e. points in the ̂ {azimuthal} direction) and thus here we need B da hence da2 daˆ also, and thus we take the closed contour C2 line-integral path around the surface S2 as shown in the side-view figure below: Side-View: © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 5 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede The induced electric field, as created by the time-varying magnetic field is: E r t d , ind C2 t t m B r t da , 2 S2 t t B r, t da = magnetic flux enclosed by contour C2 passing through surface S2 where m 2 S2 Then: C2 Eind r , t d Eind r , t d 1 Eind r , t d 2 Eind r , t d 3 Eind r , t d 4 (1) (2) (3) (4) d 1 d zˆ d 2 d ˆ ( ˆ ŷ here in yˆ zˆ plane) d 3 d zˆ d zˆ d 4 d ˆ d ˆ Now Eind r , t B r , t t tells us that if B Bˆ direction, then in cylindrical coordinates: 0 0 0 0 0 E E E E z r , t 1 1 Ez Ez ˆ ˆ ˆ z E ˆ Eind r , t z z 0 0 Thus, we see that E r , t E r , t ˆ only, for all points , , z in the gap region of ║-plate capacitor and for all times t. However, we see that due to the azimuthal / rotational symmetry associated with the cylindrical ║-plate capacitor, neither Eind r , t nor B r , t can have any explicit φ-dependence, thus E 0 and E 0 , which in turn respectively imply that E z 0 and E must also have Eind r , t B r , t . z 0 . Note further that Faraday’s Law tells us that we For d a , the electric field in the gap region of the ║-plate capacitor cannot explicitly depend on z either. Thus, E z 0 the only surviving term in Eind r , t is: E z r , t ˆ . Eind r , t Eind r , t Eind r , t zˆ i.e. the induced E -field points in the ẑ direction (must be B Bˆ ) which is satisfied because ẑ ˆ . 6 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede Thus, if the induced electric field Eind r , t Eind r , t zˆ {only}, then we see that: C2 Eind r , t d Eind r , t d 1 Eind r , t d 2 Eind r , t d 3 Eind r , t d 4 (1) E ind zˆ dzzˆ (1) (2) (3) (4) E ind zˆ d ˆ E ind zˆ dzzˆ E ind zˆ d ˆ (3) (4) (2) 0 0 zˆ ˆ ẑ ˆ E ind zˆ dzzˆ E ind zˆ dzzˆ (1) z=d z=0 (3) z=d Eind 0 dz E ind dz z=0 But Eind r , t has no explicit z-dependence, thus: C2 z d z d Eind r , t d E ind 0, t dzE nd , t dz E ind 0, t d E ind , t d Or: z o C2 z 0 Eind r , t d E ind 0, t E ind , t d where: Eind , t E ind , t zˆ E r , t d B r , t da2 where S 2 surface enclosed by contour C2 C2 ind t S2 and da2 danˆ2 daˆ (i.e. S2 lies in the y-z plane) daxˆ {here} and da dydz d dz But: i Now: B , t Bo eitˆ 2 Eo eitˆ 2c z d i it B S2 , t da2 0 z 0 2c 2 E0e ˆ d dzˆ but: ˆ ˆ 1 z d i i d 2 Eo eit d dz 2 E0 eit d and: 0 z 0 0 2c 2c 1 d 2 2 i 2 d i t B , t da S2 2 4c 2 E0e Then: And: and ˆ x̂ ( S2 lies in the y-z plane) i 2 d i 2 d it i t B r , t da E e i E e 2 0 0 2 2 t S2 t 4c 4c i i i i 2 1 2 2 {since i 1 and i 1 } © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 7 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede 2 2d B r , t da2 E0 eit . 2 S t 2 4c Then: Eind r , t d B r , t da2 yields: C2 S t 2 2 2 d d Eind 0, t Eind , t E0 eit zˆ 4c 2 Note that the d’s cancel on both sides of the above equation. Note also that because of the explicit 2 dependence on the RHS of above equation, we see that Eind 0, t 0 . 2 2 2 it it Eo e zˆ Hence: Eind , t Eo e zˆ c 4c 2 2 Thus the total E -field in the capacitor gap is: 2 2 it it it EToT , t E t Eind , t Eo e zˆ Eo e zˆ 1 Eo e zˆ 2c 2c Thus, we see here that the induced electric field caused by the time-varying magnetic field points in the direction opposite to the initial/original E -field, reducing the overall E -field for 0 , as we would expect from Lenz’s Law. However, note that we now also have an additional contribution to the B -field inside the gap-region of the parallel plate capacitor, due to the presence of the induced E -field contribution, Eind , t . Before we proceed further on this discussion, it would be best for us change our notation: Call our original time-dependent E -field, E r , t Eo eit E1 r , t . This E -field in turn creates a time-dependent B -field by Ampere’s Law: E1 r , t 1 E1 r , t B1 r , t o o 2 . t c t However, because B1 r , t also varies in time, it turn creates another induced time-dependent electric field by Faraday’s Law: B1 r , t E2 r , t . t But E2 r , t is also time-varying, and so it in turn produces another time-varying contribution to the magnetic field B2 r , t . 8 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede But because B2 r , t is also time-varying, it in turn will induce another contribution to the electric field E3 r , t and so on… i.e.: A. L. F .L. A. L. F . L. A. L. F . L. A. L. F . L. E1 r , t B1 r , t E2 r , t B2 r , t E3 r , t B3 r , t E4 r , t B4 r , t . . . . Then: ETOT r , t E1 r , t E2 r , t E3 r , t E4 r , t E5 r , t ... En r , t n 1 And: BTOT r , t B1 r , t B2 r , t B3 r , t B4 r , t B5 r , t ... Bn r , t n 1 So thus we see that: 1 C1 B1 r , t d c 2 t S1 E1 r , t da1 E r , t d B 1 r , t da2 C2 2 S t 2 1 , B r t d E 2 2 r , t da1 2 C1 S c t 1 C2 E3 r , t d t S2 B2 r , t da2 1 C1 B3 r , t d c2 t S1 E3 r , t da1 E r , t d B 3 r , t da2 C2 4 S t 2 …. etc. Algorithmically, this infinite sequence can be written as: n 1,E1 r , t Infinite Loop 1 B r t d 2 , n C1 c t En r , t da1 En 1 r , t d C2 t Bn r , t da2 S1 S2 Bn r , t En 1 r , t n n 1 Where contour C1 enclosing surface S1 and area element da1 are associated with the figure drawn on page 3 of these lecture notes, and where contour C2 enclosing surface S2 and area element da2 are associated with the figure drawn on page 5 of these lecture notes. © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 9 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede It can thus be shown for the parallel-plate capacitor with d a that: 2 4 6 1 1 1 V ... ETOT , t 1 Eo eit zˆ with: Eo o 2 2 2 d 1! 2c 2! 2c 3! 2c 2 4 6 1 1 1 i BTOT , t 1 ... Bo eitˆ with: Bo 2 Eo 2 2 2 2c 1! 2c 2! 2c 3! 2c and where: n ! n n 1 n 2 ...21 , 0! 1 , 1! 1 , 2! 2 , 3!= 6, 4! = 24, etc. 2 We also see that: Bn i i i i En and: En 1 Bn 2 En En 2 2c 2 2 2c 2c 2 i i i and: Bn 1 2 En 1 2 Bn Bn 2c 2c 2 2c Due to the cylindrical geometry / azimuthal symmetry associated with this problem, it should not come as a surprise that: x Defining: k c 2 k where k = wavenumber c c f f 2 Then the quantity in square brackets on the previous page becomes: 2 1 4 6 2 4 6 1 1 1 1 x 1 x 1 x ... 1 ... 2 2 2 2 2 2 1! 2c 2! 2c 3! 2c 1! 2 2! 2 3! 2 The so-called “ordinary” Bessel function of the first kind, of order zero has a series expansion of the form: 2 4 6 1 x 1 x 1 x J0 x 1 ... 2 2 2 1! 2 2! 2 3! 2 2k 2k 2k 1 1 x 1 x x J0 x 2 k 0 k ! k 1 2 k 0 k ! k ! 2 k 0 k ! 2 k ! k k 1 k 2 ...321 where: k 1 k ! (for k = integer) k k k In general, the series expansion of the ordinary Bessel functions of the first kind, of order n are: n2k 1 x Jn x k 0 k ! n k 1 2 10 k © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede Thus, for the cylindrical ║-plate capacitor with d a the electric and magnetic fields in the gap region are of the form: V E , t J 0 c Eo eit zˆ J 0 k Eo eit zˆ with k and Eo o d c B is 90o out-of-phase with E {here}: B , t J0 B e c it o ˆ J 0 k Bo eitˆ with Bo i Eo i 2 Eo 2 2c 2c Note that for 0 that: E 0, t Eo eit but: B 0, t 0 . x k c The {Radial} Zeroes of J0(x): n.b. the zeroes of Jn(x) are not integer related!! Since E , t J 0 k Eo eit zˆ and B , t J 0 k Bo eitˆ we see that the zeroes xn of J 0 x are physically where the electric and magnetic fields vanish (!!!), i.e. E , t 0 and B , t 0 when n xn k cxn with x1 2.4048 , x2 5.5201 , x3 8.6537 , etc.!!! So let’s now examine the frequency-dependence of the E and B fields of the ║-plate capacitor: E , t J0 E e B , t J0 B e c o c it zˆ J 0 k Eo eit zˆ it o a.) When: ω = 0, f = 0 then: k with: k c and: Eo ˆ J 0 k Bo eitˆ with: Bo c Vo d i ik Eo Eo 2 2c 2c 0 (static case). Then: x k 0 and J 0 0 1 V V E , t Eo zˆ o zˆ with: Eo o and: B , t 0 ← d d n.b. Same result as original static calculation © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 11 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede b.) When: 0 , e.g. f = 60 Hz → ω = 2πf = 120π rad/sec 2 120 rad / sec Then: k 1.257 106 rad /meter 8 c 3 10 m / sec Suppose the radius of the capacitor is a = 1 cm = 102 m (reasonable/typical diameter) Then: ka 1.257 108 x (dimensionless) And: J 0 ka J 0 1.257 108 1.0 (n.b. see/refer to above graph of J0(x) vs. x) Thus, we see that at f = 60 Hz, the E -field is ≈ that of the DC E -field, and e.g. if Vo = 10 V and V 10 V d = 0.1 mm a = 1 cm, i.e. d = 104 m a = 102 m, then: Eo o 4 105 Volts /m 10 m d 8 a ka 1.257 10 E0 and: Bo a 2 E0 105 2.11012 Tesla {i.e. is very small}. 8 2c 2c 2 3 10 Another way to see this: c Bo a 6.3 104 Volts /m Eo 105 Volts /m c.) Now suppose: f = 1 MHz = 106 Hz and 2 f 2 106 rads / sec Then: k Then: 2 106 2.1102 0.021 radians /m and if a = 1 cm = 0.01 m 8 c 3 10 (ka) = 0.021 x 0.01 = 2.1 x 104 and J0(ka) = J0 (2.1 x 104) ≈ 1 (still). 2 Vo ka (constant), and: Bo a Eo 3.5 108 Tesla 35 nT (still very small) d 2c 5 for Eo 10 Volts/meter and {still} c Bo a 10.5 Volts /m Eo 105 V /m → Eo for Vo = 10 Volts, d = 0.1 mm and a = 1 cm = 102 m. d.) Now suppose: f = 100 GHz = 1011 Hz and ω = 2πf = 6.3 x 1011 rads/sec 2 6.3 1011 2.1103 radians/m Then: k 8 c 3 10 Then: (ka) = 2.1 x 103 x 102 = 21 → J0 (kρ) has 5 zeroes in it !!! EEK!! → the E -field points in the reverse direction depending on 0 ≤ ρ ≤ a value !!! (see above graph of J 0 x vs. x on page 11 of these lecture notes) Suppose instead that we pick: ka = 2.4048 = x1 = 1st zero of J0(x) = J0(kρ) (a = 102 m = 1 cm) Then: k = 240.48 radians/m = → f = 1.15 x 1010 Hz = 11.5 GHz (in the microwave region) c 2 ik 2.61 cm , B , t J 0 k Bo eitˆ , Bo Eo Then: E , t J 0 k Eo eit zˆ and k 2c 12 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede E Eo Electric field = 0 at a when ka 2.4048 0 a c B f = 11.5 GHz a = 0.01 m cBo 12 kaEo Magnetic field = 0 at a when B 0 0 ka 2.4048 0 a f = 11.5 GHz a = 0.01 m The Inductance of a Parallel-Plate Capacitor Equate: Wm 1 2 1 LI 2 2 o Capacitance: C o A d v , I V Z V IZ ToT ToT V Vo eit 1 L ZToT ZC Z L i C for d a I Thus: Wm 2 B d Vo eit Vo e it Vo2 2 * , I then: I II 2 1 1 1 i L i L L C C C Vo2 1 1 1 2 LI L 2 2 2 o 1 2 o L C v 2 k B d J 02 k 2 Eo2 d ddz 4c 2 2 2 2 2 1 d Vo d 2 d k 2 2 a 2 Eo J 0 kp 3 d = L 2 2 0 2 2 o 4 c 1 2 C L = L Eo2 d 1 L C 2 o o k 2 Eo2 2 o a o J 02 k 3 d with 1 o o , ck , k c c2 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 13 UIUC Physics 436 EM Fields & Sources II o 2 2 2 1 2c d L C L a o 2 Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede J 02 kp 3 d A 2 2 L 1 1 1 L 2 2 L A A 2 L2 A L A 2A 2A A L C C C C C 2 2A 1 1 L A L or: A 0 C C a 2 2 b c b b 2 4ac . 2a Quadratic equation of the form: aL2 bL c 0 , solve for L: L 1 2A L C 1 2A C 2 2 2 4A 2 C2 2A 2 L 1 2A 2 2 1 4A 4A 2 4A 2 C C C C 2A 1 2A C 1 2A C 2A C 2 2A 2 1 2A C 1 4A 2A 2 2 C 2 Physically, we want L 0 when 0 must choose – (negative) sign in above formula! L 1 2A C 1 4A C 2A Now: 2 2 o 2 1 4A A 8 C 2c 2 d L 2A 2 2C 2 a 0 2 1 1 4A 1 1 2 ... for 1 , thus: 2 8 C J 02 k 3 d A 2 o 2 d 2c 2 o A2 a 0 J 02 k 3 d o d 2A 2 a 0 J 02 k 3 d d2 Where the capacitance and inductance of the parallel-plate capacitor, for d a are: C o A d and L Note that for 14 o d ka = 2A 2 a 0 4A 2 2 J 02 k 3 d for 2 C Ac 2.4048 – 1st 5.5201 – 2nd 8.6537 – 3rd 11.7915 – 4th 14.9309 – 5th 18.0711 – 6th . . . a o J 02 kp 3 d 1 . zeroes of J0(ka) © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede 2 The electric field E a 0 for these values of ka, corresponding to wavelengths k c ck and frequencies f 2 2 λ1 = 2.61 cm ↔ f1 = 1.15 x 1010 Hz = 11.5 GHz λ2 = 1.14 cm ↔ f2 = 2.64 x 1010 Hz = 26.4 GHz λ3 = 0.73 cm ↔ f3 = 4.13 x 1010 Hz = 41.3 GHz λ4 = 0.53 cm ↔ f4 = 5.63 x 1010 Hz = 56.3 GHz λ5 = 0.42 cm ↔ f5 = 7.13 x 1010 Hz = 71.3 GHz λ6 = 0.35 cm ↔ f6 = 8.63 x 1010 Hz = 86.3 GHz . . . . . . Note that because the electric field E a 0 for these specific frequencies Compare with radius a = 1.0 cm and diameter D = 2a = 2.0 cm of ║-plate cylindrical capacitor, as well as the gap dimension of d = 0.1 mm = 0.01 cm (corresponding to the zeroes of J0(x)=J0(ka)), this means that physically, we could actually short out the capacitor at ρ = a and it wouldn’t make any difference to the behavior / physics of this “capacitor” at these specific frequencies f1, f2, f3, . .!!! For ka = zero of J0(ka) {i.e. J0(ka) = 0), we can short out the capacitor by wrapping it e.g. with sheet metal at ρ = a , thus turning it into a cylindrical, fully-enclosed can with d a ! → No change in physics for frequencies f1, f2, f3, . . because E a 0 for these frequencies! Thus, at these frequencies f1, f2, f3, . . fn corresponding to the zeroes of the Bessel Function J0(ka) (i.e. J0(ka) = 0), a cylindrical conducting metal can of radius a and height d a is actually a resonant cavity with electric field: E , t J 0 kn Eo eint zˆ and magnetic field: B , t J 0 kn B0 eitˆ subject to the boundary conditions that: E a, t E z a, t 0 and also that: B a, t B a, t 0 n , n 2 f n , n = 1, 2, 3, . . . c ik V Eo and: Eo o with: Bo d 2c for: kn © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 15 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 9 Prof. Steven Errede We will see shortly in the next set of P436 Lecture Notes (# 10) that the resonant frequencies of a resonant cavity and the allowed modes of EM wave propagation in wave guides can be derived directly from the wave equation for EM waves in these structures, as determined by the boundary conditions imposed on the EM waves by the conducting walls of these devices and also the allowed polarization states of these EM waves. Here in these lecture notes, we obtained harmonic EM wave solutions for E and B in the gap region of a parallel plate capacitor (and cylindrical can capacitor, subject to boundary condition E = 0 at ρ = a) via a perturbative technique, analogous to what we did last semester in P435 for the E -field associated with a dielectric sphere immersed in an initially uniform external E -field and the B -field associated with a magnetizable sphere immersed in an initially uniform external B -field. (See/work Griffiths Problems 4.23 and 6.18). “Homework” Exercises: 1.) Calculate the electric, magnetic and total energy densities u E , , z , t , um , , z , t and uTot , , z , t and their time averages; make e.g. plots of these vs. . Investigate/plot their behavior for low frequencies 0 and at higher frequencies, when n ckn c xn a where xn kn a = zeroes of J 0 xn 0 . 1 2.) Calculate Poynting’s vector S , , z , t E , , z, t B , , z, t and its time average; o make plots of S vs. , investigate/plot its behavior for low frequencies 0 and when n ckn c xn a . 3.) Calculate the linear EM momentum density, EM , , z , t o o S , , z , t and angular momentum density, EM , , z , t r EM , , z , t , and their time averages; make plots of EM and EM vs. ; investigate/plot their behavior for low frequencies 0 and when n ckn c xn a . 16 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved.