6.642, Continuum Electromechanics Prof. Markus Zahn Lecture 1: Review of Maxwell s Equations

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1 6.64, Continuum Electromechanics Prof. Markus Zahn Lecture 1: Review of Mawell s Equations I. Mawell s Equations in Integral Form in Free pace 1. Faraay s Law C E i s = - µ H a t i Circulation of E Magnetic Flu µ -7 =4 π 1 henries/m [magnetic permeability of free space] EQ form: fiel) E i s= (Kirchoff s Voltage Law, conservative electric C MQ circuit form: i v=l t (Inuctor). Ampère s Law (with Displacement Current) H s = J a E a i i + C t i Circulation Conuction Displacement of H Current Current MQ form: EQ circuit form: C Hi s = Ji a v i=c t (capacitor) 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 1 of 9

2 3. Gauss Law for Electric Fiel i E a = ρ V V faras/m 36π 1 c= (pee of electromagnetic waves in free space) µ 4. Gauss Law for Magnetic Fiel µ Hi a = In free space: B = µ H magnetic flu ensity magnetic fiel intensity 5. Conservation of Charge Take Ampère s Law with isplacement current an let contour C lim C H i s = = J i a + E i a t C J i a + V = t ρ V V ρ V Total current Total charge leaving volume insie volume through surface 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page of 9

3 6. Lorentz Force Law ( µ H) f=q E+v II. Electric Fiel from Point Charge E i a = Er4π r = q q E = 4π r r q T sin θ =f = 4π c r T cos θ =Mg q r tan θ = = 4π r Mg l 1 3 π r Mg q= l 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 3 of 9

4 III. Faraay Cage q Ji a=i= - V = - (-q) = t t t ρ it = q 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 4 of 9

5 IV. Divergence Theorem 1. Divergence Operation Courtesy of Krieger Publishing. Use with permission. V A i = iv A V iv A = lim A i V V 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 5 of 9

6 Courtesy of Krieger Publishing. Use with permission. Courtesy of Krieger Publishing. Use with permission. 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 6 of 9

7 Φ= A, y, z yz A -, y, z yz 1 1' y + A, y + y, z z A, y, z z y ' z + A, y, z + z y A, y, z y z 3 3' ( ) A (,y+ y,z) A (,y,z) A,y,z A -,y,z y y Φ y z + y + Az,y,z+ z Az,y,z z A A A y z V + + y z A i A A A iv A = lim = + + V V y z y z Del Operator: =i +i +i y z y z A A y iv A A = + + = i A y z z. Gauss Integral Theorem Courtesy of Krieger Publishing. Use with permission. 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 7 of 9

8 N A i = A i i=1 N i i N V n N i=1 ( ) = lim i A Vi = i A V V V i A V = A i a 3. Gauss Law in Differential Form ( ) E i a = i E V = ρ V V V i ( E ) = ρ µ i i ( µ ) V i ( µ H ) = H a = H V = V. tokes Theorem 1. Curl Operation C A i s = Curl A i a n n A i s C Curl A = lim a a n 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 8 of 9

9 A i a = A i s C Courtesy of Krieger Publishing. Use with permission. 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 9 of 9

10 + y+ y i y ( ) ( ) A s = A, y + A +, y y + A, y + y C y y y+ y 4 y + A, y y ( ) A A, y - A, y+ y y +, y - Ay, y = y + y Ay A =az - y A i s A A Curl A = = - z a y y By symmetry A i s A A Curl A = = - y a z z z A i s A Curl A = = - a y z y A z y A A z y A A Az y A Curl A = i - + i - + i - y z y z z y i i i y z = et y z A A y A z = A 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 1 of 9

11 . tokes Integral Theorem Courtesy of Krieger Publishing. Use with permission. N N lim A i s = A i s i i=1 Ci C N i=1 ( ) = A i ai ( ) = A i a Courtesy of Krieger Publishing. Use with permission. 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 11 of 9

12 3. Faraay s Law in Differential Form ( ) µ E i s= E i a=- Hi a t C H = - t E µ 4. Ampère s Law in Differential Form H s= H a= J a + E a i i i C t i H= J+ E t VI. Applications to Mawell s Equations 1. Vector Ientity C C ( ) ( ) lim A i s = = A i a= i A V i ( A ) =. Charge Conservation V i H= J + E t E = i J + t 3. Magnetic Fiel ρ = i J + t i E=- µ H t =- µ H µ H = t i i 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 1 of 9

13 VII. Bounary Conitions 1. Gauss Continuity Conition Courtesy of Krieger Publishing. Use with permission. E i a = σ E - E = σ s n 1n s σ i ( 1) E -E = n E -E = σ n 1n s s. Continuity of Tangential E Courtesy of Krieger Publishing. Use with permission. C E i s = E - E l = E - E = 1t t 1t t n ( E -E 1 ) = Equivalent to Φ1 = Φ along bounary 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 13 of 9

14 3. Normal H i µ H = µ H i a= µ H - H A = an bn H = H an bn n i Ha - H = b 4. Tangential H H= J H i s = J i a C H s - H s = Ks bt at H - H = K bt at n H a - H = K b 5. Conservation of Charge Bounary Conition ρ i J+ = t J i a + V = t ρ V σ i t n J - J + s = a b 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 14 of 9

15 VIII. Poisson s an Laplace s Equations 1. Poisson s Equation E= E= - Φ ρ ρ i E= ( Φ) = Φ =- i (Poisson s Equation). Particular an Homogeneous olutions ΦP =- ρ Poisson s Equation ΦP Φ Laplace s Equation h = ΦP Φh + = - ρ ρ r' V ' r = 4 π r -r' V' Φ = Φ P + Φ must satisfy bounary conitions h 3. Uniqueness of olutions Try solutions Φ an a Φ b Φ a Φb = - ρ = - ρ Φ a Φ b - = Φ = Φ - Φ Φ = Φ = a b = + = i Φ Φ Φ Φ Φ i Φ Φ i Φ Φ V = Φ Φ i a = Φ V = V V On, Φ = or Φ i a = Φ = Φ a = Φ on b Φ Φ Φ n n E = E on a b i a = = on na nb A problem is uniquely pose when the potential or the normal erivative of the potential (normal component of electric fiel) is specifie on the surface surrouning the volume. 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 15 of 9

16 IX. Two-Dimensional olutions to Laplace s Equation in Cartesian Coorinates, Φ (, y) Φ Φ Φ, y = + = y 1. Try prouct solution: Φ (, y ) = Χ ( ) Y( y) Χ Y y Y y + X = y Multiply through by 1 XY : 1 Χ 1 Y X = - = - k Y y k=separation constant only a only a function function of of y Χ Y = - k Χ ; = k Y y. Zero eparation Constant olutions: k= Χ = Χ =a+b 1 1 Y = y Y = c y Φ (, y ) = ΧY =a +b+cy+y 3. Non-Zero eparation Constant olutions: k Χ +k Χ = X = A sin k + A cos k 1 Y - k Y = ky -ky Y = B e + B e 1 y = C 1 sinh ky + D 1 cosh ky 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 16 of 9

17 Φ Courtesy of Krieger Publishing. Use with permission. (, y ) = Χ ( ) Y ( y) ky -ky ky -ky = D 1 sin ke + D sin ke + D 3 cos ke + D 4 cos ke = E 1 sin k sinh ky + E sin k cosh ky + E 3 cos k sinh ky + E 4 cos k cosh ky 4. Parallel Plate Electroes 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 17 of 9

18 Neglecting en effects, Φ ( ). Bounary conitions are: Φ = = Φ, Φ = = Φ Try zero separation constant solution: Φ ( ) = a1 + b1 Φ ( = ) = Φ =b1 ( = ) = =a+b a = Φ Φ Φ Φ Φ ( ) = + Φ Φ Φ Φ E = - = Φ - Φ (Electric fiel is uniform an equal to potential ifference ivie by spacing) 5. Hyperbolic Electroe Bounary Conitions Courtesy of Krieger Publishing. Use with permission. 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 18 of 9

19 Φ Φ Φ Φ ( y = ab ) = V ( =, y ) = (, y = ) = (, y ) = V y ( ab) Φ Φ E = - Φ = - i - i y V ab =- yi + i y y Electric fiel lines: y E y = = E y yy = y = + C y = + y - (fiel line passes through (, y )) 6. patially Perioic Potential heet 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 19 of 9

20 Φ (, y ) = -a V sin ay e +a V sin ay e Φ Φ E = - Φ (, y ) = - i + i y y -a -Va e cos ay i - sin ay i y > = a -Va e cos ay i + sin ay i < y σ s = = E =+ -E = - = V a sin ay 7. Electric Fiel Lines: E y y = = E -cot ay > +cot ay < -a > cos ay e = constant +a < cos ay e = constant 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page of 9

21 Courtesy of Krieger Publishing. Use with permission. X. Two-Dimensional olutions to Laplace s Equation in Polar Coorinates = z 1. Prouct olution 1 Φ 1 Φ Φ Φ = r + + = r r r r φ z Φ ( r, φ) = R ( r ) F ( φ ) 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 1 of 9

22 Multiply through F ( φ) R R ( r ) F r + = r r r r r φ by R r F ( φ) r R 1 F r + = R r r F φ m -m r R R R r r r r r = m r r - m R = 1 F F =-m +mf = F φ φ Courtesy of MIT Press. Use with permission.. m= olutions (Zero eparation Constant olutions) R r = C R = C ln r + D r F = F =A φ + Β φ Potential of line charge Φ r, φ = R r F φ = A + A φ + A ln r + A φ ln r , Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page of 9

23 Φ φ = = V φ Φ 1 Φ Φ Φ ( φ ) = E = - Φ = - i + i + i r θ z α r r φ z Φ φ = α = V V ( r, = ) = E ( r, = ) = - σs φ φ φ αr V σs r, φ = α = - Eφ r, φ = = + α r V 1 Φ E φ = - = - r φ αr 3. m olutions (Non-Zero eparation Constant olutions) r R r r r -m R = F +m F = φ Try n R = Ar F = A1sin m φ + A cos mφ r n n r nar -m Ar = n n nr -mr = n= ±m m -m R ( r ) = A3r + A4r m -m ( r, ) = R ( r) F = A sin m + A cos m A r + A r Φ φ φ φ φ m -m m -m = A sin mφr + B sin mφr + C cos mφr + D cos mφr 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 3 of 9

24 4. electe olutions m=1 Φ ( r, φ) = Ar sin φ = Ay Φ ( r, φ) = Cr cos φ = C B sin φ Φ ( r, φ) = Line r ipole oriente in y irection D cos φ Φ ( r, φ) = Line r ipole oriente in irection m= Φ = Ar sin φ = Ar sin φ cos φ = Αy Generally m an integer if Φ ( φ = ) = Φ ( φ =π ) 5. Groune Perfectly Conucting Cyliner in a Uniform y Directe Electric Fiel A Φ = -Er + sin φ r R r 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 4 of 9

25 A Φ ( r=r, φ) = -ER+ = A=ER R R Φ = - E r - sin φ r R r Φ 1 Φ Φ E=- Φ = - i + i + i r θ z r r φ z R R = E 1 + sin i cos i r R φ φ > r θ r r σs ( r=r, φ) = Er ( r=r +, φ) -Er ( r=r -, φ) = E sin φ 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 5 of 9

26 Courtesy of Krieger Publishing. Use with permission. XI. Two-Dimensional olutions to Laplace s Equation in pherical Coorinates = φ 1. Prouct olution 1 Φ 1 Φ 1 Φ Φ = r + sin θ + = r r r r sin θ θ θ r sin θ φ Φ ( r, θ ) =R( r) Θ ( θ) Θ Multiply through θ R R Θ r + sin θ r r r r r sin θ θ = θ by R r Θ ( θ) 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 6 of 9

27 1 R 1 Θ R r r Θ sin θ θ θ r + sin θ = n(n+1) -n(n+1) R r - n( n + 1 ) R = r r Try p R=Ar p p Ap p + 1 r - n n + 1 Ar = p = n, - n + 1 n -n+1 R r = Ar + Br Θ sin θ +n( n+1) sin θ = θ Θ θ [Legenre s Equation] In this course, only responsible for n=1 solution Θ ( θ ) =cos θ Φ =Ar cos θ =Az is potential of uniform z irecte electric fiel B cos θ Φ = is potential of point electric ipole r n=1 B Φ ( r, θ ) = Ar + cos θ r. Groune phere in a Uniform z Directe Electric Fiel z=rcos θ 1 z=i = z ( rcos θ ) =i r ( rcos θ) + i ( rcos θ ) =i cos θ -i sin θ θ r θ r r θ r R Φ Acos θ ( r, θ ) =-Ercos θ + r 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 7 of 9

28 Φ A R 3 ( r=r, θ ) == -ER+ cos θ A = E R 3 R Φ ( r, θ ) =-E r- cos θ r R r Φ 1 Φ E=- Φ =- i + i r r r θ θ 3 3 R R =E 1+ cos θ i - 1- sin θ i 3 r 3 r r θ σ r=r, θ = E r = R, θ = 3 E cos θ s r 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 8 of 9

29 Courtesy of Krieger Publishing. Use with permission. 6.64, Continuum Electromechanics Lecture 1 Prof. Markus Zahn Page 9 of 9

12.11 Laplace s Equation in Cylindrical and

12.11 Laplace s Equation in Cylindrical and SEC. 2. Laplace s Equation in Cylinrical an Spherical Coorinates. Potential 593 2. Laplace s Equation in Cylinrical an Spherical Coorinates. Potential One of the most important PDEs in physics an engineering