6.641 Electromagnetic Fields, Forces, and Motion
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MIT OpenCourseWare http://ocw.mit.edu 6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 6.641 — Electromagnetic Fields, Forces, and Motion Spring 2009 Problem Set 4 - Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 4.1 A y x z Figure 1: Cartesian coordinate axes (Image by MIT OpenCourseWare.) For region I, we use q and q � � � q q� Φ= + 4πε1 (x2 + (y − d)2 + z 2 )1/2 4πε1 (x2 + (y + d)2 + z 2 )1/2 EI = −�Φ 1 = 4πε1 � q(xîx + (y − d)îy + z îz ) q � (xîx + (y + d)îy + z îz ) + (x2 + (y − d)2 + z 2 )3/2 (x2 + (y + d)2 + z 2 )3/2 For region II, use q �� Φ= q �� 4πε2 (x2 + (y − d)2 + z 2 )1/2 EII = −�Φ = q �� (xîx + (y − d)îy + z îz ) 4πε2 (x2 + (y − d)2 + z 2 )3/2 B Tangential E components are equal: 1 ⎤ n̂ × [EI − EII ] = 0 ⎥ ExI = ExII ⎥ ⎥ at y = 0 EzI = EzII ⎥ ⎦ 1 � Problem Set 4 6.641, Spring 2009 2 Since there is no surface charge, i.e., σs = 0. ⎤ n̂ · (εI EI − εII EII ) = 0 ⎦ at y = 0 εI EIy = εII EIIy From 1, 1 4πεI qx + q � x 2 (x + d2 + z 2 )3/2 � � = 4πεII (x2 q �� x . Therefore, + d2 + z 2 )3/2 From 2, 1 ε1 4πεI −qd + q � d (x2 + d2 + z 2 )3/2 � � = εII q �� (−d) 4πεII (x2 + d2 + z 2 )3/2 −q + q � = −q �� Therefore, q + q � q �� εI �� = ⇒q= q − q� εI εII εII q �� = q − q � Therefore, εI (q − q � ) − q � εII � � � � εII − εI εI + εII q = −q � εII εII � � εI + εII (εII − εI ) q = −q � ⇒ q � = −q εII − εI (εII + εI ) q = and εI �� q − q � εII εI �� = q − q + q �� εII � � εI + εII �� 2q = q εII εI + εII �� 2εII q q= q ⇒ q �� = 2εII (εI + εII ) q = C ¯I (x = 0, y = d, z = 0) f¯ = qE � � q � (0îx + 2dîy + 0îz ) =q 3 4πεI (02 + (2d)2 + 02 ) 2 2dq � q îy q � q îy = 3 4πεI · 8d 4πεI · 4d2 � � �� −εI q −q εεII −q 2 (εII − εI ) I +εII = î = îy y 4πεI · 4d2 16πεI d2 (εI + εII ) q 2 (εI − εII ) = îy 16πεI (εI + εII )d2 = 2 q+q � εI = q �� εII . Problem Set 4 6.641, Spring 2009 Problem 4.2 This is a charge relaxation problem, so we use, as shown in class, the equations (done in lecture 12) ∂ρf → − �· Jf + =0 ∂t → − We substitute in � · E = ρf ε → → − − and J f = σ E to get ∂ρf σ → ∂ρf − σ� · E + =0⇒ + ρf = 0 ∂t ∂t ε So t ε − ρf = ρ(→ r , t = 0)e− τe ; τe = σ Thus � ρf (r, t) = ρ0 r − τte a0 e 0 < r < a0 r > a0 0 Notice: ρf (r, t) is 0 for r > a0 . Nonetheless, there is still conduction and displacement current for a0 < r < a1 . � − � → − By Gauss, S ε E · d→ a = V ρdV . Choosing S as a cylinder with radius r � → − − ε E · d→ a = εEr r2πL S → − − where L is the length of the cylinder. Note that E · d→ a = 0 on cylinder ends. Now for RHS of Gauss � � L � r � 2π � r ρf dV = ρf (r� , t)r� dφdr� dz = 2πL ρf (r� , t)r� dr� V 0 0 r < a0 : � 0 r � ρf dV = 2πL V 0 0 ρ0 (r� )2 − τt � e e dr a0 3 = 2πL ρ0 r − τt e e a0 3 a0 < r < a1 : � � ρf dV = 2πL v a0 0 = 2πL ρ0 (r� )2 − τt � e e dr a0 a30 ρ0 − τt 2πL e e = ρ0 a20 e−t/τe a0 3 3 r > a1 : � � ρf dV = 2πL V 0 a0 ρ0 (r� )2 � dr a0 ρ0 a30 2πL = ρ0 a20 a0 3 3 (total charge on sphere is constant equal to total initial charge including surface charge at r = a1 ) = 2πL 3 Problem Set 4 6.641, Spring 2009 So: ⎧ ρ0 r2 − t τ ⎪ ⎨ 3a0 ε2 e e îr − → ρ0 a0 − τt e î E = r 3rε e ⎪ ⎩ ρ0 a20 î 3rε0 r r < a0 a0 < r < a1 r > a1 − σsf = ε0 Er (r = a+ 1 ) − εEr (r = a1 ) σsf = t ρ0 a20 (1 − e− τe ) 3a1 Problem 4.3 A → → → − − − There are no surface currents, so we have continuity of normal B and tangential H . Also, if µ → ∞, H = 0 → − inside, but B may still be nonzero. Equivalent image problem: Figure 2: Magnetic field lines due to a line current above an infinitely magnetically permeable region (Image by MIT OpenCourseWare.) Boundary conditions: Hx = Hz = 0 at y = 0 B Assume line current at origin: � I → → − − H · d l = I ⇒ Hφ = 2πr ∂Az Iµ0 → → − − �× A = B ⇒− = 2πr ∂r 0 Suggesting: Az = − Iµ 2π ln(r) + constant. Assume line current at y = d: Az = − Iµ0 � ln (y − d)2 + x2 2π Now 2 line currents; one at y = d and one at y = −d. Az = − � �� �� Iµ0 � �� 2 ln x + (y − d)2 + ln x2 + (y + d)2 2π 4 Problem Set 4 6.641, Spring 2009 C 1 → − − → �× A = H µ0 � � 1 ∂Az ∂Az = îx − îy µ0 ∂y ∂x � � I (y − d)îx − xîy (y + d)îx − xîy − → H =− + 2 x + (y + d)2 2π x2 + (y − d)2 D Field line equation: x dy Hy x2 +(y−d)2 + = = − y−d dx Hx x2 +(y−d)2 + x x2 +(y+d)2 y+d x2 +(y+d)2 = z − ∂A ∂x ∂Az ∂y ∂Az ∂Az ∂Az ∂Az dy = − dx ⇒ dx + dy = dAz = 0 ∂y ∂x ∂x ∂y � �� � Az = constant ⇒ x2 + (y − d)2 x2 + (y + d)2 = constant E → − F = unit length → − I ���� × current at y=d − → B ���� − → B field caused by image current alone at x = 0, y = −d �� � �� → − F −µ0 I (y + d)ˆix − xîy �� = (I îz ) × � unit length 2π x2 + (y + d)2 � x=0,y=d � �� � −µ0 I 2d = (I îz ) × îx 2π (2d)2 − → F µ0 I 2 =− îy unit length 4πd Problem 4.4 A → − � · J = 0; by symmetry we just have x component of J ∂Jx → − = 0 ⇒ J = J0ˆix , J0 is constant ∂x x Ex · σx = Jx ; σx = σ0 e− s Jx J0 J0 x Ex = = es x = − s σx σ0 σ0 e � s � s x J0 J0 s x ��s J0 s V0 = Ex dx = e s dx = es 0 = (e − 1) σ0 0 σ0 σ0 0 I0 = J0 · ld R= V0 = I0 J0 s σ0 (e − 1) J0 ld = s (e − 1) σ0 ld 5 Problem Set 4 6.641, Spring 2009 B � · (εE) = ρ ⇒ ρ = ε ρ= ∂Ex ∂x J0 ε x es σ0 s At x = 0: � εJ0 σs = εEx �x=0 = σ0 At x = s: � εJ0 σs = −εEx �x=s = − e σ0 C s � qV = ld ldJ0 ε (e − 1) σ0 ρdx = 0 Total surface charge qS = (σs |x=s + σs |x=0 )ld = − ldJ0 ε (e − 1) = −qV σ0 qS + qV = 0 Problem 4.5 A As no volume charge in the dielectric ¯ =0 �·D From symmetry we just have r component 1 ∂(r2 Dr ) A ε1 r = 0 ⇒ Dr = 2 , ε(r) = 2 r ∂r r a Dr Aa Aa = 2 = ε(r) r ε1 r ε1 r 3 �b � � � b � b Aa −Aa 1 �� Aa 1 1 v= Er dr = dr = = − 3 2ε1 r2 �a 2ε1 a2 b2 a a ε1 r Dr = εEr ⇒ Er = A= 2ε1 v a 1 a2 1 − 1 b2 , Er = Ē = −�Φ ⇒ Er = − � Φ= −Er dr = + 1 a2 ∂Φ = ∂r v − 1 b2 1 a2 2v 1 1 r3 − b2 1 a2 2v 1 − b12 r3 1 r2 6 Problem Set 4 6.641, Spring 2009 B σs |r=a = ε(r)Er |r=a+ = 2ε1 v 1 − b1 2 a3 1 a2 σs |r=b = −ε(r)Er |r=b− = b −2ε1 v a −2ε1 v 1 1 1 1 b3 = 1 1 ab2 − a2 b2 a2 − b2 C q = 4πa2 σs |r=a+ = −4πb2 σs |r=b− = � C= 8πε1 v � − b1 2 a 1 a2 q 8πε1 � =�1 1 v a2 − b2 a 7
