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ECE2317_Exam2_Summer_1999

# ECE2317_Exam2_Summer_1999 - Mi Namew Signature Student...

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Unformatted text preview: Mi/ Namew Signature: Student Number: ECE 231? APPLIED ELECTRICITY AND MAGNETISM EXAM 2 August 7, 1999 1. This exam is closed book and closed notes. Calculators may not be used. 2. No credit will be given if the work required to obtain the solution is not shown. 3. Perform all your work on the paper provided. The instructors will provide additional paper if needed. 4. A list of identities is provided at the end of the test. These identities may be needed in arriving at a final solution. 5. You will have a total of 90 minutes. GOOD LUCK! Do not write below this line. Prob. ‘l: .._._J25 Prob. 2: ”/25 Prob. 3: ________./25 Prob. 4; ____/25 TOTAL: [100 Problem 1 (25 points): A lossy coaxial capacitor is comprised of an inner and outer conductor of radius a and b, respectively. The conductors are each of length if and the material between the two conductors has a permittivity z»; and a conductivity 0-. Assuming a constant line charge density of p; on the inner conductor, find the a.) capacitance of the coaxial capacitor. b.) resistance of the coaxial capacitor. c.) energy stored by the coaxial capacitor. WWW” SOLUTION: a.) The capacitance for a two-conductor system is defined as the absolute value of the total charge on each conductor divided by the voitage difference between the two conductors. Thus, C:—. V We are given the charge per unit length on the inner conductor, p,. :> The total charge on the inner conductor is Q=ah (The total amount of charge on the outer conductor is equal to —Q.) 2;» From Gauss's Law, the electric field between the outer and inner conductor is given by Q: A. 2:35;) :>The voltage difference (defined from the conductor with negative charge to the conductor with positive charge) is found using {I a .01 V=- E d =-— d I ’0 p ”Zirsp p _ pf 111(b/a) _ 27%: V :5 Thus, the capacitance is _ 27rd: C ln(b/a) b.) Using the direct—current analogy, the conductance of the Eossy capacitor is given by (sea) c.) The energy stored in the capacitor can be found using the results found 1(cv2), in part a. Since UH = 2 (p!)2 111(b/a)h 4218 Ur; : Problem 2 (25 points): Two semi-infinite conducting planes are arranged at an angle (p0, as shown in the figure below. One plate is grounded and the other is charged to a potential Va. A gap at the tip of the wedge (infinitesimally small) insulates one plate from the other. Assuming that a volume charge density given by pv = )Usin(p[C/m3] occupies the region 0 < g1) < 990, a.) Find the potential in the region 0 < qa < gun, b.) Is your solution unique? Explain. ([30 ‘h; ‘ (I):0__—““—’X 5—)0 SOLUTION: a.) Given that a source exists between the conducting plates, the potential between the plates must satisfy Poisson's equation: Due to the fact that the conducting planes are infinite in p and z, we expect that there will no variation in the potential with respect to either of these variables. Thus, mag —— —0. 6p 62 The Laplacian operation becomes pz 2977 80 62 _ (/7 «90 2 2 62=_m:2+q ago 50 2 . :> (lb—n”"5’05“”0+C,¢+C2 6' 0 Next, we must appiy boundary conditions to solve for the constants C1 ENC! C2. c1)(i;a:o)=c2 =0 (BC #1) 2 . an»: re.) = WWW” =11 (BC #2) r) V. — 2 ' 3 C1 = “F“ :0: Sln \$0 Thus, the solution for the potential within the region 0 < (p < go” is given by 2 . 2 - = pop Slur/)0] Q +901) Slurp. 8' 0 \$0 80 b.) The solution is unique because it satisfies Poisson's equation in the region 0 < go < go“ and the boundary conditions. @( p,o) : o (I)(p’qpu) : V0 Problem 3 (25 points): r _.... A metal plate of area A [m2] lies a distance 0‘ above an infinitejggpund plane, as shown in the figure below. Assume that positive charge has been distributed uniformly on the surface of the metal plate. Find the capacitance between the metal plate and the ground plane. Neglect fringing effects. 42. yffiifflflll{fill/II!lifff/fftffflfft‘fTIT/J pg SOLUTION: According to image theory, the configuration given below will produce the same fields as those produced by the geometry given above, in the region z>0. VII/I’ll]!Ifllfffl’ffffff/I/fl/fklfulfill”!!! Since we can neglect fringing effects (ie. the bending of the electric field near the edges of the plates), the field can assumed to be constant between the two conductors. According to Gauss's Law, the electric field due to one of the sheets is ' given by U SE - dsmp + U SE - dsm = ﬂ p3 ds 3W ‘ ‘ of”, S ”8132;? .2613 + [Id—1332) -(—£)ds = H [3“. d5 SW. x ‘Ir Notice that the electric field is in the positive z—direction above the sheet and in the negative z—direction below the sheet. Thus, by superposition (accounting for the image), the electric field between the two conductors is given by Ez=—& for OSzSd. 8 We recognize that the electric field is equal to zero everywhere outside the plates (the electric field due to the top sheet is of equal magnitude but in opposite direction to the field due to the bottom sheet). From the electric field, the potential difference between the ground plane and the conducting sheet becomes Notice that, for convenience, the line integral is carried out along the 2- axis. However, it is important to recognize that due to the conservative nature of the electric field, the integral is path independent. Having computed the voltage drop across the conductors it is next necessary to calculate the total charge contained on the conducting plane. Recall that the capacitance is given by C=—Q». V Since the surface charge is distributed uniformly on the conducting sheet (given), the total charge becomes Q=Ipsds .3- :> Q = pS/l Substituting 0 and V into the expression for the capacitance, -15; d. C Problem 4 (25 points): An infinitely long cylindrical conductor lies along the z-axis and has an inner and outer radii equal to a and b, respectively. A current density given by J = 2pi’:iz:l m flows in the conductor. Determine the magnetic field intensity in each region. SOLUTION: Applying Ampere's Law to each region defined by the boundaries of the conductors (Amperian loops shown above): p<a: ﬁH-di sz-ds (.' S 3 §>H~dl=0 (i :3 for p<a a<p<b: p>b: §I-1-cu=ijds C 5 21,0 => fHﬁ- pdgéa = H(2p)pdpd¢ [1a for a<p<b List of Identities: 62A 62A 62A +—+ V2A= 8x2 ﬁyz 3.22 VQA 1 a 6A 1. 62A 52A =—— 10*— +7 2 1“"? pap 8p p 63¢) a; 2 16 26A 1 a _ 6A 1 63A VAx—Z—r-m—+2' “snug—+2.2 2 r 6r \ an r 8111660 06 r 811'] 06¢) ...
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ECE2317_Exam2_Summer_1999 - Mi Namew Signature Student...

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