Exam 2 - Spring 2011 - EE3340

Exam 2 - Spring 2011 - EE3340 - EE 3340 Exam 2 1 Texas...

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Unformatted text preview: EE 3340 Exam 2 1 Texas State University—San Marcos School of Engineering EE 3340 Spring 2011 Exam 2 — Mar. 3, 2011 Closed book and notes. You may use only pen or pencil, calculator, and the formula sheet provided. Express all numerical answers as a number between 1 and followed by the appropriate engineering unit (mA, A, mV, etc.) as needed. NAME—GMPIiogj-amfl #5 #6 f9 #7 "_ 1’18 #8 1’14 #9 {22 (Bonus [5) Total Multiple Choice: Put the letter corresponding to the best answer in the blank provided. 1. (4 pts.) The expression V ‘ A means (A) the vector that is the gradient of A (B) the scalar that is the divergence of A (C) the negative of the electric field associated with the voltage 141 (D) the vector that is the divergence of 121 Answer: ' 2. (4 pts.) The principle relating total charge inside a surface to the electric flux density on the surface is (A) Ampére’s Law (B) Coulomb’s Law ‘ (C) Gauss’s Law (J '; ' AT”) (D) Henry’s Law ‘““ Answer: C i EE 3340 Exam 2 2 3. (4 pts.) The ratio of current density .7 to electric field E in an ordinary isotropic conductoris J/ E = (A) permittivitye(Fm'l) - _ " 1 O. E (B) resistivity p(ohm—meters) X (C) undefined (D) conductivity cr(ohms'l meter“) Answer: 0 4. (4 pts.) Suppose a single point charge +Q is at the center of an imaginary sphere of radius r = l m. Now suppose the point charge moves from the center of the sphere to within 1 cm of the surface, but still inside the surface of the sphere. Which of the following statements is true after the point charge is moved near the surface? (A) g3 D ' (13 changes during the move, but E on the sphere's surface is the same sphere sud'ace u 'ng the move @ @D ' (15 is the same before and after the move, and E on the sphere's surface sphere sud’ace stays the same during the move (C) (5 D ' (18 changes during the move, and E on the sphere's surface changes during sphere surface the move (D) (fiD ' d5 is the same before and after the move, but E on the sphere's surface sphere surface changes during the move Answer: is Problems: When working the following problems, Show as much work as you can! I will try to give you partial credit for partially correct answers, but if all you write down is an answer and it's wrong, that is difficult. 5. (21 pts.) Forces fiom multiple point charges. A point charge A of +10 nC lies at the origin of the X—Z plane as shown in the figure on the next page. Four other point charges B, C, D, and E also lie in the X-Z plane, and have the values shown in parentheses beside each letter. EE 3340 Exam 2 3 B(+10 nC) (a) (4 pts. each) Find the force vectors PM, Fm, FDA and PM on point charge A from the charges B, C, D, and E, respectively. Write your answers as a scalar force multiplied by one or more of the unit vectors fix, 5?, or at, as appropriate. y : 61,01 1‘. fl I Z» A “k fish/(“Sc alliat‘L 1 .‘(| _ {Li A g. it.an )lelit ) ’— AM ‘5“ “l «'r (5.95% rte": )Q)’- (11.: wo"‘")00?~!0"” ') A ‘ "-—-—-—-——-—-——--— —- "8.17 Mex 5'” " Lirt' 1%%,3’1K(o“a)cr.'5)1* * 8N “ ‘ F = -$«;qs;n Md 8% . _ Us 1-16“on No") A BA £0?! 1 _‘ "M , 3 'AM 0V1- A H w Lsgamzc )L 13 pm: -%’->\%_'77 n“ m “— (Lash/o'iflto my") ,, 5 __ , h r ‘ A1? L) ka A w ~ .V C‘ F“ H a“ Lee-7’1 new 25a?“ is“? w ELEM": '41 F .__ 83ig-77 VIM 5:1 EA 9 (b) (3 pts.) Find the vector sum FTOTAL of the four forces you found in 5 (a) above. “it 5” ¥ 3’ “b A“ E c’ " PM 'L (Pi ‘ v.5 " $1.4. d _ I”: A FTOTAL: 0 Fr at“ 2 Was?) .Lu‘) 1' (-sm 77.1») r we; '27 s'IUXQLFS‘t‘SaTT m3 EE 3340 Exam 2 4 (c) (2 pts.) If the point A is now moved slightly out of the X—Z plane along the Y-axis, will it be stable (tend to move back where it came from) or unstable (tend to keep going)? Check one: CI Stable {Justable 6. (9 points) Field from sheet charges. Two infinite sheets of charge have the same charge density p3 : ~3 IMC m'z. The entire system is in a vacuum. One sheet lies on the y = -10 111 plane and the other lies on the y = +10 m plane as shown in the following figure: 3/220 m y=0 m y=+20 m Figure for Problem 6: charge density on both planes is p; = -3 “C m'z. (a) (3 points) Find the electric field intensity vector (magnitude and direction expressed as a numbe n ultiplied b e appropriate unit vector) at y = +20 to. 0 - a: .0“ ‘ "' 1 . a a - .—. _; 1:: at , L V. 132T?" Him * as, 3 ~ssaex L“ M E“ a -3 ~ 3 xx 0*" 1 .__. a ' ' — — asael - a 9&1, Rea Electrlcfield vector at y .. +20 m *m h .3 (b) (3 points) Find the electric field intensity vector (magnitude and direction expressed as a number multiplied by the appropriate unit vector) at y = -20 m. ~93 L “B l -( . I- L: 6 I" a A ‘1 ("b Electric field vector at y = #20 m = “if 3 3% - 5‘ 3- m 0‘ \fi EE 3340 Exam 2 5 (c) (3 points) Find the electric field intensity vector (magnitude and direction expressed as a number multiplied by the appropriate unit vector) at y = 0 m. Q z E ‘ ‘1 [-1 _ 2 (a 3.3% '33?!) 2Q Electricfield vector at y = 0 m = C) 7. (18 points total) Boundary-condition problem. A certain dielectric material has a relative dielectric constant er > 1. At a certain point on the surface, the electric field vector E 2 inside the dielectric makes a 30° angle with the (flat) surface of the dielectric as shown in the figure below: Vacuum £1: £0 / Diem 82?. are.) At that same point, the electric field vector E I in the vacuum above the dielectric material makes a 45° angle with the surface. There is no charge on the surface of the dielectric. Your problem is to find the value of the relative dielectric constant 8,. (a) (6 points) Consider the component vectors that compose each vector 1? l and E 2 as portrayed in the diagram below: EE 3340 Exam 2 6 Based on the relationships in each triangle, find an algebraic (not numeric) expression for the field magnitude ratios E T] / E M in terms of a trigonometric function of the angle 9;. Do the same for ET2 / EN2 in terms of 62. " .VD'I Yin" [113" 'l r 1'. E f“.'./\ \ ' Lit ‘9‘“ 1'“. i J lam." El 3 n ‘ a . , r = 8 'u Er.E~\ 7’ Z41 (1102. 1 gig—E3; ETI/Em i: L irfiér -. if LL*L : 9r EUJL (b) (4 points) Next, use the boundary conditions for tangential (ET) and normal or perpendicular (EN) electric fields to derive one algebraic equation relating ET, to ED, and another algebraic equation relating r3,w to EM, the relative dielectric constant of free space (1), and the relative dielectric constant e, in medium 2. Do this by completing the following two equations: ix 'r- ‘\ S“.J._V\ SIQIWL)1L KREE: EIK'I E11 1 “I ‘ l: u "' Et’ HT. I'ENI E r ‘ Eu E"; \ 'L (c) (8 points) Use the equations you have derived to obtain an algebraic expression for 5,, the relative dielectric constant in medium 2. It should depend only on the angles 6‘, and 62 (all the field variables should cancel out). Finally, insert the appropriate angles for 6} and 62 and solve for 5, in terms of a number (notjust a variable). 8r Lula-eht’qu—X (in 'lfl'rvvxto C91 .1 . . A -. I! “‘ '3 - , 81 'h' 1 1 {zit - L‘ Saw-‘9. -, E111L1 ‘3‘“91 1 “er 13‘? 2‘33, Eur. if‘ E-U\ ' EWL t“ E2133; Le *— 1 '3" at a A t EP‘ Lr L E 5ifle‘LAn Ef fi‘“efi’L' Esq“ t ' .' f” W'fl " L a. M‘ (r LNL 0 Ed ’fl’ - 1 r _()\{;T glsmi?‘ 1' Kit / L( (3 “'9I 1’2‘1‘" ‘ BE 3340 Exam 2 7 8. (14 points total) Conductivity problem. A certain coaxial cable was made with defective dielectric material having a conductivity of a: 10'6 S ml. The cable has a very low characteristic impedance, because its inner radius r, = 1.0 cm and its outer radius r2 = l. 2 cm. The dielectric is shown shaded gray in the following cross-section of the cable: _Q 5 (5' 1 IC (2‘ 1 \ CM The problem is to find the total resistance from the center conductor to the outer conductor of a length of cable M 2100 meters long. (a) (7 points) First, estimate the resistance in the following approximate way. The L resistance of a rectangular slab of material with conductors on opposite sides is R = —, 0A where L 2 distance from one conductor to the other through the material and A 2 cross— sectional area of the material. Using L = r2 - r, and A = total area of the dielectric coating in a IOU—meter length of cable, calculate the area by multiplying the cable length M (100 meters) by an average cable perimeter (average of r , and r2). Then apply the resistance formula. Insert numbers and find a number for R in ohms. fl _ A 1, - 7-H :55ch3 ‘11 k0 RuliL-‘A ~i M EE 3340 Exam 2 8 (b) (7 points) Next, find an exact expression for the resistance of the dielectric in the 100—meter cable by writing this expression for the radial electric field: E,(r) = E0(rI/r), where E0 is the (unknown) field at the surface of the inner conductor. Then integrate this expression to find the voltage V between the center conductor and the outer conductor. You will need the following definite integral: Then, use the expression relating electric field E , conductivity 0, and current density ,7 to find the current density in the cable. Finally, perform a surface integral at one surface (say the inner radius r,) to find the total current] in the lOOAmeter cable. Dividing V by I will give you the exact resistance R. Do all the algebra first and then insert numbers after you have found an exact expression for R. Your answer should be close to the one you got in part (a). [1,3,ka 1 Lr\\f\ .73- : Cf 1 _ AL 3 \j ‘ Ssr\ v \‘ Er. -:_. C, —. - E L‘ .f 1 r, 1 We} L w. rs ~. (33:? .1:— (1)" ( 1. ‘ 1‘ its (1'3 ‘ ‘35 U b) U u q 7H 77/”: 3; am: the flat—r : \ Efi.) 11¢“ O’ .3; EE 3340 Exam 2 9 9. (22 points total) Capacitor problem. Find the capacitance of a parallel—plate capacitor whose dimensions L = 20 cm, W: 10 cm, and T: 0.5 cm are such as to make it safe to ignore the fringing fields at the edges. (a) (7 points) Initially, the dielectric is a vacuum (3, = l): Find C in the case of a vacuum dielectric. d, 1 (i) (8-85‘K r56" [‘1 > (roam. t ‘cht-v‘\ (b) (7 points) Next, suppose the capacitor is filled with alumina (A1203) having a relative dielectric constant of e, z 9.6: o..-. (1 : fiat)“. til rib ) EE 3340 Exam 2 10 C(b): o 33% n V4 (c) (8 points) Finally, find the capacitance of the structure if the dielectric has been halfway removed from between the plates: U2fl P” To solve this problem approximately, assume all the electric field lines are vertical (go straight from one plate to the other), and the ETANGENTML boundary condition is satisfied at the edge of the dielectric that is still inside the plates. abet (1 (ti Leena e40 L 108‘“ "We-3 ____ (1l '(Z OiO’cvW- 1 1.7/7 F (1 .f GHQ-15‘! e-125_(/0m-r0m.3 7' /64,? P; C“): BE 3340 Exam 2 11 Bonus question (5 points) This question refers to Problem 9. Assume you charge the capacitor with the dielectric entirely inside it (part (0)) to a given voltage, say 10 V. CV2 Using the capacitivevenergy—storage relation U =- , you can find the energy stored in capacitor C(b). Call that energy U6. Now suppose the charged capacitor does not lose any charge on its plates when the dielectric material is removed (part b shows the halfway stage in the removal of the dielectric, and part a shows it entirely removed.) If you use the relation C : QjV and the fact that capacitance C(b) is larger than the vacuum— dielectric capacitance C(a), you will find that the energy Ua stored in the vacuum- dielectric capacitor is larger than the energy Ur stored in the capacitor with a dielectric constant greater than air. (You can check this with your results for Problem 9 to make sure). The bonus question is: where does the extra energy come from? {it}: (/rg’[(C-£rrc (it; 1‘" ...
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This note was uploaded on 09/21/2011 for the course EE 3340 taught by Professor Stephen during the Spring '11 term at Texas State.

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Exam 2 - Spring 2011 - EE3340 - EE 3340 Exam 2 1 Texas...

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