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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—GMPIiogjamﬂ #5
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(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 XZ 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
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F .__ 83ig77 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 Yaxis,
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.
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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) Boundarycondition 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 Lulaeht’qu—X (in 'lﬂ'rvvxto C91 .1 . . A . I! “‘ '3  ,
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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 crosssection 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
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: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
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.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) (885‘K r56" [‘1 > (roam. t ‘chtv‘\ (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‘“ "We3 ____ (1l '(Z OiO’cvW 1 1.7/7 F (1 .f GHQ15‘! e125_(/0mr0m.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.
 Spring '11
 Stephen

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