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Unformatted text preview: DO NOT BEGIN THIS EXAM UNTIL TOLD TO START Name:
Student Number: Instructor: ELEE 2317 Applied Electricity and Magnetism
Exam 2
Nov. 13, 1999 . This exam is closed book and closed notes. No calculators or
computers of any kind are allowed. . Show all of your work. No credit will be given if the work required
to obtain the solutions is not shown. . Perform all your work on the paper provided. . Write neatly. You will not be given credit for work that is not easiiy
legibie. . Show units in all of your final answers.
. Circle your ﬁnal answers. . if you have any questions. ask the instructors. You will not be
given credit for work that is based on a wrong assumption. . You will have a total of 90 minutes. [20 Prob.1 [20 Prob. 4
120 Prob. 2 120 Prob. 5
I20 Prob. 3 Total /1 00 Problem 1 (20 pts) A possible electric field is given as E = 2yi + l5)» + sin yg. a. Determine VxE and use it to decide whether integrate of the form dﬁ are independent of the path. ’ A n
(3 I H l! I. VxE= For this problem line integrals of E (circle one) (are, are not)
independent of the path because (give reason) b. If dt? is path independent, find a potentialmsuch that E = —V(D.
(? (1):“; . c. Evatuate the line integrai ZaM where C is the path from (0.09) to
C, (2, 1220). Answer Problem 2 (20 pts) A conducting sphere of radius a is surrounded by spherical dielectric shells of
radius b and c with permittivities of a, and 32, respectively. A conducting shell of radius c surrounds the dielectric shells. 1“ » ,.' {,3th .7 {IQ“ﬂ k) LJ r. , if) _ g . 5i g.
_1 . f :  '4'. ' , r L .l I a L? "‘
rTIw ‘ . 
J a...“ , 1 I _ I_ F _ . ,_ _ I
I  ' f. .fg!§§31}':‘ '3'
Plane win \ r 'r {.F ‘
’l
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: f; ' J ’i —
i i , g i , ___, t") , I '4“ 2 /
.1 "\_ "’ .
’ _ F 5r r Pl“. f l‘t' /
J‘ﬂf—M t“ i 75." r
I’M,” a. Assume a appropriate charge on the conductors and find the electric
fields in the region between the two conductors. Charge assumed on which conductors: r" ' n E in region 1: E in region 2: b. Determine the potential difference between the inner and outer
conducting shells. A 1 r
. . "_H_ i 7 {':,»/\ ' f .‘y.. A
Potential drfference: in ~ 2‘. '4 52 i (7 l/ (/ c'. Use the definition of capacitance to determine the totai capacitance of
the structure. Capacitance: The capacitance can be written as the (circie one)(series, para/tel) combination of
two capacitors. d. Find the total energy stored in the system (by using any method). Stored energy: /? if?) " 3 “We? 3
ﬂ \‘h ‘_.‘r a /
l PW
if / M;
£0 “Hf .1
I’llﬁt.
1'! i
27/
A _;. if. .
f I " f r ’V F‘ L t {hfPHI” fh~p/}/‘/ (F/
; r» .
a r 
{a  L r? x? r‘. " 5. (’ 5’11 :’ ’ r’: h ”
"1‘ " r I» (Pi) ! " I;
,. .
C y I 1 a (a t
L I)‘ .II' 1 f I}! I i l)
y ‘ a \ IL {f z I,
q . ff» : (r '"I Problem 3 (20 pts) A spehrical conductor of radius a is suspended at a height h above an inﬁninte conducing ground plane (I: is the distance from the ground plane to the center of the sphere). The air
outside the sphere has a dielectric breakdown of 3.0 [MV/m]. The height h is much larger
than the radius a, so that the charge density on the sphere may be assumed to be uniform. 1. Determine the maximum charge Qmax that can be placed on the sphere before
dielectric breakdown will occur. (Do not neglect the effects of the ground plane!) 2. Determine the voltage drop between the sphere and the ground plane when a charge
Qmax is on the Sphere.  _ r I 7 %,‘T/.“_\ Problem 4 (20 pts) A hollow Spherical shell of uniform charge density p5 [C/mz], having a radius a, is in
free space. 1. Determine the potential function both inside and outside the shell, assuming that the
potential at the center of the Shell is zero Volts. 2. What is the voltage dr0p between the center of the sphere and a point outside the
sphere at a distance r = 2a from the center? (That is, calculate VAB, where A is at
the center of the sphere and B is at r = 20 .) /
\ 10 Problem 5 (20 pts) A slab of uniform charge density pv [C/rn3] is situated in free space above an inﬁnite perfectly conducting ground plane (at x = 0), as shown below. The thickness of the
charge slab is h. The potential at the ground plane is taken as zero Volts. 1. GM: a convincing argument as to why the electric ﬁeld above the slab (x 211) should be zero. 2. Use the result from the previous part, along with the boundary c0ndition on the ground plane, to solve for the potential function inside the slab ( 0 < x < h) using
Poisson’s equation. . /‘  ' r  . ~ 0‘ /J\ "'c
f » a {rt f I) FF.
 i, .(_ J _ j j a
/I . if. I
me i ~ ‘ ‘ ~ . Z (1) '3 ? if \/ Hi: I: ;m " t T) 4/ J '
_. .M In, I
I“ Pp!) ' _ I
~ ‘> .f r n r : t ". .
r E x . ,. . n . . ._ 1.
.“I {‘de .. _ “(an i )de f f {MUN (tiff, ; Kt (\qu / . ...
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