This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: UNIVERSITY OF CALIFORNIA College of Engineering
Department of Electrical Engineering and Computer Scences
EECS117
Midterm 2 Examination Spring 2010 T.K. Gustafson ; Kedar Patel
April 19 2010 PRINT YOUR NAME:
SID. : SIGNATURE: Do your work on the exam. If you do need to use extra sheets
attach these to the exam so that these are considered as well. Make your methods clear so that partial credits is possible. There are two problems
THE PROBLEMS ARE EACH WORTH THE SAME 2 Problem Number One)
Consider a point charge Q located a distance h from the origin along the z—axis. Z Point r Char e
at zg=% a) What is the electric potential at point P in terms of 71 ? Vm = Q.
“mew—k b) Express the electric potential of part a) in terms of r and 6 using the cosine law ( r and
6 are the spherical coordinates ) ,‘
/"b wt: Lr‘l—W‘ — 1H4 Lege\ \ILP\= Q lI
‘mec (WW: wk c036) 1’ A second charge —Q is now placed at Z : —h.
c) What is the total potential at P due to the two charges in terms of 71 and 7‘2, the distance
from the second charge to the point P “a 3 Now the Charges are made to oscillate by having a sinusoidal current ﬂow between the two
Charges. ( I : [elm/2 + 0.0.) ~
d) If the Charge is represented in phasor forni ( q : (jam/2 + 0.0 ) how is (1 related to I? Com—Hm v C’E‘ﬂ EcLUanovx CW <7 » I «— 645 : o
Q N i 3 t
Qﬁ J E “kc'9 \A “4": 2,15de OVQV‘ CW.“ $PLLQ.V‘{
t cw: 4V Xq : aw =g1Ag a; Q, dcQJ.
a t ‘K‘ C  G .7, a \ 1,
Ua.’
—— Q 3 K‘s“ Lméaiiaﬂut L; W
‘ —v .2 "' g {— C   c 3’: 'x l : ~
e) Obtain the ﬁrst non—zero term in the potential (the dipole approximation) (new for the oscillating Charge pair. gout—kn. unobkmj \lC\°\‘ 3 tie, _Le,
‘Eﬁée / \'.VV\e. e. au W” Va; e'dﬁue. dALLL e. ‘lco OQUJTP t mm
_L Qicti—hwgeQ 3 a:  LCV— lease)
(1, V— v 'b r
' _ e — \r .
g‘bkv“; 3:)‘ka $w§\ﬂ&LKC\+J~L\\/\COSG7
 \u
SKMfKOLV I kw _
QTLKW"; 2, Q’)“ L ( —— Co$6\
' t ._ 'kv‘ .
\lLC’lg g L bk” 9),” K gr. hLﬁSewC\'\LK<\’\ (1396‘
«en V W x» cm“? — U —E toga“ “1” “‘36”
Vz/A—f/ — 9.. v L {1330}: 2:);kv L °‘\’_\. («9%93 + (30%)
\ ~ ‘E’Cxe \C \C
M— 4 Problem Number Two) 7
Consider an inﬁnite line current [1 : Iéeiwt’kz Take to : 0 and k: : 0 for parts a), b), P
M Mw/z
I1 “’X I2
<——— d —> Currents are in
z direction 0), and d).
a) Using Amperes law, What is the magnetic ﬁeld as a function of 71, the radial distance
from the line current, [1 7 Ecgu:y§_\& l.
1L,er ix b) A second parallel line current f2 : —Ié’eiwt+kz is located a distance d from the ﬁrst. What
is the total magnetic ﬁeld as a function of X in the plane of the two Wires for w : 0 and ‘= ﬁe .1. — Ix "
T5 5} (sleigh n: d/ +50 3) wk: 3—39 5 c) Calculate the magnetic flux between the Wires for a differential distance A2 and —L <
at<+_L_3WhereL<d. Z ’1.
gangsQ“ Ly L4) — w (‘14—‘13) ; 5&1 m {our r. A;
Tl L (3: L4,) @563 "/Q 1T cl
d) Using c), obtain an expression for the inductance per unit length F\Ux: L 4‘1 I ._I .. L: \V\ OLE:
tr M e) Does the result in d) change When w 71$ 0. State a reaon for your answer. D663 W01? WWIta; . Two wKVeQ‘C\*eMA
gvvfuvhk TE M Nome Se AcVCLVLaVeN‘Ke,
Quick tbquhw {3% a. JroJrCc. $0 \u'li'cwrg Tku$ L Vemem—s Hue. gamma. f) Is there an electric ﬁeld When w 71$ 0 7 If so give its value ( and direction ) along X for
y=0and —L<a:<+L. Cowk{wui'tﬁ q t T. f @573 :0 Gian/munch
* A: m _
Do \Me dmavcse $3 = E E :— 1/“?Qlia—usgygl
Tke. ‘6 v CTOLARQS Law,
W \ m A 95th ~Jikz. E II
<16?
b
I, I
I, ‘r’ [0 Problem Number Three
A thin Wire carrying a current I is located parallel to a conducting plane and a distance d from
it. Determine the direction and force per meter on the Wire. [Hintz use the method of images] \\’\C)‘kJC€—‘l v QJV Q {ck ‘auV 2:19 Lomdek‘l‘Ag
QUUUCW " /' $mr$<1££ Qumrde 16* e—ﬁ :H A We .c
H E vaoJeud“ 4w V‘e.$u\."ck‘.vx QU‘OM q_vx
“Mg, 2, QUthwk TI wbxcdn w\ keg Us»: 3 I
gow+\\A\/ou&
1" 4
x ggq R—mce Quwexfﬁ
Q,quva QM," Force, \$ 'HMJS KL, 1 won‘tko V‘L‘QUKSK‘UL
(aftr 164 ‘M ...
View
Full Document
 Spring '09

Click to edit the document details