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Unformatted text preview: " 7 FINAL EXAM 7 c. GENERAL PHYSICS II SPRING QUARTER 1W ; _? June 7, 1999 Ke¢‘°,{;,.\ 4 StudentID# 4 ﬂ 9M6“ Lecture Instructor Section Number 1/41teo = 9 x 109 Nm2/Cz E0 = 8.85 x 10‘” €le m2
1.10 = 4 TC x 10‘7 Tm/A
mp = 1.67 x 10'27 kg
me = 9.11 x 10"l kg I.“ C = 1.6 X 1019 C 1
Full credit will only be awarded to answers with all work shown. 3 7 . V \f ,5;3+J2;;—0 s? ;,= 5/;2’6‘5
(I m e ' , ) 2
ggrcﬂ. :6) = 1.1+;3 : 9/124?) A3 1/12 3 ~ Q
S $9,. «SJ‘5, 4 jib‘5: "ATS a: for )e‘Pr l’gifﬂ" d
U 1.‘ ME;
 ' “iv :0 .
“5' '7 )2/2 453 ' _ a ___
43 H 9: _ LN, M: WM} 9.3:“;
 ‘3 T " ﬂ  — ‘  . A i ® 23 = 0.2é3A’ 41 ' 5/12"9 0 Hz 1 V53 3 EL) ‘12?) K‘ +82.
f (5) V”, : W,— [9.262A>[$n§ +2v
: ; V.
K?) : £2 : 3V' w? For the circuit shown below, Rl = 5 Q, R: = 2 Q, 81 = 8:: 3 v, 83 .__ 84: 4 v. a) Find the current through the battery 84.
b) What is the voltage across AB?
c) If a l pf capacitor were connected between B and C, how much charge would accumulate on it at equilibrium? ma . ‘ " {21.141.12031
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a, R,
mfgiw—l' L1
R2. 3 81 K] 83 521+;L3 fl “5%; +"/V+ BVZIUJ ‘3V— 52. " —5;3 ~25, +Hv=0 i‘Po‘r ‘rs'rth Lag! cvsa : ()«DiBVJ = ’ 344,; g: 3 0‘ {at “1’5
‘—————_’"—/
2. For the circuit shown below, answer the questions providing the reasons for your
conclusions. At t = O, the switch is closed.
a) At t = 0, determine the current in the inductor,
b) the current in the 100 Q resistor, c) the current in the 200 Q resistor,
d) the potential difference across the 100 Q resistor, e) the potential difference across the 200 Q resistor,
D the potential difference across the inductor,
g) fort — no, determine the quantities listed in a) thrOugh D. ,W 4 «3 lmtovo
H as Lenore
H c) 1130(0) = (00 _ A‘ 1:30 \\\w(03 = Rico—L\O°(°}: O .§A H e) “10°(o\: \OOV
. a “dam: \oo\l a. 1‘ yﬁwﬁmswcr
= \A l 1:) :wa’B: \A it) Xm(“>:‘SA W \1,.,eteo= “3°” m \luot‘u\= ‘°°" pa \LLJCN\1 O : t 3. An LC circuit includes a 0.025 mF capacitor and a 340 ml! inductor. ‘——"' a) If the peak voltage on the capacitor is 190 V, what is the peak current in the inductor?
‘ b) How long after the voltage peak does the current peak occur? c) At an instant when the magnetic and electric energies are equal, the current is 540 mA. What is the instantaneous voltage across the capacitor?
d) What fraction of a cycle passes before the energy in the inductor falls to 1/4 of its peak value? L 2
L333“ 9,0
C
i— L: .L
LC. E \ L
A _ .02: 'l“ 0 ._\‘G3l\
LN" (ml ‘5‘ 7 2
7 ._ 'Es‘koiuo 3 3"" 940 M53 : Q 3 \/
6/ V  eagle‘s C B
"‘ .. \ . \ J L — "ii—'13:»; L‘ g L — BabeVt) : . 7 5 99703 4. A Certain 2 dimensional electric field has an electrical potential which depends on position
according to the relation V(x,y)=3x3+2xy4
where x and y are the coordinates in meters of the point at which the potential is V Volts. I: a) What is the electrical potential at the origin? . b) How much work in Joules is required to move a charge of 1 nC from the origin to the point (x, y) = (0.5, 1.5)? . c) With the charge in place at this point, what is the potential at the origin now? a d) What is the potential at the origin if thel nC charge is moved to the point (x.y) = (1.5. 0.5)? Slﬁce f‘ in a? SCUM— aArweu' is ? Mlle.) L c)
Q . " .I _
K V ~ v “that: J ’J" W 6311». 29 Fol'rJi‘S 5. A parallel plate capacitor has air between the plates. It is ﬁrst connected to a 12V battery
and charged up with 6 [JG It is then disconnected from the battery. a) What is the capacitance? b) How much energy is stored by it at this point? c) While still holding the charge, the capacitor is n0w physically altered by having its plates
drawn apart to twice their original separation. What is the new capacitance? d) For part c) what is the new voltage across the capacitor?
(Remember, the charge was not allowed to leave the plates even though the plates were
moved farther apart.) e) For part c), how much energy is now stored in the capacitor after the plates have been
moved farther apart?
0 In a few words, explain any difference in energy between part b) and part e).
a (. Q) E ' £QVI 124 ero—exlélr I ® l1\\CE"J.‘3G‘1. Lilia“)! RJL (MgJ; (“My {’0 y‘ukcl) ﬁd‘f‘ﬂ‘rj ‘V‘W‘ invalid dam3'» So Hoke/i t t‘jrvwf Cf (“3%. t 6. An insulating sphere of radius a and charge +3 Q, uniformly distributed throughout its
volume, is concentric with a solid spherical conducting shell of inner radius b and outer g radius c. The shell has a net charge of 5 Q. Find the expressions for the electric ﬁeld as a ' function of the radius r for: (a) O<r<a ‘ ' (b) a < r < b
7 (c) b < r < c
(d) c < r.
l ?.‘r§ / (e) What charges reside on the inner and outer surfaces of the conducting shell? on) gin: 9% Eﬂnrr“) : (3%:ij 4
r ) ’7 3 Q r 3 M Whigs, ., \T\
\\ ., ’
 7.A a» ' @ '30 Pa '5 “T5 The ﬁgure below shows the cross section of a long coaxial cable. The inner conductor
consists of a solid conducting cylinder of radius c. The outer conductor consists of a
cylindrical conducting shell of inner radius b and outer radius a. The two conductors carry
uniform currents I that are equal, but flow in Opposite directions. Derive an expression for
the magnitude of the magnetic ﬁeld B(r) for the following regions: I'<C ..e
c<r<b
b<r<a
r>a. :’ ’u'O km!
7.
5(11'1‘) » MOVI( 51>
\
: 452.er i:
E zrrq‘
5(2WYB 3 m1
5 z: 19;,
27W ,5?)
F[¢7H’> = MoI.,_.f<°l.°1.,__(t_4»::"‘_
 " 143‘] ;
;§ﬁ5 A¢i§ﬁY‘Zil ' a’byl Three identical point charges of q = +6 uC are ﬁxed at three ends of a square with sides of length L = 20 cm. As shown in the ﬁgure. What is the magnitude and direction of the electric ﬁeld at the center of the square?
If a point charge of +2 uC, having a mass of 50 g, were released from rest at the center of
the square, what would its speed be when it reached a very large distance from the square? 'E ;' E+E+ga = E: anca i=3;
95' “lag: “yak—«A = 55.0 g 1‘
. a q 6 F b 1
)El = 4/ = ’09—‘33 “009) :;’z.’+x1°y/
[m go (02“) :— L‘____W \ Vzk .. e
r e .1: 9
: {mo "f; 13%???“ C)  inns/o y mp Jmtofﬂ) WJES(
g‘V 1 J71)“ N1 5—,.
M I —. A, ‘ zfzxzﬁbcmrwbwo “4) 2
\ l w y‘ a. 050 59. 9. s7 #75 @ust after S is opened or closed W; C 5' 72375:)5/ 520657731.) Circle the correct answer for each of the following questions. A straight length of currentcarrying wire is in a uniform magnetic ﬁeld. If the wire does not
experience a force, / a) that is as it should be :0 he wire is parallel to B c) this is impossible d) the wire is perpendicular to B e) there must be an E ﬁeld in the opposite direction of B If the current through a long solenoid is doubled while the coil’s length is also doubled, keeping the
number of turns constant, the magnetic ﬁeld inside thﬁolenoid is a) four times larger b) twice as large nchanged 5 = MD M e:
d) half as large e) onefourth as large E ; M0} L ,2,
In Ampere’s Law the integration must be performed over / A, 1;
a) any surface b any closed surface 5 '3 M o /
c) any path y closed path M
c) any closed path comp etely surrounding all the current In an overhead straight wire the current is north. The B ﬁeld due to this current at Our point of _: servation is
Q est b) east c) south d) north e) down The lines of B inside the solenoid shown in the ﬁgure are;
a) clockwise circles as one looks down from the top of the page oward the top of the page
toward the bottom of the page
e) no direction, B = 0 b: counterclockwise circles as one looks down from the top of the page The bar magnet in the ﬁgure is moving at a constant speed toward the coil. The current measured
across the resistor a) ﬂows from A to B ﬂows from B to A c) is zero 4= A B In the circuit shown, there will be a nonzero reading in the galvanometer G only
a) just after S is closed ‘ b) just after S is opened c) while S is kept closed d) never l ...
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 Spring '08
 Kogan
 Physics

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