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Unformatted text preview: Physics 171.102 Exam 1 solutions Sept. 30, 2004
Prof. Barnett Take 1/(471’60)= 9 ><109 N  m2/C2, 60 = 9 x 1012 02/(N  m2).
Use a pen. You may use material written on one 3” x 5” index cards. 1. Three charges, Q1— — 1/10, 622—  2,110, Q3 # 300, are  .
at R1— — 1. 02, R2: —5 01' + 3. 03', R3: —1. 03' meters. Q2 2,2”.0 ‘ l .' ; _
Charge Q3 has mass M3: 59 . Q . . .. .. A) (10 pts) What is the force, 131, of Q1 on Q3? R2 = *501 + 30], Fi— — ,1 1%er — ——6.75 x 103 2 N R3 = —1.02 m
mo ~ ' r — 59
B) (10 pts) What is the force, F2, of Q2 on Q3? Q3 — 3130 A 3
F2 :4—7rleoQ122g3f2 A A F, Y};
:(1.73><10‘3i—1.30><10—3j)N ‘ 1%,};
C) (10 pts) What is the acceleration, d3, of M3? F
Tom L 53 = Fwd/1143 = (—1.004 E — 0.26 3) m/sa;2 2. The ﬁgure below shows a thin inﬁnite ﬂat plane of charge and a solid, aluminum inﬁnite
rectangular plate, both oriented perpendicular to the x axis. The ﬂat plane is at at : 0.0 m
and the aluminum plate extends from x : 1.5 m to a: = 3.0 m. The surface charge density
on the plane is 0(320) = +2 pC/m2. The aluminum plate has no net charge. The electric potential is deﬁned to be zero at m = 0, i.e. at the ﬂat plane, V(a: : 0) = 0.0 volts. A) (15 pts) What are the values of the electric ﬁeld, E, at the points x : —1.0, 1.0, 2.0, and 3.5 meters? I:
3 <
4 ll 4?
El for inﬁnite plane is (7/260 and points away 3 + H
from the plane for positive 0. There will be H 0
induced charges on the surfaces of the aluminum 1; +7?
plate such that inside the aluminum ]E]— — 0 1: + ,0
E(—1.0)=—1.11><1051'N/C S 3‘:
E(1. 0:) +1.11x1052N/C 3015;; E’(2 0)— _ 0 N/C + E(3.5 5:) +1.11 x 105 i N/C a .1ch B) (15 pts) What are the surface charge densities on the 3 sides of the aluminum plate, 0(z:1_5), 00323.0),
and the volume charge density inside the aluminum
at :1: z 2.0 m, p(x = 2.0)? To get [El— — 0 inside the conductor there must be the same amount of charge to the
right and to the left, includeing the ﬂat plane at :I:— — 0. That means that there 13 a total
of 1 [JO/7712 to the right and 1 11C’/m2 to the left. Therefore, 0(1. 5): —1 nC/mZ, a.(2 5): +1 110/7132.
p(2.0) : 0 because there is no net charge 1n a conductor. C) (10 pts) Given that V(a: = 0) : 0.0 volts, ﬁnd the
potentials at :1: = —1.0, 1.0, 2.0, and 3.5 meters. Vf—%:—fifEdf >09
Don’t simply say V 2 Ed. v;1 — V0 = —f01E’ 03F: —1.11x105 v zu I 5
V0=0—>V_1:—1.11><105V
Vl—Vb=—f01EdF=—1.11x105 v
V0=0—>V1=—1.11><105V The potential is constant inside the aluminum
plate because there is no electric ﬁeld there.
V15 — VD = — 01513  air": —1.665 x 105 v
V2 2 V15 because E = 0 inside the aluminum.
V2 2 —1.665 ><105 V V3 : —1.665 ><105 V V35 — V3 = — 335 E  dF: —0.555 x 105 v
V35 : ~2.220 x 105 V A gragh of V(x) is at right. 3. The adjacent circuit has four capacitors,
Cl =1/J,F,Cz = QMF, 03 = 3MF, C4 = 4/,LF,
and a battery supplying a potential of V = 10 volts. A) (15 pts) Find the charges, Q1, Q2, Q3, Q4,
on the four capacitors. Q1 and Q2 are in series with each other.
They each have the same charge. The sum
of their potential differences is 10 volts. tau '1 " The equivalent capacitor for 01 and 02 is V = 10 volts ' Q12 = V012 = (10 V)(0.667,u.F) = 6.67 X 10“6 Coulombs Q12 = Q1 = Q2 2 6.67 X 10‘6 Coulombs C3 and 04 each have a potential of 10 volts on them.
Q3 2 3 x 10“5 coulombs. Q4 2 4 X 10‘5 coulombs. B) (15 pts) Find the capacitance of a single
capacitor which would be equivalent to
these four capacitors in this circuit. 01234 : 012 ‘i‘ 03 + C4 = 7.667 pF. A L ‘J F" [[U 1/ r4
P L. H Yr? V :' Co.“ DTPNT ...
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 Barnett

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