Exam 1 Fall 2004 - Physics 171.102 Exam 1 solutions Sept 30...

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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 10-12 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 = *5-01 + 3-0], Fi— — ,1 1%er — ——6.75 x 10-3 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—%:—fifE-df >09 Don’t simply say V 2 Ed. v;1 — V0 = —f0-1E’- 03F: —1.11x105 v zu- I 5 V0=0—>V_1:—1.11><105V Vl—Vb=—f01E-dF=—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 = — 33-5 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.“ DTP-NT ...
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