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Unformatted text preview: E & M SOLUTIONS #' 1999 Physics C Solutions Distribution
of points
E & M 1 (15 points)
(a) 4 points
For [Ring the relationship between potzmial and charge 1 pain:
_ 1 2.
'— 47t£° 7'
Solving for Q:
Q = 41reoVr
For correct substitutions for the potential and radius 1 poim 90 = Mum—2000 V)(0.20 m) or (2000 vx0.20 m)/( 9 x109 Nnf/c1 )
Q1 2 —16001r€°C or — 4.4 x10" C Forth: correct magnitude of go 1 point
For the negative sign 1 point
(b) Spoims
LFoa'indicatingthatthcdecu'icﬁeldiswo Ipoim
ii. Theehargeonﬁnsphqccanbctrcamdasapoimchargeatitscm
1 Q0
E = ——
47560 ,r1 a.
E = (9 x 10" Nnf/c2 {ii$43] 3965 599.5 
E— r2 c or r1 C whuensmmetus
Formyoftheaboveaqnessionsme lpoim
ﬁi.ForindicaLingthatthe¢lecu'icﬁddism lpoint
iv.Fmindicatingthatthedearicﬁddiszao lpoim ForhavinganfananswasomrectORfcrsmnemmﬁonofusingthcmclosed
chargeORfusomemmﬁonofGauss’law Ipoim E 8: H SOLUTIONS #1 1999 Physics C Solutions Distribution
of points
E & M I (continued) (c) 3 points 5
AV=D;—V‘=j.Edr For recognition of the need to take the difference of the potentials at radii a and b, or for writing the deﬁnite integral (with limits) 1 point
b
90 dr
[Ar/I ' 41:60 I r1 o
47:60 : a [AV] = Q“ _ b 42150 17 a
For correct substitution of variables or numerical values for Q, a, and b 1 point
For the correct answer 1 point
59.
[AV] — 81mg 01' 1000V
(Alternate solution) (Alternate points)
For recognition of the need to take the diﬁ‘erenee of the potentials at radii a and b 1 point
AV = V‘  V.
99 1 Qo 1
M  [a]  (a)
For correct substitution of Q, , a, and b I point
90 I 1
AV " 47:60 (E — E]
For the correct answer 1 point
590
(Alternate solution) (Alternate points)
V = g
C
For using the above relationship I point
For substituting QL1 from part (a) and C from part (d) alternate solution I point
For the correct answer 1 point
lAV = SQ” or 1000 V 81m, E I M SOLUTIONS #‘I 1999 Physics C Solutions Distribution
of points
E & M I (continued) (d) 2 points 0 C=='—°— V
For using the above relationship 1 point
For substituting Q,0 from part (a) and AV from part (c) 1 point
C = 4.4 x104 C
1000 V
C = 4.4x10‘" F {Alternate solution) (Alternate points}
For writing the equation for the capacitance of the spherical capaCitor 1 point h 4369a!)
C ‘ b  a (0.02 mX0.04 m)
(9 x 109 NmI/c2 )(o.04 m — 0.02 m)
For the correct answer I point
C = 4.4x10‘“ F C: For correct units on two answers and no incorrect units 1 point ...
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 Spring '09
 Park
 Physics

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