1999 E&M C1 Solution

1999 E&M C1 Solution - E & M SOLUTIONS #' 1999...

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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 N-nf/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'icfieldisw-o Ipoim ii. Theehargeonfinsphqccanbctrcamdasapoimchargeatitscm 1 Q0 E = —— 47560 ,r1 a. E = (9 x 10" N-nf/c2 {ii-$43] -3965 599.5 -- E— r2 c or r1 C whuensmmetus Formyoftheaboveaqn-essionsme lpoim fii.ForindicaLingthatthe¢-lecu'icfiddism lpoint iv.Fmindicatingthatthedearicfiddisza-o 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 421-50 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 N-mI/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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This note was uploaded on 01/22/2012 for the course PHYS 2425 taught by Professor Park during the Spring '09 term at Cy-Fair College.

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1999 E&M C1 Solution - E & M SOLUTIONS #' 1999...

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