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### UnitExam1Dsolutions

Course: PHYS 2B, Summer 2008
School: UCSD
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Word Count: 793

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D UCSD ID: Physics 2B Summer Session Answer Section MULTIPLE CHOICE 1. ANS: C Unit Exam 1D Charges, Force &amp; Fields By definition, E = N F 6.9 102 N = = 2.1 10 5 -3 C q 3.3 10 C TOP: ELECTRIC FIELD 2. ANS: A ^ ~ ~ ~ -4 ~ charge Q ~ 6.60 10 C 3 ~ = 5.29 C ~ = = = 3 ~ volume 4 4 (3.14) 3.10 10 -2 m^ ~ 3 m3 ~ ~ R ~ ~ ~ ~ 3 TOP: VOLUME CHARGE DENSITY 3. ANS: B For a sphere with...

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D UCSD ID: Physics 2B Summer Session Answer Section MULTIPLE CHOICE 1. ANS: C Unit Exam 1D Charges, Force & Fields By definition, E = N F 6.9 102 N = = 2.1 10 5 -3 C q 3.3 10 C TOP: ELECTRIC FIELD 2. ANS: A ^ ~ ~ ~ -4 ~ charge Q ~ 6.60 10 C 3 ~ = 5.29 C ~ = = = 3 ~ volume 4 4 (3.14) 3.10 10 -2 m^ ~ 3 m3 ~ ~ R ~ ~ ~ ~ 3 TOP: VOLUME CHARGE DENSITY 3. ANS: B For a sphere with radial symmetry, we can take the volume element to be the product or the surface of a spherical shell times its thickness dV = A surface dr = 4 r 2 dr. We integrate the charge elements dq = (r)dV (r) as spherical shells since the density depends only on the radius r Q = (r)dV = (r)4 r 2 dr = = R 0 o 4 o r 4 r 2 dr = R R R r 3 dr 0 4 o R 4 ~ C ^ ^3 ~ = R 3 o = (3.14) 1.31 10 2 m~ 7.29 10 -5 3 ~ = 5.15 10 2 C ~ ~ R 4 m ~ TOP: VOLUME CHARGE DENSITY 4. ANS: A Fundamental properties of conductors: Electric field is zero inside Electric field is perpendicular at surface Excess charge must reside at surface only TOP: CONDUCTOR FIELD CONCEPT 5. ANS: D This is a conversion problem, so use the identity e + = 1.60 10 -19 C to create "unit fraction" # e+ = ^ ~ total charge ~ 1e + ^ 12 ~ ~ ~ = 985 10-9 C ~ ~ 1.60 10 -19 C ~ = 6.16 10 charge/proton ~ TOP: ELEMENTARY CHARGE 4 ID: D 6. ANS: C N m2 ^ 2 ~ E = E A = EA cos = 73.1 10-6 N / C ~ (26.2m) cos (61.1) = 2.43 10 -2 C TOP: Electric Flux 7. ANS: A For a single charged sheet, the field points perpendicular to the plane - away if positive and towards if negative. 1.80 10 -9 C / m2 N E= = = 1.02 10 2 -12 2 2^ 2 0 2 8.85 10 C / N m ~ C ~ Note that the distance r doesn't matter! TOP: FIELD 1 SHEET 8. ANS: A The electric field from a point charge is E = kQ r2 . Outside a charged spherical shell, the field looks the same as if the whole charge were located in a point at the center. Halfway between the shells, the distance from their center is 3 R. Hence the field is 2 kQ kQ 4 kQ E= 2 = = 2 9 R 3 r R^ ~ 2 ~ ~ 3 ^2 ~ The other way to do this is to construct a Gaussian Surface at r = 3R / 2. The enclosed charge is +Q and the surface area A = 4 r 2 = 4 R ~ = 9 R 2 . The electric field is 2 ~ ~ ~ +Q 1 4 kQ 9 2 ^ = 9 R 2 where we used k = 4 0 ~ 4 0 R ~ 4 ~ ~ ~ Q enc E= = 0A TOP: ELECTRIC FIELD SPHERICAL 2 ID: D 9. ANS: C This is analogous to the case inside the non-conducting sphere. Instead of using the "canned" formula k E=2 for which we'd need to calculate the linear charge density, let's construct a Gaussian Cylinder R with R radius = 4.9 cm and arbitrary length L. The surface area A = 2 RL while the volume VOL = R 2 L. The enclosed charge is Q ENC = R 2 L Now apply Gauss's Law: 4.9 10-2 m^ 0.200 10 -6 C / m3 ^ ~ ~ ~ ~ Q ENC R R 2 L N E= = = = = 554 C 0 (2 RL) 2 0 0A 8.85 10-12 C 2 / N m2 ^ ~ 2 ~ Note: the outside radius of the cylinder is not used. TOP: ELECTRIC FIELD INSIDE CYLINDER 10. ANS: D ^ -9 ^ -9 ~ ~ 2 ^ 63.0 10 C ~ -35.0 10 C ~ ~ q1q2 ~ 9.00 10 9 Nm ~ ~ F=k 2 = = -3.18 10 -4 N ~ 2 ~ 2 ~ -2 ^ r C 25.0 10 m~ ~ The minus sign shows that the force is attractive, as you'd expect between unlike charges. TOP: COULOMB FORCE 11. ANS: B To apply the line/cylinder formula, we need the linear charge density . Pick some length L for a sample section of cylinder. Then the lateral surface area of that section will be A = 2 RL and the total charge is the area times the surface charge density Q = 2 RL . Q = 2 R so that L 2k 2k2 R 4k R E= = = r r r Now = ^ ^ ^ 4 9.00 10 9 Nm2 / C 2 ~ (3.14) 19.00 10-2 m~ 27.0 10 -9 C / m2 ~ ~ ~ ~ kN = = 1.87 10 3 N = 1.87 -2 ^ C ~ 31.0 10 m~ The much easier way is to apply Gauss's Law from scratch: Charged Cylinder Q = 2 RL Gaussian Cylinder Area A = 2 rL Q 2 RL R E= = = 0 A 2 rL 0 r 0 TOP: ELECTRIC FIELD CYLINDER 3 ID: D 12. ANS: A Make a ratio of the two forces using the fact that the only change is that the new distance is r new = 1.95 * r k F new = F q1q2 r ^ 2 2 new ~ r ^2 ^2 ~ ~ ~ F = F r ~ = (33.0N ) 1 ^ = 8.68N ~ ~ ~ ~ ~ = ~ ~ 1.95 ~ new r ~ r ~ ~ q1q2 new ~ new ~ k 2 (r) TOP: COULOMB FORCE CONCEPT 13. ANS: B ^ -6 -2 ^ ~ ~ Q ENC D 1.61 10 C / m~ (3.14) 26.3 10 m~ N m2 E = = = = 1.50 10 5 0 0 C 8.85 10 -12 C 2 / N m2 TOP: ELECTRIC FLUX 14. ANS: C The charge for each species is the number of particles times the charge per particle. The total net charge is the sum of these: ^ ~ ~ ~ Q net = n p q p + n e q e = n p e + ~ + n e e - ^ = n p - n e ^ e + ~ ^ ^ ~ = 2.90 10 22 - 9.01 10 23 ~ +1.60 10 -19 C ~ ~ ^ ^ ~ ~ = -8.72 10 23 ~ +1.60 10 -19 C ~ = -1.40 10 5 C TOP: ELEMENTARY CHARGE 15. ANS: A kxQ kx (2 a ) 1 E= = = 3/2 3/2 4 0 2 2 2^ 2^ x + a ~ x + a ~ ~ ~ x (2 a ) xa = 3/2 2 2 ^3/2 2^ x + a ~ ~ ~ 2 0 x + a 2 ~ 63.0 10 -2 m^ 72.0 10 -2 m^ 2.13 10 -4 C / m^ ~ ~ ~ ~ ~ ~ = 3/2 ~ ^ ^2 ^2^ 2 8.85 10 -12 C 2 / Nm2 ~ 63.0 10 -2 m~ + 72.0 10-2 m~ ~ ~ ~ ~ ~ ~ ~ = 6.23 106 N / C TOP: ELECTRIC FIELD RING 16. ANS: A This is the most basic stuff TOP: CHARGE CONCEPT 4
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