ID: B
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ID: B UCSD Physics 2B Answer Section Unit Exam 1B Charges, Force & Fields MULTIPLE CHOICE 1. ANS: B N ˆ˜ Ê ˆ ÊÁ τ = p × E = pE sin θ = ÁÁ 35 × 10 −9 C ⋅ m˜˜ ÁÁÁÁ 2.2 × 10 −3 ˜˜˜˜ sin165° = +2.0 × 10 −11 N ⋅ m ¯ ÁË Ë C ˜¯ TOP: ELECTRIC DIPOLE TORQUE 2. ANS: A To apply the line/cylinder formula, we need the linear charge density λ . Pick some length L for a sample section of cylinder. Then the lateral surface area of that section will be A = 2π RL and the total charge is the area times the surface charge density Q = 2π RLσ . Q 2π RLσ = 2π Rσ so that Now λ = = L 2k λ 2k2π Rσ 4k π Rσ E= = = r r r L Ê ˆ ˆ Ê ˆÊ 4 ÁÁ 9.00 × 10 9 Nm2 / C 2 ˜˜ (3.14) ÁÁ 19.00 × 10 −2 m˜˜ ÁÁ 27.0 × 10 −9 C / m2 ˜˜ Ë ¯ ¯ Ë ¯Ë = ÊÁ −2 ˆ Á 31.0 × 10 m˜˜ Ë ¯ = 1.87 × 103 N = 1.87 kN C The much easier way is to apply Gauss's Law from scratch: Charged Cylinder Q = 2π RLσ Gaussian Cylinder Area A = 2π rL ˆ Ê Q 2π RLσ R σ 19 ÁÁÁÁ 27.0 × 10 −9 ˜˜˜˜ E= = = = ε 0 A 2π rLε 0 r ε 0 31 ÁÁÁ 8.85 × 10 −12 ˜˜˜ ¯ Ë Note the extra advantage of not having to convert centimeters to meters since the units cancel TOP: ELECTRIC FIELD CYLINDER 3. ANS: A This is one of the basic facts about charges - from grade school, I think. TOP: CHARGE CONCEPT 4. ANS: A The electric field inside any ideal conductor, even if it’s hollow & empty inside, is always zero. TOP: CONDUCTOR FIELD CONCEPT 1 ID: B 5. ANS: D 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 3 R. Hence the field is 2 kQ kQ 4 kQ E= 2 = = 2 9 R ÊÁ 3 ˆ r ÁÁ R ˜˜˜ Ë 2 ¯ distance from their center is The other way to do this is to construct a Gaussian Surface at r = 3R / 2. The enclosed charge is +Q and 2 ÁÊ 3 ˆ˜ the surface area A = 4π r 2 = 4π ÁÁÁÁ R ˜˜˜˜ = 9π R 2 . The electric field is 2 Q enc E= = ε 0A Ë ¯ +Q 1 4 kQ ÊÁ 9 2 ˆ˜ = 9 R 2 where we used k = 4πε 0 4π ε 0 ÁÁÁÁ R ˜˜˜˜ Ë4 ¯ TOP: ELECTRIC FIELD SPHERICAL 6. ANS: B In case you don’t recall the formula (the hint given in Thursday lecture), just start with the formula for a single kQ point charge E = 2 since all point on the ring are equidistant from the point P. Then take the x-component (since r x the others cancel by symmetry) and the factor is r kQ ÊÁÁ x ˆ˜˜ kxQ kxQ kx (2π a λ ) E = 2 ÁÁÁ ˜˜˜ = 3 = = 3/2 3/2 r Ê ˆ Ê r r Ë ¯ ÁÁ x 2 + a 2 ˜˜ ÁÁ x 2 + a 2 ˆ˜˜ Ë ¯ Ë ¯ ÊÁ 2 ˆ ˜˜ ÁÁ ÁÊ ˜ˆ ÁÁ 9.00 × 10 9 Nm2 ˜˜˜ ÊÁÁ 63.0 × 10−2 mˆ˜˜ 2 (3.14) ÊÁÁ 72.0 × 10 −2 mˆ˜˜ ÁÁÁ 2.13 × 10 −4 C ˜˜˜ ˜ ÁÁ Á m ˜¯ ¯ Ë ¯Ë C ˜¯ Ë Ë = ÊÁ Ê ˆ3/2 ÁÁ ÁÁ 63.0 × 10 −2 m˜ˆ˜ 2 + ÁÊÁ 72.0 × 10 −2 m˜ˆ˜ 2 ˜˜˜ ÁÁ Ë ¯ Ë ¯ ˜˜¯ Ë = 6.24 × 10 6 N C TOP: ELECTRIC FIELD RING 2 ID: B 7. ANS: A This is a conversion problem, so use the identity e + = 1.60 × 10 −19 C to create "unit fraction" ˆ˜ ÊÁ total charge ˜˜ ÁÁ 1e + ÊÁ −9 ˆ ˜ ˜ = 6.16 × 10 12 Á #e = = Á 985 × 10 C ˜ ÁÁ charge/proton Ë ¯ Á 1.60 × 10 −19 C ˜˜˜ ¯ Ë + NOTE: the error in the original text (micro Coulombs) was corrected on the board. TOP: ELEMENTARY CHARGE 8. ANS: B 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 Ê ˆ = ÁÁ n p e + ˜˜ + ÊÁÁ n e e − ˆ˜˜ ¯ Ë ¯ Ë Ê ˆÊ ˆÊ ˆ Ê ˆ = ÁÁ 2.90 × 10 22 ˜˜ ÁÁ +1.60 × 10 −19 C ˜˜ + ÁÁ 9.01 × 10 23 ˜˜ ÁÁ −1.60 × 10 −19 C ˜˜ ¯Ë ¯Ë Ë ¯ Ë ¯ Ê ˆ ˆÊ = ÁÁ 2.90 × 10 22 − 9.01 × 1023 ˜˜ ÁÁ +1.60 × 10 −19 C ˜˜ Ë ¯ ¯Ë Ê ˆÊ ˆ = ÁÁ −8.72 × 10 23 ˜˜ ÁÁ +1.60 × 10 −19 C ˜˜ = −1.40 × 10 5 C ¯Ë Ë ¯ Yes... another typo, but answer ‘B’ is by far the closest. TOP: ELEMENTARY CHARGE 9. ANS: D F 712N N = 4.75 × 10 5 By definition, E = = q 1.50 × 10−3 C C NOTE: The original typo 25.0 mC was corrected on the board. TOP: ELECTRIC FIELD 10. ANS: C F=k q1q2 r2 ÊÁ ˆ˜ −9 ˆ −9 ˜ ÊÁ ÊÁ 2 ˆ ˜ ÁË 63.0 × 10 C ˜¯ ÁË −53.0 × 10 C ˜¯ ÁÁ 9 Nm ˜ ˜˜ = ÁÁÁ 9.00 × 10 = −4.81 × 10 −4 N 2 ˜ 2 ˜ Á C ˜¯ ÁÊÁ 25.0 × 10 −2 m˜ˆ˜ Ë Ë ¯ The minus sign shows that the force is attractive, as you’d expect between unlike charges. TOP: COULOMB FORCE 11. ANS: C N ⋅ m2 ˆ Ê 2 Φ E = E • A = EA cos θ = ÁÁ 31.7 × 10−9 N / C ˜˜ (2.66m) cos (11.6°) = 2.20 × 10 −4 ¯ Ë C TOP: Electric Flux 3 ID: B 12. ANS: B For a single charged sheet, the field points perpendicular to the plane - away if positive and towards if negative. 3.60 × 10 −9 C / m2 σ N E= = Ê = 2.03 × 10 2 2ε 0 2 ÁÁ 8.85 × 10−12 C 2 / N ⋅ m2 ˆ˜˜ C ¯ Ë Note that the distance r doesn’t matter! TOP: FIELD 1 SHEET 13. ANS: D Make a ratio of the two forces using the fact that the only change is that the new distance is r new = 0.85 * r k F new = F q1q2 2 ÁÊ r new ˜ˆ ÊÁ r ˆ˜ 2 ÊÁ r ˆ˜ 2 ÊÁ 1 ˆ˜ 2 Ë ¯ ˜ ˜ Á Á ˜˜ = 16.7 N ˜ ˜ Á Á = ÁÁ ˜ ⇒ F new = F ÁÁ ˜ = (12N ) ÁÁÁÁ ˜˜ ÁË r new ˜˜¯ ÁË r new ˜˜¯ q1q2 0.85 ¯ Ë k 2 (r) NOTE: Answers ‘C’ and ‘D’ are so close that EITHER ONE was accepted as correct TOP: COULOMB FORCE CONCEPT 14. ANS: A We apply Gauss’ Law by finding the total enclosed charge Q ENC = λ × circumference = λπ D ΦE = Q ENC ε0 Ê ˆ Ê ˆ ÁÁ 6.11 × 10 −6 C / m˜˜ (3.14) ÁÁ 26.3 × 10 −2 m˜˜ 2 ¯ Ë ¯ λπ D Ë 5 N ⋅m = = = 5.70 × 10 ε0 C 8.85 × 10 −12 C 2 / N ⋅ m2 TOP: ELECTRIC FLUX 15. 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 = 4π r 2 dr. We integrate the charge elements dq = ρ (r)dV (r) as spherical shells since the density depends only on the radius r ∫ ∫ Q = ρ (r)dV = ρ (r)4π r 2 dr = = ∫ R 0 ρo 4π ρ o r 4π r 2 dr = R R ∫ R r 3 dr 0 4π ρ o R 4 C ˆ˜ Ê ˆ 3 ÊÁ = π R 3 ρ o = (3.14) ÁÁ 3.11 × 10 2 m˜˜ ÁÁÁÁ 2.97 × 10 −6 3 ˜˜˜˜ = 2.81 × 10 2 C Ë ¯ Ë R 4 m ¯ TOP: VOLUME CHARGE DENSITY 16. ANS: D Ê ˆÊ ˆ By definition, p = qd = ÁÁ 35 × 10 −9 C ˜˜ ÁÁ 8.0 × 10−2 m˜˜ = 2.8 × 10 −9 C ⋅ m Ë ¯Ë ¯ The vector points towards the positive charge, i.e., down. TOP: ELECTRIC DIPOLE 4 ID: B 17. ANS: A ˆ˜ ÊÁ ˜˜ ÁÁ −4 ˜˜ Á charge Q Á 6.60 × 10 C 3Á ˜˜ = 5.29 C ρ= = = ÁÁÁ ˜ 3 volume 4 4 ÁÁ Ê ˜ −2 ˆ ˜ m3 πR3 ÁÁ (3.14) ÁÁ 3.10 × 10 m˜˜ ˜˜˜ Ë ¯ ¯ 3 Ë TOP: VOLUME CHARGE DENSITY 5
