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Name: _________________________________________ Code # _________ UCSD Physics 2B Practice Exam 1 Summer Session 2008 ID: X Charges, Force & Fields READ ME !!! Helpful Hints: Do this exam a few times... Once without the answers, Once after seeing the solutions & Once again with your formula sheet This is a good time to add formulas/constants to your sheet. Remember, this is a timed exam... work swiftly so you have time to check your answers. 1. A positive charge q 1 = +63.0 nC is placed at a distance r = 15.0cm from a negative charge q 2 = −45.0 nC . What is the magnitude of the force on one charge due to the other? A. 1.13 × 10 −3 N B. 1.70 × 10 −4 N C. 1.02 × 10 −7 N D. 1.25 × 10−13 N 2. How many electrons must be transferred to an electrically neutral body to produce a total net charge q tot = −125nC ? A. 1.25 × 10 −7 B. 7.81 × 10 11 C. 1.28 × 10 12 D. 3.45 × 10 11 3. An initially neutral empty vacuum chamber designed to contain a hot plasma is injected with n p = 1.67 × 10 23 protons and n e = 7.99 × 10 22 electrons. What is the net charge inside the chamber? A. +3.95 × 10 4 C B. +1.39 × 10 4 C C. −3.95 × 10 4 C 1 D. −1.39 × 10 4 C Name: ________________________ ID: A 4. Three charges q 1 = +22.0 μC , q 2 = −44.0 μC and q 3 = +10.0 μC are placed in a line as shown below where the distance r 1 = 1.00 cm and r 2 = 0.750 cm. What is the magnitude and direction of the force on the center charge q 2 ? Take the plus direction to the right. A. +1.67 × 10 4 N B. −1.58 × 10 5 N C. −1.67 × 10 4 N D. +1.58 × 10 5 N 5. A charge q = 23.7mC feels a force F = 525N at a certain point in space. What is the magnitude of the electric field at that point? A. 2.22 × 10 1 N C B. 2.22 × 10 4 N C C. 4.51 × 10−2 N C D. 4.51 × 10−5 N C 6. Two charges q 1 = 40.0 μC and q 2 = 20.0 μC are separated by a distance r = 2.00mm and located along a line as shown below. What is the magnitude and direction of the electric field at point x? Take right as the plus direction. A. −6.75 × 1010 N C B. +6.75 × 1010 N C C. +1.35 × 1010 2 N C D. −1.35 × 1010 N C Name: ________________________ ID: A 7. A thin rod of infinite length carries a linear charge density λ = 1.7 × 10 −6 C / m. What is the magnitude of the electric field at a distance r = 4.2 cm from the rod? A. 7.3 × 10 5 N C B. 1.7 × 10 11 N C C. 1.4 × 10 7 N C D. 2.3 × 10 6 N C 8. An infinitely long straight wire carries a linear charge density λ = 2.2 × 10 −9 C / m. A cardboard tube from an exhausted bathroom tissue roll is placed as shown in the diagram below. If the tube length L = 15 cm and radius R = 6.6 cm, what is the electric flux passing through the tube? A. 37 k N ⋅ m2 C2 B. 68 k N ⋅ m2 C2 C. 37 N ⋅ m2 C D. 68 N ⋅ m2 C 9. Two infinite sheets of charge are separated by a distance d = 1.20mm and carry opposite charge C densities σ = 1.30 × 10 −10 2 . What is the electric field between the sheets? m A. 14.7 N C B. 29.4 N C C. 1.15 × 10 −21 3 N C D. 1.08 × 10−7 N C Name: ________________________ ID: A 10. Two concentric spheres carry charges +Q on the inner sphere and −Q on the outer. If the inner sphere has radius R and the outer sphere 2R, what is the electric field at a point just outside the outer sphere? Take the direction pointing radially out as positive. A. kQ R B. − 2 kQ R C. − 2 kQ (2R) 2 D. Zero 11. A perfectly spherical balloon of radius R = 22.0 cm has no net charge. If a point charge q = 1.60 × 10−9 C is placed at it center, what is the electric flux passing through the balloon? A. 181 N ⋅ m2 C B. 65.0 N ⋅ m2 C C. 297 N ⋅ m2 C D. 466 N ⋅ m2 C 12. A non-conducting sphere of radius r = 3.00 cm carries a uniform volume charge density C ρ = 5.00 × 10−7 3 . What is the magnitude of the electric field at its surface R = 3.00 cm. m A. 36.0 N C B. 232 N C C. 147 N C D. 565 N C 13. A point charge +q = +35nC is directly above a point charge −q = −35nC separated by distance d = 4.0cm. What is the magnitude and direction of the dipole moment p? A. 1.4 × 10 −9 C ⋅ m ↓ B. 1.4 × 10 −9 C ⋅ m ↑ C. 2.8 × 10 −9 C ⋅ m ↓ 4 D. 2.8 × 10 −9 C ⋅ m ↑ Name: ________________________ ID: A 14. The force between two very small particle is found to be F. If the separation distance r between them is doubled without altering the charges, the force between them becomes A. 4F B. 2F C. 1 F 2 D. 1 F 4 15. Electric charges of the opposite sign A. have the same magnitude B. attract each other C. repel each other D. exert no force on each other 16. A non-conducting sphere of radius R = 1.30 × 10 −2 m carries a uniformly distributed total charge Q = 6.60 × 10 −5 C . What is the volume charge density ρ inside the sphere? A. 12.4 C m3 B. 7.18 C m3 C. 0.311 mC m3 D. 22.7 kC m3 17. A solid non-conducting cylinder of radius R = 1.33 × 10 −2 m and length L = 8.74 × 10 −1 m is given a r non-uniform electric charge density which varies with distance r from the axis as ρ (r) = ρ 0 where R C ρ 0 = 5.88 × 10 −3 3 . What is the total net charge contained in the cylinder? Hint: try making your m differential charge elements in the shape of nested cylindrical shells of length L and radius r and then integrate the charge density times the volume elements as r → 0,R. A. 2.85μ C B. 1.90μ C C. 5.71μ C 5 D. 1.43μ C Name: ________________________ ID: A 18. An electric field of strength E = 31.7 μ N / m passes through a square sheet of side s = 2.66 m at an angle θ = 11.6 ° from the sheet normal as shown in the diagram below. How much electric flux passes through the sheet? A. 8.26 × 10 −5 B. 2.20 × 10 −4 N ⋅ m2 C N ⋅ m2 C C. 1.70 × 10 −5 D. 4.51 × 10 −5 N ⋅ m2 C N ⋅ m2 C 19. An electric dipole vector p = 25.0nC ⋅ m makes an angle θ = 61.0° with a uniform electric field vector N E = 3.00 × 10−6 . What is the torque on the dipole? C A. −7.25 × 10 −14 N ⋅ m B. −1.94 × 10 −14 N ⋅ m C. 3.64 × 10 −14 N ⋅ m D. 6.56 × 10 −14 N ⋅ m 6 ID: A UCSD Physics 2B Answer Section Practice Exam 1 Charges, Force & Fields MULTIPLE CHOICE 1. ANS: A F=k q1q2 r2 Ê ˆÊ ˆ ˆ ÁÁ 63.0 × 10 −9 C ˜˜ ÁÁ −45.0 × 10 −9 C ˜˜ 2 ˜ ÁÊÁ ˜ Ë Ë Nm ¯ ¯ = ÁÁÁÁ 9.00 × 10 9 2 ˜˜˜˜ = −1.13 × 10 −3 N 2 ÊÁ Á ˜ −2 ˆ C Á 15.0 × 10 m˜˜ Ë ¯ Ë ¯ The minus sign shows that the force is attractive, as you’d expect between unlike charges. On the other hand, the problem didn’t ask for this, but if it had, ... PTS: 1 2. ANS: B TOP: COULOMB FORCE 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 ˜ ˜ = 7.81 × 10 11 Á #e = = Á −125 × 10 C ˜ ÁÁ charge/electron Ë ¯ Á −1.60 × 10 −19 C ˜˜˜ ¯ Ë − PTS: 1 3. ANS: B TOP: ELEMENTARY CHARGE 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 1.67 × 10 23 ˜˜ ÁÁ +1.60 × 10 −19 C ˜˜ + ÁÁ 7.99 × 10 22 ˜˜ ÁÁ −1.60 × 10 −19 C ˜˜ ¯Ë ¯Ë Ë ¯ Ë ¯ Ê ˆ ˆÊ = ÁÁ 1.67 × 10 23 − 7.99 × 1022 ˜˜ ÁÁ +1.60 × 10 −19 C ˜˜ Ë ¯ ¯Ë Ê ˆÊ ˆ = ÁÁ 8.71 × 10 22 ˜˜ ÁÁ +1.60 × 10 −19 C ˜˜ = +1.39 × 10 4 C Ë ¯Ë ¯ PTS: 1 TOP: ELEMENTARY CHARGE 1 ID: A 4. ANS: C Since the positive direction is to the right, the unit vector for F 12 is r8 = 8i ÊÁ ˆÊ ˆ ÊÁ Á 22.0 × 10 = 6 C ˜˜ ÁÁ −44.0 × 10 = 6 C ˜˜ 2 ˆ ˜ ÁÁ ˜ Ë Ë Nm ¯ ¯8 F 12 = k 2 r8 = ÁÁÁ 9.00 × 109 2 ˜˜˜˜ i = −8.71 × 10 4 N 8i 2 Á ˜ ÊÁ −2 ˆ r1 C Á 1.00 × 10 m˜˜ Ë ¯ Ë ¯ 8 The unit vector for F 32 is r8 = −i since the source is on the other side (the vector always points away q1q2 from the source charge), so that Ê ˆÊ ˆ ˆ ÁÁ −44.0 × 10−6 C ˜˜ ÁÁ 10.0 × 10 −6 C ˜˜ 2 ˜ ÁÊÁ ˜ Ë Ë Nm ¯ ¯ ÊÁ 8 ˆ˜ 4 F 32 = k 2 r8 = ÁÁÁÁ 9.00 × 109 2 ˜˜˜˜ ÁË −i ˜¯ = +7.04 × 10 N 8i 2 Á ˜ ÊÁ −2 ˆ r2 C Á 0.750 × 10 m˜˜ Ë ¯ Ë ¯ 4 = −1.67 × 10 N i8 which points (negative) to the left. Adding, we get the net force F q3q2 NET PTS: 1 5. ANS: B TOP: COULOMB FORCE By definition, E = F 525N N = = 2.22 × 10 4 −3 q 23.7 × 10 C C PTS: 1 6. ANS: B TOP: ELECTRIC FIELD The unit vector at x points to the right (away from q 1 and q 2 ) for both charges, ˆ˜ ˆ Ê ˜˜ k ÁÁÁ q 1 q 2 ˜˜˜ ˜ ˜ Á = + ˜˜ Á 2 1 ˜˜˜ ˜ r 2 ÁÁ 4 (2r) ¯ ¯ Ë Ê ˆ 9 2 2 Á −6 −6 ˜ N 9.00 × 10 Nm / C ÁÁÁ 40.0 × 10 C 20.0 × 10 C ˜˜˜ E= + = +6.75 × 10 10 ÁÁ ˜ 2 ˜ 4 1 C ÊÁ ˆ Á ˜ Á 2.00 × 10 −3 m˜˜ Ë ¯ ¯ Ë E = E1 + E2 = k PTS: 1 7. ANS: A q1 +k ÊÁ q q2 ÁÁ 1 ÁÁ = k + 2 2 2 Á Á (2r) (r) (r) Ë q2 TOP: ELECTRIC FIELD If you have the formula handy, apply it. If not, see “Gauss’s Law” for cylindrical geometry. ˆ ÁÊÁ ˆ N ⋅ m2 ˜˜˜˜ ÊÁÁÁ −6 C ˜ ˜˜ 2 ÁÁÁÁ 9.0 × 109 1.7 × 10 Á ˜˜ ˜ 2 ˜˜ Á m Á C ¯ Ë N 2k λ ¯Ë E= = = 7.3 × 10 5 ˆ Ê −2 C r ÁÁ 4.2 × 10 m˜˜ ¯ Ë PTS: 1 TOP: ELECTRIC FIELD LINE CHARGE 2 ID: A 8. ANS: C Here we’re not asked for the electric field, but the flux. If we remember the whole formula, we can immediately write ΦE = Q ENC ε0 Ê ˆÊ ˆ ÁÁ 2.2 × 10 −9 C / m˜˜ ÁÁ 15 × 10 −2 m˜˜ N ⋅ m2 ¯Ë ¯ λL Ë = = = 37 ε0 C 8.85 × 10 −12 C 2 / N ⋅ m2 Note that two of the answers have the wrong units! PTS: 1 9. ANS: A TOP: ELECTRIC FLUX CYLINDER For twin sheets, the field outside is zero, while between the sheets, σ 1.3 × 10 −10 C / m2 N E= = = 14.7 −12 2 2 ε 0 8.85 × 10 C / N ⋅ m C Note that the separation distance d doesn’t matter! PTS: 1 10. ANS: D TOP: FIELD 2 SHEETS Two ways to solve this 1. Draw a Gaussian Surface just outside the outer sphere. Since +Q & -Q cancel, Q ENC = 0 so the electric field is zero. 2. Recall that, outside uniform spherical shells, the field is the same as if all charge is located at the center. Again ÁÊË +Q ˜ˆ¯ + ÁÊË −Q ˜ˆ¯ = 0. QED. PTS: 1 11. ANS: A TOP: ELECTRIC FIELD SPHERICAL Since the balloon completely surrounds the charge, we don't need the radius & little calculation is required: ΦE = Q ENC ε0 PTS: 1 = 1.60 × 10−9 C N ⋅ m2 = 181 C 8.85 × 10 −12 C 2 / N ⋅ m2 TOP: ELECTRIC FLUX SPHERE 3 ID: A 12. ANS: D Construct a Gaussian Sphere whose surface is at the radius of interest. The charge enclosed is the volume times the charge density - in this case the totasl charge of the sphere Q ENC = VOL × DENSITY = ÊÁ 4 ˆ ÁÁ π R 3 ˜˜˜ ρ ÁÁ 3 ˜˜ Ë ¯ Meanwhile, the surface area is A = 4π R 2 . Now apply Gauss's Law ÊÁ 4 ˆ ÁÁ π R 3 ˜˜˜ ρ ÁÁ 3 ˜˜ Q ENC Rρ Ë ¯ E= = Ê = = ˆ 2 ε 0A ε 0 ÁÁ 4π R ˜˜ 3ε 0 Ë ¯ ˆÊ ˆ ÊÁ Á 3.00 × 10 −2 m˜˜ ÁÁ 500 × 10 −9 C / m3 ˜˜ ¯Ë ¯ Ë N = 565 2 C C 8.85 × 10 −12 2 N ⋅m Note the wisdom of solving the algebra first... lots o’ stuff cancels! PTS: 1 13. ANS: B TOP: ELECTRIC FIELD INSIDE SPHERE Ê Ë ˆÊ ¯Ë ˆ ¯ By definition, p = qd = ÁÁ 35 × 10 −9 C ˜˜ ÁÁ 4.0 × 10−2 m˜˜ = 1.4 × 10 −9 C ⋅ m The vector points towards the positive charge, i.e., up. PTS: 1 TOP: ELECTRIC DIPOLE 14. ANS: D You might be able to do this in your head, using the fact that the Coulomb force is an inverse-square law. However, to approach it more systematically, construct a ratio using the given information, i.e., that the new distance r new = 4r k F new = F q1q2 2 ÁÊ r new ˜ˆ ÊÁ r ˆ˜ 2 ÊÁ r ˆ˜ 2 1 Ë ¯ ˜˜ = ÁÁ ˜˜ = = ÁÁÁÁ ˜ ÁÁ 2r ˜˜ q1q2 4 ÁË r new ˜˜¯ Ë ¯ k 2 r or F new = F / 4 PTS: 1 TOP: COULOMB CONCEPT 15. ANS: B This is a experimental basic fact... sorry, you just have to memorize it. PTS: 1 16. ANS: B TOP: CHARGE CONCEPT ˆ˜ ÊÁ ˜˜ ÁÁ −5 ˜˜ Á charge Q 6.6 × 10 C 3 ÁÁ ˜˜ = 7.18 C ρ= = = ÁÁÁ 3 ˜ volume 4 4 ÁÁ (3.14) ÊÁÁ 1.3 × 10 −2 m˜ˆ˜ ˜˜˜ m3 πR3 ˜˜ ÁÁ Ë ¯ ¯ 3 Ë PTS: 1 TOP: VOLUME CHARGE DENSITY 4 ID: A 17. ANS: C For a cylinder where there is radial symmetry dV = 2π Lrdr. In other words, we integrate the charge elements dq = ρ (r)dV (r) as cylindrical shells since the density depends only on the radius r ∫ ∫ Q = ρ (r)dV = ρ (r)2π r L dr = = ∫ R 0 ρo 2π Lρ o r 2π r L dr = R R ∫ R r 2 dr 0 2π Lρ o R 3 2 = π R 2 Lρ o = 5.71μ C R 3 3 PTS: 1 18. ANS: B TOP: VOLUME CHARGE DENSITY 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 PTS: 1 19. ANS: D TOP: Electric Flux Ê ˆÊ ˆ τ = p × E = pE sin θ = ÁÁ 25.0 × 10 −9 C ⋅ m˜˜ ÁÁ 3.00 × 10 −6 N / C ˜˜ sin65° = +6.56 × 10−14 N ⋅ m Ë PTS: 1 ¯Ë ¯ TOP: ELECTRIC DIPOLE TORQUE 5