Version: A. Earth s gravitational field g = 9.81 N/kg Vacuum Permeability µ 0 = 4π 10 7 T m/a
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1 PHYS 2212 GJ Quiz and Exam Formulæ & Constants Fall 2015 Fundamental Charge e = C Mass of an Electron m e = kg Coulomb constant K = N m 2 /C 2 Vacuum Permittivity ϵ 0 = C 2 /N m 2 Earth s gravitational field g = 9.81 N/kg Vacuum Permeability µ 0 = 4π 10 7 T m/a Unless otherwise directed, friction, drag, and gravity should be neglected, and all batteries and wires are ideal. All integrals in free-response problems must be evaluated. k = 1 4πϵ 0 V = E d s V = k q r U = q V I = dq/dt P = I V R = V I Series : 1 = 1 C eq C i R eq = R i Parallel : 1 R eq = 1 R i C eq = C i ϵ 0 E = k q r 2 ˆr F = k q 1q 2 r 2 ˆr F = q E p = q d τ = p E U = p E E p r 3 Φ E = E da E d A = q enclosed E d l = dφ B dt C = Q V A C = ϵ 0 d U = 1 2C [ V ]2 R = ρ l A τ C = RC u E = 1 2 ϵ 0E 2 B = µ 0q 4π db = µ 0I 4π F = q v B F = I l B µ = NI A τ = µ B v ˆr r 2 d l ˆr r 2 U = µ B Φ B = B da B d A = 0 B d l = µ 0 (I c + I d ) L = Φ B I L = µ 0 N 2 A l U = 1 2 LI2 B = µ 0 ni τ L = L/R u B = 1 2µ 0 B 2 q = q max (1 e t/τ C q = q 0 e t/τ C I = I max (1 e t/τ L I = I 0 e t/τ L I = J da J = σ E E = N dφ B dt I d = ϵ 0 dφ E dt E = L di dt c = fλ = E B S = 1 µ 0 E B I = P 4πr 2 P r = (2)I c I trans = I 0 cos 2 θ ) ) Please remove this sheet from your Quiz or Exam Version: A
2 Recitation Sections YOUR form number is 221
3 Version Quiz #2 Form #221 Name: A Physics 2212 GJ Fall 2015 Recitation Section: Print your name, quiz form number (3 digits at the top of this form), and student number (9 digit Georgia Tech ID number) in the section of the answer card labeled Student Identification. Bubble the Quiz Form Number in columns 1 3, skip column 4, then bubble your Student Number in columns Free-response questions are numbered I III. For each, make no marks and leave no space on your card. Show all your work clearly, including all steps and logic. Box your answer. Multiple-choice questions are numbered For each, select the answer most nearly correct, circle this answer on your quiz, and bubble it on your answer card. Do not put any extra marks on the card. Turn in your quiz and answer card as you leave. Your score will be posted when your quiz has been graded. Quiz grades become final when the next quiz is given. You may use a calculator that cannot store letters, but no other aids or electronic devices. I. (17 points) Four charges (-1.0 nc, +2.0 nc, -3.0 nc, and -4.0 nc), each with mass 0.10 g, are brought together and fixed at adjacent corners of a cube of edge length 1.0 mm, as shown. The -4.0 nc charge breaks free and accelerates away. How fast does it ultimately end up going? Form #221 Page 1 of 7
4 II. (16 points) Three point charges, q 1 = 3.5 µc, q 2 = +12 µc, and q 3 = +4.5 µc, are situated at three corners of a rectangle as shown. What is the electric potential at the free corner where there is no charge? Announced during quiz: Potential is zero at infinite distance. 1. (5 points) In the problem above, in what quadrant, if any, is the direction of the potential at the free corner? (a) The potential has no direction. (b) The potential is directed in quadrant I. (c) The potential is directed in quadrant III. (d) The potential is directed in quadrant II. (e) The potential is directed in quadrant IV. Form #221 Page 2 of 7
5 III. (17 points) An infinitely long, straight non-conducting cylinder (radius R) has a uniform charge per unit volume ρ. Use Gauss Law to find an expression for the electric field at a point outside the cylinder, a distance r from the cylinder s axis. Express your answer as a function of the variable r and the fixed parameters R and ρ (as well as any necessary constants such as ϵ 0 and π). 2. (5 points) Suppose one wanted to graph the magnitude of the electric field as a function of r, for all values of r (that is, including points inside the charged cylinder). Which of the following sketches best represents the situation? (a) (b) (c) (d) (e) Form #221 Page 3 of 7
6 3. (5 points) Two effectively infinite parallel plates, 6 mm apart, form a capacitor with the negative plate on the left, as shown. Point A is 1 mm from the negative plate and point B is 2 mm from the positive plate. Letting the potential be zero at the negative plate, what is V A /V B, the ratio of the potential at point A to that at point B? (a) V A /V B = 1/3 (b) V A /V B = 1 (c) V A /V B = 1/4 (d) V A /V B = 4 (e) V A /V B = 3 4. (5 points) Where on the horizontal axis, other than at ± infinity, is the potential equal to zero? Define zero potential as the potential at infinity. (a) At a point in region ii and at a point in region iii. (b) Only at a point in region i. (c) Only at a point in region iii. (d) Only at a point in region ii. (e) At a point in region i and at a point in region ii. Form #221 Page 4 of 7
7 5. (5 points) A chunk of metal has two hollow cavities (see figure). The metal itself carries net charge of -5 nc. The right cavity contains a point charge +2 nc, while the left cavity contains a point charge -3 nc. What is the net charge on the outer surface of the metal? (a) -1 nc (b) -6 nc (c) zero (d) -4 nc (e) -5 nc 6. (5 points) Two positive charges, with masses m 1 = 1.0 g and m 2 = 2.0 g, are held a distance d apart. They are released simultaneously and accelerate away from each other, ending up with speeds v 1 and v 2, respectively. How are v 1 and v 2 related? (a) v 1 = v 2 (b) It cannot be determined, because the relationship between v 1 and v 2 depends on d. (c) v 1 < v 2 (d) v 1 > v 2 (e) It cannot be determined, because the relationship between v 1 and v 2 depends on the charges q 1 and q 2. Form #221 Page 5 of 7
8 7. (5 points) A thin rod of length L lies on the x axis with its center at the origin. It has a linear charge density λ that varies with position x according to λ = λ 0 x L where λ 0 is a constant. What is the electric potential due to the rod, with respect to zero at infinity, at a point located at d on the +y axis? (a) Kλ 0 L/d (b) zero (c) KλL/d (d) Kλ 0 /d (e) Kλ/d 8. (5 points) The potential at the center of a uniformly charged ring is V 0 (relative to zero at infinity). If 1/4 of the ring is removed, as shown below, what will be the new potential at the same position? (a) V 0 / 2 (b) V 0 /2 (c) V 0 (d) V 0 /4 (e) 3V 0 /4 Form #221 Page 6 of 7
9 9. (5 points) A positively charged particle lies on the axis of a cylindrical surface, a distance d from one end, as shown. The cylinder has height h. Compare the magnitude of the electric flux through the top and bottom surfaces of the cylinder. (a) Φ top < Φ bottom if d < h, but Φ top > Φ bottom if d > h (b) Φ top > Φ bottom (c) Φ top = Φ bottom (d) Φ top < Φ bottom (e) Φ top > Φ bottom if d < h, but Φ top < Φ bottom if d > h 10. (5 points) A positively charged particle and a negatively charged particle, each having charge magnitude q, lie on the axis of a cylindrical surface, equidistant from the ends, as shown. Rank the flux through the top, side, and bottom of the cylinder, from greatest to least. Remember that electric flux can be positive or negative. (a) Φ bottom > Φ side > Φ top (b) Φ top = Φ side = Φ bottom (c) Φ top > Φ side > Φ bottom (d) Φ top = Φ bottom > Φ side (e) Φ side > Φ top > Φ bottom Form #221 Page 7 of 7
Form #221 Page 1 of 7
Version Quiz #2 Form #221 Name: A Physics 2212 GH Spring 2016 Recitation Section: Print your name, quiz form number (3 digits at the top of this form), and student number (9 digit Georgia Tech ID number)
More informationVersion: A. Earth s gravitational field g = 9.81 N/kg Mass of a Proton m p = kg
PHYS 2212 G & J Quiz and Exam Formulæ & Constants Fall 2017 Fundamental Charge e = 1.602 10 19 C Mass of an Electron m e = 9.109 10 31 kg Earth s gravitational field g = 9.81 N/kg Mass of a Proton m p
More informationVersion: A. Earth s gravitational field g = 9.81 N/kg Mass of a Proton m p = kg
PHYS 2212 K Quiz and Exam Formulæ & Constants Summer 2016 k = 1 4πϵ 0 V = E d s V = k q r U = q V I = dq/dt P = I V R = V I Series : 1 = 1 C eq C i R eq = R i Parallel : 1 R eq = 1 R i C eq = C i ϵ 0 E
More informationVersion: A. Earth s gravitational field g = 9.81 N/kg Vacuum Permeability µ 0 = 4π 10 7 T m/a
PHYS 2212 GJ Quiz and Exam Formulæ & Constants Fall 2015 Fundamental Charge e = 1.602 10 19 C Mass of an Electron m e = 9.109 10 31 kg Coulomb constant K = 8.988 10 9 N m 2 /C 2 Vacuum Permittivity ϵ 0
More informationVersion: A. Earth s gravitational field g = 9.81 N/kg Mass of a Proton m p = kg
PHYS 2212 G & J Quiz and Exam Formulæ & Constants Fall 2017 Fundamental Charge e = 1.602 10 19 C Mass of an Electron m e = 9.109 10 31 kg Earth s gravitational field g = 9.81 N/kg Mass of a Proton m p
More informationForm #425 Page 1 of 6
Version Quiz #4 Form #425 Name: A Physics 2212 G Spring 2018 Recitation Section: Print your name, quiz form number (3 digits at the top of this form), and student number (9 digit Georgia Tech ID number)
More informationVersion: A. Earth s gravitational field g = 9.81 N/kg Vacuum Permeability µ 0 = 4π 10 7 T m/a
PHYS 2212 GJ Quiz and Exam Formulæ & Constants Fall 2015 Fundamental Charge e = 1.602 10 19 C Mass of an Electron m e = 9.109 10 31 kg Coulomb constant K = 8.988 10 9 N m 2 /C 2 Vacuum Permittivity ϵ 0
More informationω = k/m x = A cos (ωt + ϕ 0 ) L = I ω a x = ω 2 x P = F v P = de sys J = F dt = p w = m g F G = Gm 1m 2 D = 1 2 CρAv2 a r = v2
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PHYS 2211 A & B Final Exam Formulæ & Constants Fall 2016 Unless otherwise directed, use the gravitational definition of weight, all problems take place on Earth, drag is to be neglected, and all pulleys
More informationForm #231 Page 1 of 6
Version Quiz #3 Form #231 Name: A Physics 2211 A & B Fall 2016 Recitation Section: Print your name, quiz form number (3 digits at the top of this form), and student number (9 digit Georgia Tech ID number)
More informationPhysics 2212 GH Quiz #2 Solutions Spring 2015
Physics 2212 GH uiz #2 Solutions Spring 2015 Fundamental Charge e = 1.602 10 19 C Mass of an Electron m e = 9.109 10 31 kg Coulomb constant K = 8.988 10 9 N m 2 /C 2 Vacuum Permittivity ϵ 0 = 8.854 10
More informationω = k/m x = A cos (ωt + ϕ 0 ) L = I ω a x = ω 2 x P = F v P = de sys J = F dt = p w = m g F G = Gm 1m 2 D = 1 2 CρAv2 a r = v2
PHYS 2211 A & B Final Exam Formulæ & Constants Fall 2016 Unless otherwise directed, all problems take place on Earth, and drag is to be neglected. A v = d r ω = dθ a = d v α = d ω v sf = v si + a s t ω
More informationω = k/m x = A cos (ωt + ϕ 0 ) L = I ω a x = ω 2 x P = F v P = de sys J = F dt = p w = m g F G = Gm 1m 2 D = 1 2 CρAv2 a r = v2
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