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(jg47854) gilvin - 8: The Electric Potential - meyers - (21235) This print- out should have 22 questions. Multiple-choice questions may continue on the next column or page - find all choices before answering. This homework is due Tuesday, March 1, at midnight Tucson time. 001 10.0 points To recharge a 12 V battery, a battery charger must move 2.4 105 C of charge from the negative terminal to the positive terminal. How much work is done by the battery charger? Correct answer: 2.88 106 J. Explanation: Given : q = 2.4 105 C and V = 12 V . The potential difference is V =W q, W =qV = (2.4 105 C) (12 V) = 2.88 10 J .
6
1
so the magnitude of the potential difference is V = E d = d 0 ( = 1.1 1010 C/m2) (0.011 m) 8.85419 1012 C2/N m2 = 0.136658 V . 003 10.0 points A voltmeter indicates that the difference in potential between two plates is 53 V. The plates are 0.36 m apart. What electric field intensity exists between them? Correct answer: 147.222 N/C. Explanation:
Let : V = 53 V and d = 0.36 m . The potential difference is V =Ed
002 10.0 points Two flat conductors are placed with their inner faces separated by 11 mm. If the surface charge density on one of the inner faces is 110 pC/m2 and the other inner face 110 pC/m2, what is the magnitude of the electric potential difference between the two conductors? Correct answer: 0.136658 V. Explanation: Let: 0 = 8.85419 1012 C2/N m2 , = 110 pC/m2 = 1.1 1010 C/m2 , and d = 11 mm = 0.011 m . The electric field between two flat conductors is E= 0
E=V d = 53 V 0.36 m = 147.222 N/C .
004 10.0 points Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 2022 N/C. If the voltage is quadrupled and the distance between the plates is reduced to 1 5 the original distance, what is the magnitude of the new electric field? Correct answer: 40440 N/C. Explanation:
gilvin (jg47854) - 8: The Electric Potential - meyers - (21235) Q is the total Let : E = 2022 N/C , V = 4 V , and d = 1 5d. The electric field between two parallel conducting plates is E = V d , where V is the voltage between the plates, and d is the distance between the plates, so the new electric field has a magnitude of ( ) E = V = 4 V = 20 V d d d 5 = 20 E = 20 (2022 N/C) = 40440 N/C . 005 10.0 points When you touch a friend after walking across a rug on a dry day, you typically draw a spark of about 2 mm. The magnitude of the electric field for which dielectric breakdown occurs in air is about 3 M V/m. Estimate the potential difference between you and your friend before the spark. Correct answer: 6000 V. Explanation: Let : r = 2 mm = 0.002 m and Eb = 3 MV/m = 3 106 V/m . The potential difference is V = Eb r = (3 106 V/m) (0.002 m) = 6000 V . 006 (part 1 of 4) 10.0 points The volume of the sphere shown in the figure below is uniformly charged (with positive charge). charge inside the sphere. p R
2
Using Gausss Law, you can show that for r r < R the electric field is given by E = , 3 0 where is the charge density. The charge density is 1. = Q R3 . Q 2. = 0 . 3. = Q 2R. 4. = Q 4 R3 . 5. = Q 0 . 6. = 4Q 3 R3 . 7. = Q R2 . 8. = Q 4 R2 . Q
9. = R3 . 3Q 10. = correct 4 R3 . Explanation: The charge density is the charge per unit volume, so = Q = 3Q 4 R3 4 R3 . 3
007 (part 2 of 4) 10.0 points Compare the potentials between the point p at the center and a point on the surface of the sphere where r = R (p is located at the center of the sphere).
1. Vp < V
R
gilvin (jg47854) - 8: The Electric Potential - meyers - (21235) R r = 0 3 dr 0 = R2 6 0 .
3
2. Vp = V R 3. Vp > V R correct Explanation: Vp = = R
0 R 0
VR
Er dr r 3 0 dr
009 (part 4 of 4) 10.0 points Q is 4 C and R is 2.1 m. What is the potential at the surface, assuming that the potential is zero at infinity? Correct answer: 17123.8 V. Explanation: Let : k = 8.99 109 N m2/C2 , Q = 4 C = 4 106 C and R = 2.1 m . The potential on the surface of the sphere can be calculated as Vs = ( R kQ
= R2 6 0 . This implies that V R < Vp. 008 (part 3 of 4) 10.0 points The magnitude of the potential difference between (r = R) and p is given by R2 1. V
R
Vp = Vp = Vp = Vp =
2. V
R
0 R2 3 0 . 4 R2 3 0 R2 6 0 . correct
dr = k Q R 2.1 m
r2
3. V
R
=
8.99 109 N m2/C2) (4 106 C)
4. V
R
= 17123.8 V .
5. V
R
6. V
R
Vp = 0 . R Vp = 3 0 . Vp = 3 0 . R2 Vp = Vp = 2 0 . 4 R2 0 R
010 10.0 points An electric field is given by Ex = (4 kN/C) x3. Find the potential difference between the points on the x-axis at x = 1 m and x = 5 m. Correct answer: 624 kV. Explanation: Let : Ex = (4 kN/C) x3 , x1 = 1 m , x2 = 5 m . and
7. V 8. V
R
R
9. V
R
R V p = 2 . Explanation: 0
10. V
V R Vp =
R
E r dr
Ex = d V
0
dx , so
gilvin (jg47854) - 8: The Electric Potential - meyers - (21235) dV = Ex dx x2 V 2 V1 = Ex dx
x1
4
8. Q2 = Q1 Q2 9. Q2 = Q2 Q1 2 Explanation: Basic Concepts: Gauss Law Sketch a concentric Gaussian surface S (dashed line) within the shell. r
= (4 kN/C) = (4 kN/C)
x2
x 3 dx
( x1 ) 1( ( 4 1 4 x 2 x1) )
= (4 kN/C) [
(5 m)4 (1 m)4] = 624 kV . 011 (part 1 of 3) 10.0 points Consider a solid conducting sphere with a radius a and charge Q1 on it. There is a conducting spherical shell concentric to the sphere. The shell has an inner radius b (with b > a) and outer radius c and a net charge Q2 on the shell. Denote the charge on the inner surface of the shell by Q 2 and that on the outer surface of the shell by Q2 . Q1 , a P b , Q 2 Q2
Since the electrostatic field in a conducting medium is zero, according to Gausss Law, S = Q1 + Q2 = 0 0 Q 2 = Q1 But the net charge on the shell is Q2 = Q 2 + Q2 , so the charge on the outer surface of the shell is Q2 = Q2 Q 2 = Q2 + Q1 . 012 (part 2 of 3) 10.0 points Find the (magnitude )f the electric field at point P E P EP , where the distance from P to the center is r = a + b 2 1. E P = 2 ke(aQ1 b)2Q2) + 2. EP = 2 ke(aQ1 b)2Q2) + 3. E P = 4 ke Q2 (a + b)2
Q 2 , c Find the charge Q2. 1. Q2 = Q1 + Q2 correct 2. Q2 = 2 (Q2 Q1) 3. Q2 = 2 (Q1 + Q2) 4. Q2 = Q2 Q1 5. Q2 = Q1 + Q2 2 6. Q2 = 2 (Q1 Q2) 7. Q2 = Q1 Q2 2
Q1
4. EP = 4 ke(aQ1 b)2Q2) +
gilvin (jg47854) - 8: The Electric Potential - meyers - (21235) 4 ke Q1
P
5
5. EP =
(a + b) correct
2
6. E P = 4 ke(aQ1 b)2Q2) + 7. E P = 0 2 ke Q1 8. E P = (a + b)2 9. E P = 2 ke Q22 (a + b) Explanation: Choose as your Gaussian surface concentric with the spherical surface S, which passes through P . Here, A = 4 r2 E E = Q1 0 E
P
c Explanation: Using the superposition principle, adding the 3 concentric charge distributions; i.e., Q1 at a, Q1 at b and Q1 + Q2 at c, gives Er VP = dr , by symmetry,
9. V
= 2ak e Q1 b ? k e Q2 +
= Er dr
=
c c
c
b
0 dr
b
(a+b)/2
=
Er dr k (Q1 + Q2)
b c
Er dr
P
r2 0 dr
dr
(a+b)/2 b c
k (Q1 + Q2) r2
dr
= k (Q1 + Q2)
= 4 Q1 0 r2 = ke Q 1 r2 = 4 ke Q1 (a + b)2 .
(1) i ri
i i (a+b)/2 0 k (Q1 + Q2) v r b e Q1 = 2 ke Q1 a + b ? k + ke (Q1 + Q2) b c 014 (part 1 of 5) 10.0 points Consider two solid conducting spheres with radii r1 = 2 R and r2 = 5 R ; i.e., r2 r1 = 2 R = 2 . The two spheres are separated by a large distance so that the field and the potential at the surface of sphere #1 only depends on the charge on #1 and the corresponding quantities on #2 only depend on the charge on #2. Place an equal amount of charge on both spheres, q1 = q2 = Q . r2 r1
(
1
)
013 (part 3 of 3) 10.0 points Assume: The potential at r = is zero. Find the potential V P at point P . 1. V 2. V 3. V 4. V 5. V
P P
= 2ak e Q1 b + k e Q2 + c
P
= 2 k e (Q1 ? Q2) a+b = ae Q1 b ? k e Q2 + b = 2ak e Q1 +b = 2ak e + 1b ? 2 k e Q2 Q b
P
P
6. V P = 0 7. V 8. V rect
P
= 2ak e Q1 b + k e Q1e (Q1 Q2) + k b c = 2ak e+ b ? k e+Q1e (Q1 + Q2) corQ1 k b c
#1
q1 q2 #2
P
gilvin (jg47854) - 8: The Electric Potential - meyers - (21235) After the electrostatic equilibrium on each sphere has been established, what is the ratio of the potentials V2 at the centers of the V1 two solid conducting spheres? 1. V2 V1 = 2 2. V2 V1 = 5 correct = r1 r2 =2R 5R =2 5. 015 (part 2 of 5) 10.0 points What is the ratio of the electric fields the surfaces of the two spheres? 1. E2 E1 = 2 2. E2 E1 = 1 3. E2 E1 = 252 4. E2 E1 = 258 5. E2 E1 = 245 correct
6
3. V2
E2 at E1
V1 = 1
4. V2 V1 = 4 5. V2 V1 = 245 6. V2 V1 = 258 7. V2 V1 = 254 8. V2 V1 = 252 9. V2 V1 = 5 Explanation: For a solid conducting sphere, the charge is uniformly distributed at the surface. From Gauss Law, the electric field outside the sphere is given by E(r) = k Q r2 , where Q is the total charge on the sphere and r is the distance from the center of the sphere. By integration with respect to r, the potential can be expressed as V (r) = k Q r , so the potential at the surface of the sphere is V (r) = k Q (1)
6. E2
E1 = 5
7. E2 E1 = 4 8. E2 E1 = 254 9. E2 E1 = 5 Explanation: For a conducting sphere, the charge is uniformly distributed at the surface. Based on Gauss law, the electric field on the surface of a conducting sphere of radius R with charge Q is E(r) = ke Q r2 , where r R . (2)
r, where R is radius of the sphere and r R . For the electrostatic case, the potential is constant throughout a conducting body, so the potential at the center is the same as anywhere on the conductor. Thus at two centers k q2 V2 r2 V1 = k q 1 r1
Thus on the surface r = R of the two spheres, k q2 E2 r2 E1 = k q2 r1
gilvin (jg47854) - 8: The Electric Potential - meyers - (21235) 7 ( )2 charge will flow from one to the other until = r1 the potential on both spheres is the same. r (2) 2 As noted, V2 =2 V1 = 1 , defines equilibrium. R The spheres are connected by a wire and no 5 R ( )2 current is flowing (at equilibrium), therefore 2 the ends of the wire are at the same potential = V2 = V1 . (3) 5 =4 25 . 017 (part 4 of 5) 10.0 points E2 Now, what is the ratio of the electric fields 016 (part 3 of 5) 10.0 points E1 Now connect the two spheres with a wire. at the surfaces of the two spheres? r2 r1 #1 q1 q2 #2 1. E2 E1 = 2 2. E2 E1 = 4 3. E2 E1 = 254 4. E2 E1 = 252 5. E2 E1 = 5 6. E2 E1 = 5 correct
There will be a flow of charge through the wire until equilibrium is established. V2 What is the ratio of the potentials V at 1 the centers of the two spheres? 1. V2 V1 = 2 2. V2 V1 = 254 3. V2 V1 = 4 4. V2 V1 = 252 5. V2 V1 = 245 6. V2 V1 = 1 correct
7. E2
E1 = 258
8. E2 E1 = 1 9. E2 E1 = 245 Explanation: For a conducting sphere, the charge is uniformly distributed at the surface. Based on Gauss law, on the surface of a conducting sphere of radius R with charge Q is E(r) = ke Q r2 , where r R , and V (r) = k Q r , where r R . Thus on the surface r = R of the two spheres, E2 E1 = k q2 r2 k q2 r1 (4)
7. V2
V1 = 5
8. V2 V1 = 258 9. V2 V1 = 5 Explanation: When the spheres are connected by a wire,
gilvin (jg47854) - 8: The Electric Potential - meyers - (21235) 1 k q2 r2 r2 k q1 1 r1 r1 1 V2 r 2 = 1, V1 r1 = = r1 r2 =2R 5R =2 5. 018 (part 5 of 5) 10.0 points Now, what is the charge q1 on sphere #1? 1. q1 = 5 7Q 2. q1 = 7 2Q 7 3. q1 = 10 Q 4. q1 = 2 7Q 5. q1 = 7 4Q =5 2 q1 .
8 (5)
since V1 = V2
The total charge of the system remains constant; i.e., from the initial condition q1 = q2 = Q, the total change on both spheres is q1 + q2 = 2 Q. Using q2 from Eq. 5, we have q1 + q2 = 2 Q q1 + 5 2 q1 = 2 Q 7 2 q1 = 2 Q 4 q1 = 7 Q . And the charge on sphere # 2 is q2 = 10 7 Q, since q1 + q2 = 4 7 Q + 107 Q = 2 Q . Check Eq. 4: On the surfaces of the two spheres, E2 E1 = ( q2 q1 )( r1 r2 )2
Explanation:
6. q1 = 7 5Q 7. q1 = 10 7Q 8. q1 = Q 9. q1 = 4 7 Q correct
i i i 10 =i Q( )2 7 i 2R i (4 5R 7Q ) ( )2 5 2 = 2 5 =2 5. Fourth of eighteen versions. 019 10.0 points The uniformly charged sphere gives a reasonable model of the proton charge distribution in an atomic nucleus. A nucleus of lead-208 has a radius of 2.6691015 m and contains 82 protons, each with a charge of 1.6 1019 C. Calculate the electric potential at the surface of this nucleus. Correct answer: 4.41801 107 V. Explanation:
When the spheres are connected by a wire, charge will flow from one to the other until the potential on both spheres is the same. In this case, this implies that ke q1 r1 = ke r2 , or q2 = r2 r1 q 1 =5R 2 R q1
gilvin (jg47854) - 8: The Electric Potential - meyers - (21235)
9
Let : r = 2.669 1015 m , e = 1.6 1019 C , Z = 82 , and ke = 8.98755 109 N m2/C2 . The potential is V = ke dq r
021 (part 1 of 2) 10.0 points A uniformly charged sphere has a potential on its surface of 430 V. At a radial distance of 20 cm from this surface, the potential is 120 V. The Coulomb constant is 8.99 9 N m2/C2. 10 What is the radius of the sphere? Correct answer: 0.0774194 m. Explanation: Let : Vs = 430 V , Vd = 120 V , and
Gausss law tells us that the expression for the electric field outside of a sphere (charge Q) is the same as that of a point charge located at the center of the sphere with the same charge Q. Thus we can calculate the electric potential at the surface of the sphere by assuming all the charge is concentrated at the center: V = k e Qtot r = ke Z e r = 8.98755 109 N m2/C2 ( 82 1.6 1019 C) 2.669 1015 m = 4.41801 107 V . 020 10.0 points A 13 V battery is connected across two parallel metal plates separated by 0.26 cm. Find the magnitude of the electric field. Correct answer: 5000 V/m. Explanation: Let : V = 13 V and d = 0.26 cm .
V 1
d = 20 cm = 0.2 m . is
R , so the ratio of the potentials Vs +d Vd = R R
Vs R = Vd (R + d) R = Vd d
Vs V d = (120 V) (0.2 m) 430 V 120 V = 0.0774194 m .
022 (part 2 of 2) 10.0 points What is the charge of the sphere? Correct answer: 3.70304 nC. Explanation: Let : k = 8.99 109 N m2/C2 .
The potential on surface is Vs = k Q R Q = Vs R k 8.99 * 109 N E m 2/C2 E 109 nC = (430 V) (0.0774194 m) 1C = 3.70304 nC . keywords:
V = E d | E | = V d 13 V = 0.0026 m = 5000 V/m .

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HW1 Solutions1. In 1995, the IBM AS/400 line of computers transitioned from a CISC instruction set to a RISC instruction set. Due to the simpler instruction set, the realizable clock frequency for a given technology generation and the CPI metric improved

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ESD.36J System and Project Management HW1 UAV Development projectMaster Solution Due: 9/18/031. Construct a task table from the UAV description. Clearly designate each task with its tag, description and identify immediate predecessors and expected task

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Math 220a - Hw1 solutions (Revised) 1. Linear - (b) Semilinear - (a),(d) Quasilinear -(e) Fully nonlinear - (c) 2. The equations for the characteristic curves are dt(r, s) ds dx(r, s) ds du(r, s) ds =1 =x = t3The boundary conditions can be re-written in

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C S/MA320 HW1 Solutions - Parts 1 & 21 .1 Reading. All of this is relatively important. A lot of it is how to translate between Logic and English sentences - I will not spend much time on this in class, but will answer questions. When in doubt as to impo

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ECE302 Spring 2006HW1 SolutionsJanuary 16, 20061Solutions to HW1Note: These solutions were generated by R. D. Yates and D. J. Goodman, the authors of our textbook. I have added comments in italics where I thought more detail was appropriate.Problem

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HW1SolutionsProblem 1: Each of ten people simultaneously requests a distinct movie from a new service that broadcasts 3D movies. These 3D movies are broadcast as a left and a right stream, which are distinct. (So 10 distinct movies, 20 distinct streams

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Suggested solutions for HW2 Econ 3070February 5, 2009Carefully show how you derive you answer and be sure to interpret your answer where necessary. 1. (a) The utility function is U (H) = 10H - H 225 20 15 U(H) 10 5024H6810Marginal utility is: U

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Principles of MRI EE290T/BIOE265, Spring 2010 Principles of MRIEE225E/BIO265, Spring 2011Miki LustigHW2 Solution1.Nishimura 3.3 Answer1232.Precession in Tops (Double Points: brain teasers)(a) Explain why only spinning tops precess. That is, exp

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HW1 SOLUTION1. Chapter 2 Problem 2: Suppose you have algorithms with the six running times listed below.(Assume these are the exact number of operations performed as a function of the input size n.) Suppose you have a computer that can perform 1010 opera

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HW2 Solutions1. (P5.7) Assume that an one-bit (history bit) state machine is used as the prediction algorithm for predicting the execution of the two branches in this loop. Indicate the predicted and actual branch directions of the b1 and b2 branch instr

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ESD.36J System and Project Management HW2 Task-based Design Structure MatrixMaster Solution Due: 9/18/031. Construct a Design Structure Matrix (DSM), with dimensions 13x13, including all UAV tasks b through n (all tasks except "start" and "finish"). Thi

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HW2 MAT 267 Advanced Ordinary Differential EquationsDue: October 18, 20101. An n-th order homogeneous linear differential equation is an ODE of the form y (n) + p1 (t)y (n1) + p2 (t)y (n2) + + pn2 (t)y + pn1 (t)y + pn (t)y = 0, where y (k) denotes the k

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CS152, Spring 2011, Assignment 2 Due: Thursday 24 February 2011, 10:00AMLast updated: February 8 Ensure you understand the course policies for assignments. hw2.tar, available on the course website, contains several Caml les you need. This assignment cons

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Math103B Homework Solutions HW2Jacek Nowacki April 29, 2007Chapter 14Problem 6. Find all maximal ideals in a. Z8 , b. Z10 , c. Z12 , d. Zn . Proof. We begin by noticing that each ring in this problem considered as a group is cyclic (this has to do with

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Srping 2007 Math 510 HW2 SolutionSection 2.4.3 16. Prove that in a group of n > 1 people there are two who have the same number of 2 acquaintances in the group. (it is assumed that no is acquainted with himself or herself.)Proof. Let a1 , a2 , . . . ,

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Math 103A Homework Solutions HW2Jacek Nowacki January 28, 2007Chapter 2Problem 8. Show that the set cfw_5, 15, 25, 35 is a group under multiplication modulo 40. What is the identity element of this group? Can you see any relationship between this group

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[SOLUTION] HW2:SQL CIS330 Due: Feb. 8th, 2011This problem set involves using the relational database system Oracle, which is available on eniac. The query language for this system is SQL. Instructions for creating an Oracle account and starting the Oracl

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HW2 SolutionsDilip Raghavan September 13, 20081Section 5.2Problem 28. Though the problem does not make this clear, I'm assuming that the teams are not ordered. That is, if I take all the members of Team 1 and swap them with all the members of Team 2,