Chapter 21: Electric Potential Solutions Questions: 2, 7, 14 Exercises & Problems: 1, 6, 10, 12, 16, 21, 25, 27, 71 Q21.2: Charge q is fired through a small hole in the positive plate of a capacitor. a) If q is a positive charge, does it speed up or slow down inside the capacitor? Answer this question twice: (i) Using the concept of force. (ii) Using the concept of energy. b) Repeat part (a) if q is negative charge. Q21.2. Reason: (a) (i) When a positive charge enters the region between the plates at the site of the positive plate, the positive plate will exert a repulsive force that will accelerate it and the negative plate will exert an attractive force that will also accelerate it. Since both forces accelerate the charge, it will speed up. (ii) When a positive charge enters the region between the plates at the site of the positive plate it has a maximum electric potential energy. As the particle travels from the positive to the negative plate the electric potential energy decreases and the kinetic energy increases. As the kinetic energy of the particle increases, it speeds up. (b) (i) When a negative charge enters the region between the plates at the site of the positive plate, the positive plate will exert an attractive force that will decelerate it and the negative plate will exert a repulsive force that will also decelerate it. Since both forces decelerate the charge, it will slow down. (ii) When a negative charge enters the region between the plates at the site of the positive plate it has a minimum electric potential energy. As the particle travels from the positive to the negative plate the electric potential energy increases and the kinetic energy decreases. As the kinetic energy of the particle decreases, it will slow down. Assess: One of the principle lessons from this problem is that one can frequently answer a question using different approaches. We have answered the question using force considerations and then again using energy considerations. Getting the same answer by both methods gives us confidence that our logic is correct (for both cases). It should be noted that when a particle with the opposite charge is used, the result is reversed. Q21.7: As shown in the figure, two protons are launched with the same speed from point 1 inside a parallel-plate capacitor. One proton moves along the path from 1 to 2, the other from 1 to 3. Points 2 and 3 are the same distance from the positive plate. a) Is U 12, the change in potential energy along the path 1 2, larger than, smaller than, or equal to U13? Explain. b) Is the proton s speed v2 at point 2 larger than, smaller than, or equal to the proton s speed v 3 at point 3? Explain.
Q21.7. Reason: (a) The electric field between the plates is uniform and since both protons move the same distance in a direction opposite that of the electric field, they both experience the same electric potential difference ( = E d) and hence the same change in electric potential energy ( U = q = qe d). (b) At point 1, the protons have the same electric potential energy and the same kinetic energy (since they are at the same position and have the same mass and speed). Since they both experience the same increase in electric potential energy, they will both experience the same decrease in kinetic energy. This means that the kinetic energy of the proton at point 2 will be equal to the kinetic energy of the proton at point 3, hence that have the same speed. Assess: This question does a nice job of showing that potential energy and kinetic energy are tied together in a conservative force system. Q21.14: Rank in order, from largest to smallest, the electric field strengths E 1, E2, E3, and E4 at the four labeled points in the figure. Explain. Q21.14. Reason: Since the figure gives information about the equipotential lines and shows the distance between these lines, we E = / d can use to get a qualitative value for E at each site. A qualitative value is adequate for comparison purposes. In order for us to talk about the different potential differences ( ) and distances ( d) let s label them as shown in the figure. Notice that each Looking at the d Using E / d, (between the first equipotential line on either side) is 20 volts: for each case we see that 1 = 2 = 3 = 4 = 20. d = d > d > d 2 4 1 3 = this allows us to conclude that E3 < E1 < E2 = E4 Assess: Since the electric field is not uniform, the expression value for E when the value for d is small. E / d = gives a better P21.1: Moving a charge from point A, where the potential is 300, to point B, where the potential is 150, takes 4.5 10-4 J of work. What is the value of the charge?
P21.1. W = U = q Prepare: The work done is equal to the change in electric potential energy: elec. = 150 300 = 150. We are given that Solve: Solve the first equation for q: 4 W 4.5 10 J q= = = 3.0 10 150 Assess: The answer is negative because it requires positive work to move a negative charge to a lower potential. As for units, J/ = C. 6 C P21.6: An electron has been accelerated from rest through a potential difference of 1000. a) What is its kinetic energy, in electron volts? b) What is its kinetic energy, in joules? c) What is its speed? P21.6. Prepare: Mechanical energy is conserved. The potential energy is determined by the electric potential and is given by U = q. The figure shows a before-and-after pictorial representation of an electron moving through a potential difference of 1000. A negative charge speeds up as it moves into a region of higher potential (U K). We also note that 1 e = 1.60 10 19 J. Solve: (a) Conservation of energy, expressed in terms of the electric potential, is K + q = K + q K = K q( ) (b) K f f f i i f i f i K = ( e)( ) = e(1000 ) = 1000 e f 19 1.60 10 J 16 (1000 e) 1.60 10 J = = 1e (c) Because K 2 f = 1/2 mvf, so 1 2 16 f 1.60 10 J 2 mv = v = 16 2(1.6 10 J) f 31 9.11 10 kg = 1.9 10 7 m/s Assess: The electric potential difference of 1000 already existed in space due to other charges or sources. The electron of our problem has nothing to do with creating the potential. P21.10: A proton with an initial speed of 800,000 m/s is brought to rest by an electric field. a) Did the proton move into a region of higher potential or lower potential?
b) What was the potential difference that stopped the proton? c) What was the initial kinetic energy of the proton, in electron volts? P21.10. Prepare: Energy is conserved. The potential energy is determined by the electric potential. The figure shows a before-and-after pictorial representation of a proton moving through a potential difference. Solve: (a) Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential. (b) Using the conservation of energy equation, (c) Assess: 1 1 1 2 Kf + Uf = Ki + Ui Kf + qf = Ki + qi f i = ( Ki Kf) = mvi 0J q () e 2 2 27 5 2 mvi (1.67 10 kg)(8.0 10 m/s) = = = 3340 = 3000 19 2e 2(1.60 10 C) 1 2 Ki = mv i = q = ( e)(3340) = 3000 e 2 A positive confirms that the proton moves into a higher potential region. P21.12: A 2.0 cm 2.0 cm parallel-plate capacitor has a 2.0 mm spacing. The electric field strength inside the capacitor is 1.0 10 5 /m. a) What is the potential difference across the capacitor? b) How much charge is on each plate? P21.12. Prepare: The electric potential difference between the plates is determined by the uniform electric field in the parallel-plate capacitor and is given by Equation 21.6. Solve: (a) The potential difference C across a capacitor of spacing d is related to the electric field inside by Equation 21.6: = = = = d C 5 E C Ed (1.0 10 /m)(0.002 m) 200 (b) The electric field of a capacitor is related to the charge by Equation 20.7: Q = ε = = 12 2 2 4 2 5 10 0 AE (8.85 10 C /(N m ))(4.0 10 m )(1.0 10 /m) 3.5 10 C Assess: A charge of 0.35 nc on the positive plate and an equal negative charge on the negative plate create a significant potential difference across the parallel. P21.16: a) What is the electric potential at points A, B, and C in the figure? b) What is the potential energy of an electron at each of these points?
c) What are the potential differences AB and BC? P21.16. Prepare: Please refer to Figure P21.16. The charge is a point charge. We will use Equations 21.10 and 21.1 to calculate the potential and the potential energy of the charge. Solve: (a) The electric potential of the point charge q is q = = (9.0 10 N m /C ) = 4πε 0 r r r For points A and B, r = 0.01 m. Thus, A 9 2 1 9 2 2 2.0 10 C 18.0 N m /C 2 18.0 N m /C N m = B = = 1800 = 1800 m 1800 0.01m C = m For point C, r = 0.02 m and C = 900. (b) The potential energy of a charge q at a point where the electric potential is is U = q. The expression for the potential in part (a) assumes that we have chosen = 0 to be the potential at r =. So, we are obtaining potential/potential energy relative to a zero of potential/potential energy at infinity. Thus, U U q e (c) The potential differences are 19 16 A = B = ( ) = ( )( ) = (1.60 10 C)(1800 ) = 2.9 10 J 19 16 U C ( 1.60 10 C)(900 ) 1.4 10 J = = AB = B A = 1800 1800 = 0 BC = C B = 900 1800 = 900 Assess: Clearly A = B and C and A, so, as expected, AB = 0 and BC is negative. P21.21: In the figure, the electric potential at point A is 300. What is the potential at point B, which is 5.0 cm to the right of A? P21.21. Prepare: In order to use Equation 21.17, E = / d, we need to know the shortest distance between two equipotential surfaces. We are given the distance between point A which is on one surface and point B which is on another surface. However this is not the shortest distance, as you can see from the figure. The shortest distance, d, is given by d = (5.0 cm)cos30 = 2.5 3 cm.
Solve: Since the electric field vectors point from the surface containing A to the surface containing B, A is at a higher potential and A B will be a positive potential difference. From Equation 21.17, the potential difference is given by: Ed 2 A B = = (1200 /m)(2.5 3 10 m) = 52 Given that A = 300, we conclude that B = 352. Assess: Even though the electric field vectors do not point directly from point A to point B, the potential is less at B than at A because the displacement vector from A to B forms an acute angle, 30, with the electric field. That means that to get from A to B, one component of your motion must be in the direction of the field. P21.25: One standard location for a pair of electrodes during an EKG is shown on the right. The potential difference 31 = 3 1 is recorded. For each of the three instants a, b, and c during the heart s cycle shown below, will 31 be positive or negative? Explain. P21.25. Prepare: Examine Figure 21.29. In parts (a) and (c) of the problem electrode 3 is at a positive potential and electrode 1 is at a negative potential, so the difference 31 = 3 1 is positive. In part (b) electrode 1 is at a positive potential and electrode 3 is at a slightly negative potential, so 31 is negative.
Solve: (a) positive (b) negative (c) positive Assess: Think of the positive potentials as up on a hill and negative potentials as in a valley. P21.27: Two 2.0 cm 2.0 cm square aluminum electrodes, spaced 0.50 mm apart, are connected to a 100 battery. a) What is the capacitance? b) What is the charge on the positive electrode? P21.27. Prepare: We will assume that the battery is ideal and that the capacitor is a parallel-plate capacitor. We will use Equations 21.18 and 21.20. Solve: (a) From Equation 21.20, the capacitance is 12 2 2 ε 0 A (8.85 10 C /(N m ))(0.02 m 0.02 m) C = = = = d 0.0005 m (b) The charges on the electrodes are Assess: 12 7.1 10 F 7.1pF 12 10 Q C C (7.1 10 F)(100 ) 7.1 10 C 0.71nC =± =± =± =± Practical capacitors are usually measured in units of µ F or pf. P21.71: In proton-beam therapy, a high-energy beam of protons is fired at a tumor. The protons come to rest in the tumor, depositing their kinetic energy and breaking apart the tumor s DNA, thus killing its cells. For one patient, it is desired that 0.10 J of proton energy be deposited in a tumor. To create the protons beam, the protons are accelerated from rest through a 10 M potential difference. What is the total charge of the protons that must be fired at the tumor to deposit the required energy? P21.71. Prepare: From conservation of energy, the increase in kinetic energy of the K = U = q. protons equals their decrease in potential energy. In other words, In the following figure, the source of the potential difference is a capacitor, with the negatively charged plate having a hole cut in the center to allow protons through. Notice that while the protons are between the plates, they are accelerating, but once they pass through the plate on the right, they maintain a constant speed. This is because there is no electric field outside an ideal capacitor. Solve: If we solve the equation for q, we get: K K 0.10 J q = = = = = 6 10 10 8 1.0 10 C 10 nc
Here we have used the fact that since the protons are accelerating, they must be traveling =. through a decreasing potential. Consequently, is negative and Assess: Since the protons have a charge of e, their kinetic energy after passing the potential difference is = = 1 e 19 1.60 10 J 6 12 10 Me (10 10 e) 1.60 10 J. Since the total energy N = = 12 10 (0.10 J)/(1.60 10 J) 6.3 10. was 0.10 J, we can determine the number of protons, N : 10 19 (6.3 10 )(1.60 10 C) = 10 nc, As a check, the charge of this quantity of protons is as we find in the statement of the problem.