21 ELECTRIC CHARGE AND ELECTRIC FIELD 21.1 (a) IDENTIFY and SET UP: Use the charge of one electron (−1.602 × 10−19 C) to find the number of electrons required to produce the net charge EXECUTE: The number of excess electrons needed to produce net charge q is q −3.20 × 10−9 C = = 2.00 × 1010 electrons −e −1.602 × 10−19 C/electron (b) IDENTIFY and SET UP: Use the atomic mass of lead to find the number of lead atoms in 8.00 × 10−3 kg of lead From this and the total number of excess electrons, find the number of excess electrons per lead atom EXECUTE: The atomic mass of lead is 207 × 10−3 kg/mol, so the number of moles in 8.00 × 10−3 kg is n= 8.00 × 1023 kg mtot = = 0.03865 mol N A (Avogadro’s number) is the number of atoms in mole, M 207 × 10−3 kg/mol so the number of lead atoms is N = nN A = (0.03865 mol)(6.022 × 1023 atoms/mol) = 2.328 × 1022 atoms 21.2 2.00 × 1010 electrons = 8.59 × 10−13 2.328 × 1022 atoms EVALUATE: Even this small net charge corresponds to a large number of excess electrons But the number of atoms in the sphere is much larger still, so the number of excess electrons per lead atom is very small IDENTIFY: The charge that flows is the rate of charge flow times the duration of the time interval SET UP: The charge of one electron has magnitude e = 1.60 × 10−19 C The number of excess electrons per lead atom is EXECUTE: The rate of charge flow is 20,000 C/s and t = 100 μs = 1.00 × 10−4 s Q = (20,000 C/s)(1.00 × 10−4 s) = 2.00 C The number of electrons is ne = 21.3 Q = 1.25 × 1019 1.60 × 10−19 C EVALUATE: This is a very large amount of charge and a large number of electrons IDENTIFY: From your mass estimate the number of protons in your body You have an equal number of electrons SET UP: Assume a body mass of 70 kg The charge of one electron is −1.60 × 10−19 C EXECUTE: The mass is primarily protons and neutrons of m = 1.67 × 10−27 kg The total number of protons and neutrons is np and n = 70 kg 1.67 × 10−27 kg = 4.2 × 1028 About one-half are protons, so np = 2.1 × 1028 = ne The number of electrons is about 2.1 × 1028 The total charge of these electrons is Q = (−1.60 × 10−19 C/electron )(2.10 × 1028 electrons) = −3.35 × 109 C EVALUATE: This is a huge amount of negative charge But your body contains an equal number of protons and your net charge is zero If you carry a net charge, the number of excess or missing electrons is a very small fraction of the total number of electrons in your body © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-1 21-2 21.4 Chapter 21 IDENTIFY: Use the mass m of the ring and the atomic mass M of gold to calculate the number of gold atoms Each atom has 79 protons and an equal number of electrons SET UP: N A = 6.02 × 1023 atoms/mol A proton has charge +e EXECUTE: The mass of gold is 17.7 g and the atomic weight of gold is 197 g/mol So the number of atoms is ⎛ 17.7 g ⎞ 22 N A n = (6.02 × 1023 atoms/mol) ⎜ ⎟ = 5.41 × 10 atoms The number of protons is ⎝ 197 g/mol ⎠ np = (79 protons/atom)(5.41×1022 atoms) = 4.27 ×1024 protons Q = (np )(1.60 × 10−19 C/proton) = 6.83 × 105 C (b) The number of electrons is ne = np = 4.27 × 1024 21.5 EVALUATE: The total amount of positive charge in the ring is very large, but there is an equal amount of negative charge IDENTIFY: Each ion carries charge as it enters the axon SET UP: The total charge Q is the number N of ions times the charge of each one, which is e So Q = Ne, where e = 1.60 × 10−19 C EXECUTE: The number N of ions is N = (5.6 × 1011ions/m)(1.5 × 10−2 m) = 8.4 × 109 ions The total charge Q carried by these ions is Q = Ne = (8.4 × 109 )(1.60 × 10−19 C) = 1.3 × 10−9 C = 1.3 nC 21.6 EVALUATE: The amount of charge is small, but these charges are close enough together to exert large forces on nearby charges IDENTIFY: Apply Coulomb’s law and calculate the net charge q on each sphere SET UP: The magnitude of the charge of an electron is e = 1.60 × 10−19 C EXECUTE: F = q2 4π ⑀0 r This gives q = 4π ⑀0 Fr = 4π ⑀ (4.57 × 10−21 N)(0.200 m) = 1.43 × 10−16 C And therefore, the total number of electrons required is n = q /e = (1.43 × 10−16 C)/(1.60 × 10−19 C/electron) = 890 electrons 21.7 EVALUATE: Each sphere has 890 excess electrons and each sphere has a net negative charge The two like charges repel k q1q2 IDENTIFY: Apply F = and solve for r r2 SET UP: F = 650 N EXECUTE: r = 21.8 k q1q2 (8.99 × 109 N ⋅ m /C2 )(1.0 C) = = 3.7 × 103 m = 3.7 km F 650 N EVALUATE: Charged objects typically have net charges much less than C IDENTIFY: Use the mass of a sphere and the atomic mass of aluminum to find the number of aluminum atoms in one sphere Each atom has 13 electrons Apply Coulomb’s law and calculate the magnitude of charge q on each sphere SET UP: N A = 6.02 × 1023 atoms/mol q = n′ee, where n′e is the number of electrons removed from one sphere and added to the other EXECUTE: (a) The total number of electrons on each sphere equals the number of protons ⎛ ⎞ 0.0250 kg 24 ne = np = (13)( N A ) ⎜ ⎟ = 7.25 × 10 electrons 0.026982 kg/mol ⎝ ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field (b) For a force of 1.00 × 104 N to act between the spheres, F = 1.00 × 104 N = q2 4π ⑀0 r 21-3 This gives q = 4π ⑀ (1.00 × 104 N)(0.800 m) = 8.43 × 10−4 C The number of electrons removed from one sphere and added to the other is n′e = q /e = 5.27 × 1015 electrons (c) n′e /n e = 7.27 × 10−10 21.9 EVALUATE: When ordinary objects receive a net charge the fractional change in the total number of electrons in the object is very small IDENTIFY: Apply Coulomb’s law SET UP: Consider the force on one of the spheres (a) EXECUTE: q1 = q2 = q F= q1q2 4π ⑀0 r = q2 4π ⑀0r so q = r F 0.220 N = 0.150 m = 7.42 × 10−7 C (on each) (1/4π ⑀0 ) 8.988 × 109 N ⋅ m /C2 (b) q2 = 4q1 F= q1q2 4π ⑀0 r = 4q12 4π ⑀0r so q1 = r F F = 12 r = (7.42 × 10−7 C) = 3.71 × 10−7 C 4(1/4π ⑀0 ) (1/4π ⑀ ) And then q2 = 4q1 = 1.48 × 10−6 C 21.10 EVALUATE: The force on one sphere is the same magnitude as the force on the other sphere, whether the spheres have equal charges or not IDENTIFY: We first need to determine the number of charges in each hand Then we can use Coulomb’s law to find the force these charges would exert on each hand SET UP: One mole of Ca contains N A = 6.02 × 1023 atoms Each proton has charge e = 1.60 × 10−19 C The force each hand exerts on the other is F = k q2 r2 EXECUTE: (a) The mass of one hand is (0.010)(75 kg) = 0.75 kg = 750 g The number of moles of Ca is n= 750 g = 18.7 mol The number of atoms is 40.18 g/mol N = nN A = (18.7 mol)(6.02 × 1023 atoms/mol) = 1.12 × 1025 atoms (b) Each Ca atom contains positive charge 20e The total positive charge in each hand is N e = (1.12 × 1025 )(20)(1.60 × 10−19 C) = 3.58 × 107 C If 1.0% is unbalanced by negative charge, the net positive charge of each hand is q = (0.010)(3.58 × 107 C) = 3.6 × 105 C 21.11 (c) The repulsive force each hand exerts on the other would be (3.6 × 105 C) q2 F = k = (8.99 × 109 N ⋅ m /C2 ) = 4.0 × 1020 N This is an immense force; our hands (1.7 m) r would fly off EVALUATE: Ordinary objects contain a very large amount of charge But negative and positive charge is present in almost equal amounts and the net charge of a charged object is always a very small fraction of the total magnitude of charge that the object contains qq IDENTIFY: Apply F = ma, with F = k 22 r SET UP: a = 25.0 g = 245 m/s An electron has charge − e = −1.60 × 10−19 C EXECUTE: F = ma = (8.55 × 10−3 kg)(245 m/s ) = 2.09 N The spheres have equal charges q, so F =k q2 r and q = r 2.09 N F = (0.150 m) = 2.29 × 10−6 C k 8.99 × 109 N ⋅ m /C2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-4 Chapter 21 q 2.29 × 10−6 C = = 1.43 × 1013 electrons The charges on the spheres have the same sign so the e 1.60 × 10−19 C electrical force is repulsive and the spheres accelerate away from each other EVALUATE: As the spheres move apart the repulsive force they exert on each other decreases and their acceleration decreases IDENTIFY: We need to determine the number of protons in each box and then use Coulomb’s law to calculate the force each box would exert on the other SET UP: The mass of a proton is 1.67 × 10−27 kg and the charge of a proton is 1.60 × 10−19 C The N= 21.12 distance from the earth to the moon is 3.84 × 108 m The electrical force has magnitude Fe = k where k = 8.99 × 109 N ⋅ m /C2 The gravitational force has magnitude Fgrav = G m1 m2 r2 q1 q2 r2 , , where G = 6.67 × 10−11 N ⋅ m /kg EXECUTE: (a) The number of protons in each box is N = 1.0 × 10−3 kg 1.67 × 10−27 kg = 5.99 × 1023 The total charge of each box is q = Ne = (5.99 × 1023 )(1.60 × 10−19 C) = 9.58 × 104 C The electrical force on each box is Fe = k q2 r2 = (8.99 × 109 N ⋅ m /C2 ) (9.58 × 104 C) (3.84 × 108 m) = 560 N = 130 lb The tension in the string must equal this repulsive electrical force The weight of the box on earth is w = mg = 9.8 × 10−3 N and the weight of the box on the moon is even less, since g is less on the moon The gravitational forces exerted on the boxes by the earth and by the moon are much less than the electrical force and can be neglected mm (1.0 × 10−3 kg) (b) Fgrav = G 2 = (6.67 × 10−11 N ⋅ m /kg ) = 4.5 × 10−34 N r (3.84 × 10 m) 21.13 EVALUATE: Both the electrical force and the gravitational force are proportional to 1/r But in SI units the coefficient k in the electrical force is much greater than the coefficient G in the gravitational force And a small mass of protons contains a large amount of charge It would be impossible to put 1.0 g of protons into a small box, because of the very large repulsive electrical forces the protons would exert on each other IDENTIFY: In a space satellite, the only force accelerating the free proton is the electrical repulsion of the other proton SET UP: Coulomb’s law gives the force, and Newton’s second law gives the acceleration: a = F/m = (1/4π ⑀0 )(e /r )/m EXECUTE: (a) a = (9.00 ×109 N ⋅ m /C )(1.60 ×10−19 C)2 /[(0.00250 m) (1.67 ×10−27 kg)] = 2.21×104 m/s (b) The graphs are sketched in Figure 21.13 EVALUATE: The electrical force of a single stationary proton gives the moving proton an initial acceleration about 20,000 times as great as the acceleration caused by the gravity of the entire earth As the protons move farther apart, the electrical force gets weaker, so the acceleration decreases Since the protons continue to repel, the velocity keeps increasing, but at a decreasing rate Figure 21.13 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field 21.14 21-5 IDENTIFY: Apply Coulomb’s law SET UP: Like charges repel and unlike charges attract EXECUTE: (a) F = q1q2 4π ⑀0 r This gives 0.200 N = (0.550 × 10−6 C) q2 4π ⑀ (0.30 m)2 and q2 = +3.64 × 10−6 C The force is attractive and q1 < 0, so q2 = +3.64 × 10−6 C (b) F = 0.200 N The force is attractive, so is downward 21.15 EVALUATE: The forces between the two charges obey Newton’s third law IDENTIFY: Apply Coulomb’s law The two forces on q3 must have equal magnitudes and opposite directions SET UP: Like charges repel and unlike charges attract G qq EXECUTE: The force F2 that q2 exerts on q3 has magnitude F2 = k 2 and is in the +x-direction r2 G q q q q F1 must be in the −x-direction, so q1 must be positive F1 = F2 gives k = k 2 r1 r2 2 ⎛r ⎞ ⎛ 2.00 cm ⎞ q1 = q2 ⎜ ⎟ = ( 3.00 nC ) ⎜ ⎟ = 0.750 nC ⎝ 4.00 cm ⎠ ⎝ r2 ⎠ EVALUATE: The result for the magnitude of q1 doesn’t depend on the magnitude of q2 21.16 IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on Q SET UP: The force that q1 exerts on Q is repulsive, as in Example 21.4, but now the force that q2 exerts is attractive EXECUTE: The x-components cancel We only need the y-components, and each charge contributes (2.0 × 10−6 C)(4.0 × 10−6 C) sin α = −0.173 N (since sinα = 0.600) equally F1 y = F2 y = − 4π ⑀0 (0.500 m) Therefore, the total force is F = 0.35 N, in the − y -direction EVALUATE: If q1 is −2.0 μC and q2 is +2.0 μC, then the net force is in the + y -direction 21.17 IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on q1 G G SET UP: Like charges repel and unlike charges attract, so F2 and F3 are both in the + x -direction EXECUTE: F2 = k q1q2 r122 = 6.749 × 1025 N, F3 = k q1q3 r132 = 1.124 × 10−4 N F = F2 + F3 = 1.8 × 10−4 N F = 1.8 × 10−4 N and is in the + x -direction G G EVALUATE: Comparing our results to those in Example 21.3, we see that F1 on = − F3 on 1, as required 21.18 by Newton’s third law IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on q2 G SET UP: F2 on is in the + y -direction EXECUTE: F2 on = (9.0 × 109 N ⋅ m /C2 )(2.0 × 1026 C)(2.0 × 1026 C) (0.60 m) = 0.100 N ( F2 on 1) x = and ( F2 on 1) y = +0.100 N FQ on is equal and opposite to F1 on Q (Example 21.4), so ( FQ on ) x = −0.23 N and ( FQ on ) y = 0.17 N Fx = ( F2 on 1) x + ( FQ on 1) x = −0.23 N Fy = ( F2 on ) y + ( FQ on ) y = 0.100 N + 0.17 N = 0.27 N The magnitude of the total force is G 0.23 = 40°, so F is 40° counterclockwise from the +y-axis, F = (0.23 N) + (0.27 N) = 0.35 N tan −1 0.27 or 130° counterclockwise from the +x- axis EVALUATE: Both forces on q1 are repulsive and are directed away from the charges that exert them © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-6 21.19 Chapter 21 IDENTIFY and SET UP: Apply Coulomb’s law to calculate the force exerted by q2 and q3 on q1 Add these forces as vectors to get the net force The target variable is the x-coordinate of q3 G EXECUTE: F2 is in the x-direction F2 = k q1q2 = 3.37 N, so F2 x = +3.37 N r122 Fx = F2 x + F3 x and Fx = −7.00 N F3 x = Fx − F2 x = − 7.00 N − 3.37 N = − 10.37 N For F3 x to be negative, q3 must be on the − x -axis F3 = k q1q3 x , so x = k q1q3 F3 = 0.144 m, so x = −0.144 m EVALUATE: q2 attracts q1 in the +x-direction so q3 must attract q1 in the −x-direction, and q3 is at 21.20 negative x IDENTIFY: Apply Coulomb’s law G SET UP: Like charges repel and unlike charges attract Let F21 be the force that q2 exerts on q1 and let G F31 be the force that q3 exerts on q1 EXECUTE: The charge q3 must be to the right of the origin; otherwise both q2 and q3 would exert forces in the +x-direction Calculating the two forces: q1q2 (9 × 109 N ⋅ m /C2 )(3.00 × 10−6 C)(5.00 × 10−6 C) F21 = = = 3.375 N, in the +x-direction 4π ⑀0 r122 (0.200 m) F31 = (9 × 109 N ⋅ m /C2 )(3.00 × 10−6 C)(8.00 × 10−6 C) r132 We need Fx = F21 − F31 = −7.00 N, so 3.375 N − r13 = = 0.216 N ⋅ m 0.216 N ⋅ m r132 r132 , in the −x-direction = −7.00 N 0.216 N ⋅ m = 0.144 m q3 is at x = 0.144 m 3.375 N + 7.00 N EVALUATE: F31 = 10.4 N F31 is larger than F21, because q3 is larger than q2 and also because r13 is less than r12 21.21 IDENTIFY: Apply Coulomb’s law to calculate the force each of the two charges exerts on the third charge Add these forces as vectors SET UP: The three charges are placed as shown in Figure 21.21a Figure 21.21a EXECUTE: Like charges repel and unlike attract, so the free-body diagram for q3 is as shown in Figure 21.21b © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field F1 = F2 = 21-7 q1q3 4π ⑀0 r132 q2q3 4π ⑀0 r23 Figure 21.21b F1 = (8.988 × 109 N ⋅ m /C2 ) (1.50 × 10−9 C)(5.00 × 10−9 C) F2 = (8.988 × 109 N ⋅ m /C2 ) = 1.685 × 10−6 N (0.200 m) (3.20 × 10−9 C)(5.00 × 10−9 C) G G G The resultant force is R = F1 + F2 (0.400 m) = 8.988 × 10−7 N Rx = R y = −( F1 + F2 ) = −(1.685 × 10−6 N + 8.988 × 10−7 N) = −2.58 × 10−6 N The resultant force has magnitude 2.58 × 10−6 N and is in the − y -direction EVALUATE: The force between q1 and q3 is attractive and the force between q2 and q3 is replusive 21.22 IDENTIFY: Apply F = k qq′ r2 to each pair of charges The net force is the vector sum of the forces due to q1 and q2 SET UP: Like charges repel and unlike charges attract The charges and their forces on q3 are shown in Figure 21.22 EXECUTE: F1 = k F2 = k q2q3 r22 q1q3 r12 = (8.99 × 109 N ⋅ m /C2 ) = (8.99 × 109 N ⋅ m /C2 ) (4.00 × 10−9 C)(6.00 × 10−9 C) (0.200 m) (5.00 × 10−9 C)(6.00 × 10−9 C) (0.300 m) = 5.394 × 10−6 N = 2.997 × 10−6 N Fx = F1x + F2 x = + F1 − F2 = 2.40 × 10−6 N The net force has magnitude 2.40 × 10−6 N and is in the +x-direction EVALUATE: Each force is attractive, but the forces are in opposite directions because of the placement of the charges Since the forces are in opposite directions, the net force is obtained by subtracting their magnitudes Figure 21.22 21.23 IDENTIFY: We use Coulomb’s law to find each electrical force and combine these forces to find the net force SET UP: In the O-H-N combination the O− is 0.170 nm from the H + and 0.280 nm from the N − In the N-H-N combination the N − is 0.190 nm from the H + and 0.300 nm from the other N − Like charges repel and unlike charges attract The net force is the vector sum of the individual forces The force due to qq e2 each pair of charges is F = k 2 = k r r © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-8 Chapter 21 EXECUTE: (a) F = k q1 q2 r2 =k e2 r2 O-H-N: O − - H + : F = (8.99 × 109 N ⋅ m /C2 ) O − - N − : F = (8.99 × 109 N ⋅ m /C2 ) (1.60 × 10−19 C)2 (0.170 × 10−9 m) (1.60 × 10−19 C) (0.280 × 10 −9 m) = 7.96 × 10−9 N, attractive = 2.94 × 10−9 N, repulsive N-H-N: N − - H + : F = (8.99 × 109 N ⋅ m /C2 ) N − - N − : F = (8.99 × 109 N ⋅ m /C2 ) (1.60 × 10−19 C) (0.190 × 10−9 m) (1.60 × 10−19 C)2 (0.300 × 10−9 m)2 = 6.38 × 10−9 N, attractive = 2.56 × 10−9 N, repulsive The total attractive force is 1.43 × 10−8 N and the total repulsive force is 5.50 × 10−9 N The net force is attractive and has magnitude 1.43 × 10−8 N − 5.50 × 10−9 N = 8.80 × 10−9 N 21.24 e2 (1.60 × 10−19 C)2 = 8.22 × 10−8 N (0.0529 × 10−9 m) r EVALUATE: The bonding force of the electron in the hydrogen atom is a factor of 10 larger than the bonding force of the adenine-thymine molecules IDENTIFY: We use Coulomb’s law to find each electrical force and combine these forces to find the net force SET UP: In the O-H-O combination the O− is 0.180 nm from the H + and 0.290 nm from the other O− (b) F = k = (8.99 × 109 N ⋅ m /C2 ) In the N-H-N combination the N − is 0.190 nm from the H + and 0.300 nm from the other N − In the O-H-N combination the O − is 0.180 nm from the H + and 0.290 nm from the other N − Like charges repel and unlike charges attract The net force is the vector sum of the individual forces The force due to qq e2 each pair of charges is F = k 2 = k r r q1 q2 e2 EXECUTE: Using F = k = k , we find that the attractive forces are: O− - H + , 7.10 × 10−9 N; r r N − - H + , 6.37 × 10−9 N; O− - H + , 7.10 × 10−9 N The total attractive force is 2.06 × 10−8 N The repulsive forces are: O − - O − , 2.74 × 10−9 N; N − - N − , 2.56 × 10−9 N; O − - N − , 2.74 × 10−9 N The total repulsive 21.25 force is 8.04 × 10−9 N The net force is attractive and has magnitude 1.26 × 10−8 N EVALUATE: The net force is attractive, as it should be if the molecule is to stay together IDENTIFY: F = q E Since the field is uniform, the force and acceleration are constant and we can use a constant acceleration equation to find the final speed SET UP: A proton has charge +e and mass 1.67 × 10−27 kg EXECUTE: (a) F = (1.60 × 10−19 C)(2.75 × 103 N/C) = 4.40 × 10−16 N (b) a = F 4.40 × 10−16 N = = 2.63 × 1011 m/s m 1.67 × 10−27 kg (c) vx = v0 x + axt gives v = (2.63 × 1011 m/s )(1.00 × 10−6 s) = 2.63 × 105 m/s 21.26 EVALUATE: The acceleration is very large and the gravity force on the proton can be ignored q IDENTIFY: For a point charge, E = k r G SET UP: E is toward a negative charge and away from a positive charge © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field 21-9 EXECUTE: (a) The field is toward the negative charge so is downward 3.00 × 10−9 C E = (8.99 × 109 N ⋅ m /C2 ) = 432 N/C (0.250 m) kq (8.99 × 109 N ⋅ m /C2 )(3.00 × 1029 C) = = 1.50 m E 12.0 N/C EVALUATE: At different points the electric field has different directions, but it is always directed toward the negative point charge IDENTIFY: The acceleration that stops the charge is produced by the force that the electric field exerts on it Since the field and the acceleration are constant, we can use the standard kinematics formulas to find acceleration and time (a) SET UP: First use kinematics to find the proton’s acceleration vx = when it stops Then find the electric field needed to cause this acceleration using the fact that F = qE (b) r = 21.27 EXECUTE: vx2 = v02x + 2ax ( x − x0 ) = (4.50 × 106 m/s) + 2a (0.0320 m) and a = 3.16 × 1014 m/s Now find the electric field, with q = e eE = ma and E = ma/e = (1.67 × 10−27 kg)(3.16 × 1014 m/s )/(1.60 × 10−19 C) = 3.30 × 106 N/C, to the left (b) SET UP: Kinematics gives v = v0 + at , and v = when the electron stops, so t = v0 /a EXECUTE: t = v0 /a = (4.50 × 106 m/s)/(3.16 × 1014 m/s ) = 1.42 × 10−8 s = 14.2 ns (c) SET UP: In part (a) we saw that the electric field is proportional to m, so we can use the ratio of the electric fields Ee /Ep = me /mp and Ee = ( me /mp ) Ep EXECUTE: Ee = [(9.11 ×10 −31 kg)/(1.67 ×10 −27 kg)](3.30 ×106 N/C) = 1.80 ×103 N/C, to the right 21.28 EVALUATE: Even a modest electric field, such as the ones in this situation, can produce enormous accelerations for electrons and protons IDENTIFY: Use constant acceleration equations to calculate the upward acceleration a and then apply G G F = qE to calculate the electric field SET UP: Let +y be upward An electron has charge q = − e EXECUTE: (a) v0 y = and a y = a, so y − y0 = v0 yt + 12 a yt gives y − y0 = 12 at Then a= E= 2( y − y0 ) t = 2(4.50 m) (3.00 × 10−6 s) = 1.00 × 1012 m/s F ma (9.11 × 10−31 kg)(1.00 × 1012 m/s ) = = = 5.69 N/C q q 1.60 × 10−19 C The force is up, so the electric field must be downward since the electron has negative charge (b) The electron’s acceleration is ~ 1011 g , so gravity must be negligibly small compared to the electrical force 21.29 EVALUATE: Since the electric field is uniform, the force it exerts is constant and the electron moves with constant acceleration (a) IDENTIFY: Eq (21.4) relates the electric field, charge of the particle, and the force on the particle If the particle is to remain stationary the net force on it must be zero SET UP: The free-body diagram for the particle is sketched in Figure 21.29 The weight is mg, downward For the net force to be zero the force exerted by the electric field must be upward The electric field is downward Since the electric field and the electric force are in opposite directions the charge of the particle is negative mg = q E Figure 21.29 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-10 Chapter 21 mg (1.45 × 1023 kg)(9.80 m/s ) = = 2.19 × 10−5 C and q = −21.9 μC E 650 N/C (b) SET UP: The electrical force has magnitude FE = q E = eE The weight of a proton is w = mg EXECUTE: q = FE = w so eE = mg mg (1.673 × 10−27 kg)(9.80 m/s ) = = 1.02 × 10−7 N/C e 1.602 × 10−19 C This is a very small electric field EVALUATE: In both cases q E = mg and E = ( m/ q ) g In part (b) the m/ q ratio is much smaller EXECUTE: E = 21.30 (∼ 10−8 ) than in part (a) (∼ 102 ) so E is much smaller in (b) For subatomic particles gravity can usually be ignored compared to electric forces IDENTIFY: The net electric field is the vector sum of the individual fields SET UP: The distance from a corner to the center of the square is r = ( a/2)2 + (a/2) = a/ The magnitude of the electric field due to each charge is the same and equal to Eq = kq r =2 kq a2 All four y-components add and the x-components cancel EXECUTE: Each y-component is equal to Eqy = − Eq cos 45° = − is 21.31 2kq a2 Eq = −2kq 2a =− 2kq a2 The resultant field , in the − y -direction EVALUATE: We must add the y-components of the fields, not their magnitudes q IDENTIFY: For a point charge, E = k The net field is the vector sum of the fields produced by each r G G G charge A charge q in an electric field E experiences a force F = qE SET UP: The electric field of a negative charge is directed toward the charge Point A is 0.100 m from q2 and 0.150 m from q1 Point B is 0.100 m from q1 and 0.350 m from q2 EXECUTE: (a) The electric fields at point A due to the charges are shown in Figure 21.31a q 6.25 × 10−9 C = 2.50 × 103 N/C E1 = k 21 = (8.99 × 109 N ⋅ m /C2 ) (0.150 m) rA1 q2 rA22 q2 = (8.99 × 109 N ⋅ m /C2 ) 12.5 × 10−9 C = 1.124 × 104 N/C (0.100 m) Since the two fields are in opposite directions, we subtract their magnitudes to find the net field E = E2 − E1 = 8.74 × 103 N/C, to the right (b) The electric fields at point B are shown in Figure 21.31b q 6.25 × 10−9 C = 5.619 × 103 N/C E1 = k 21 = (8.99 × 109 N ⋅ m /C2 ) (0.100 m)2 rB1 E2 = k = (8.99 × 109 N ⋅ m /C2 ) 12.5 × 10−9 C = 9.17 × 102 N/C rB22 (0.350 m)2 Since the fields are in the same direction, we add their magnitudes to find the net field E = E1 + E2 = 6.54 × 103 N/C, to the right E2 = k (c) At A, E = 8.74 × 103 N/C, to the right The force on a proton placed at this point would be F = qE = (1.60 × 10−19 C)(8.74 × 103 N/C) = 1.40 × 10−15 N, to the right EVALUATE: A proton has positive charge so the force that an electric field exerts on it is in the same direction as the field © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-38 Chapter 21 G G G G EXECUTE: F1 + F3 = 0, so the net force is F = F2 F= q (3q ) 4π ⑀0 ( L/ 2) = 6q 4π ⑀0 L2 , away from the vacant corner Figure 21 77a (b) SET UP: The forces are sketched in Figure 21.77b EXECUTE: F2 = F1 = F3 = q (3q ) 4π ⑀0 ( L) q (3q ) 4π ⑀0 L = = 3q 4π ⑀0 (2 L2 ) 3q 4π ⑀0 L2 The vector sum of F1 and F3 is F13 = F12 + F32 Figure 21 77b G 2q G F13 = F1 = ; F13 and F2 are in the same direction 4π ⑀0 L 3q ⎛ 1⎞ ⎜ + ⎟ , and is directed toward the center of the square 2⎠ 4π ⑀0 L2 ⎝ EVALUATE: By symmetry the net force is along the diagonal of the square The net force is only slightly larger when the −3q charge is at the center Here it is closer to the charge at point but the other two F = F13 + F2 = 21.78 forces cancel IDENTIFY: Use Eq (21.7) for the electric field produced by each point charge Apply the principle of superposition and add the fields as vectors to find the net field (a) SET UP: The fields due to each charge are shown in Figure 21.78a cosθ = x x + a2 Figure 21.78a EXECUTE: The components of the fields are given in Figure 21.78b © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field E1 = E2 = E3 = 21-39 ⎛ q ⎞ ⎜ ⎟ 4π ⑀0 ⎝ a + x ⎠ ⎛ 2q ⎞ ⎜ ⎟ 4π ⑀0 ⎝ x ⎠ Figure 21.78b E1 y = − E1 sin θ , E2 y = + E2 sin θ so E y = E1 y + E2 y = E1x = E2 x = + E1 cosθ = ⎛ q ⎞⎛ x ⎜ ⎟⎜ 4π ⑀0 ⎝ a + x ⎠ ⎜⎝ x + a ⎞ ⎟ , E3 x = − E3 ⎟ ⎠ ⎛ ⎛ q ⎞⎛ x E x = E1x + E2 x + E3 x = ⎜ ⎜ ⎜ 4π ⑀ ⎜⎝ a + x ⎟⎠ ⎜ x + a ⎝ ⎝ Ex = − ⎞⎞ 2q ⎟⎟ − ⎟ ⎟ 4π ⑀ x ⎠⎠ ⎞ ⎞ 2q ⎛ x 2q ⎛ ⎜⎜ − ⎟⎟ = − ⎜⎜ − ⎟ 3/ 2 2 3/2 4π ⑀0 ⎝ x (a + x ) ⎠ 4π ⑀0 x ⎝ (1 + a /x ) ⎟⎠ Thus E = ⎛ ⎞ , in the − x-direction ⎜⎜1 − 2 3/2 ⎟ 4π ⑀0 x ⎝ (1 + a /x ) ⎟⎠ 2q (b) x a implies a /x and (1 + a /x ) −3/2 ≈ − 3a /2 x Thus E ≈ 21.79 ⎛ ⎛ 3a ⎞ ⎞ 3qa ⎜1 − ⎜1 − ⎟ ⎟ = 4π ⑀0 x ⎜⎝ ⎜⎝ x ⎟⎠ ⎟⎠ 4π ⑀0 x 2q EVALUATE: E ∼ 1/x For a point charge E ∼ 1/x and for a dipole E ∼ 1/x3 The total charge is zero so at large distances the electric field should decrease faster with distance than for a point charge By G symmetry E must lie along the x-axis, which is the result we found in part (a) IDENTIFY: The small bags of protons behave like point-masses and point-charges since they are extremely far apart SET UP: For point-particles, we use Newton’s formula for universal gravitation (F = Gm1m /r ) and Coulomb’s law The number of protons is the mass of protons in the bag divided by the mass of a single proton EXECUTE: (a) (0.0010 kg)/(1.67 × 10−27 kg) = 6.0 × 1023 protons (b) Using Coulomb’s law, where the separation is twice the radius of the earth, we have Felectrical = (9.00 × 109 N ⋅ m /C2 )(6.0 × 1023 × 1.60 × 10−19 C) /(2 × 6.38 × 106 m) = 5.1 × 105 N Fgrav = (6.67 × 10−11 N ⋅ m /kg )(0.0010 kg) /(2 × 6.38 × 106 m) = 4.1 × 10−31 N (c) EVALUATE: The electrical force (≈200,000 lb!) is certainly large enough to feel, but the gravitational 21.80 force clearly is not since it is about 1036 times weaker IDENTIFY: We can treat the protons as point-charges and use Coulomb’s law SET UP: (a) Coulomb’s law is F = (1/4π ⑀0 ) q1q2 /r EXECUTE: F = (9.00 × 109 N ⋅ m /C2 )(1.60 × 10−19 C) /(2.0 × 10−15 m) = 58 N = 13 lb, which is certainly large enough to feel (b) EVALUATE: Something must be holding the nucleus together by opposing this enormous repulsion This is the strong nuclear force © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-40 21.81 Chapter 21 IDENTIFY: Estimate the number of protons in the textbook and from this find the net charge of the textbook Apply Coulomb’s law to find the force and use Fnet = ma to find the acceleration SET UP: With the mass of the book about 1.0 kg, most of which is protons and neutrons, we find that the number of protons is (1.0 kg)/(1.67 × 10−27 kg) = 3.0 × 1026 EXECUTE: (a) The charge difference present if the electron’s charge was 99.999, of the proton’s is Δq = (3.0 × 1026 )( 0.00001)(1.6 × 10−19 C ) = 480 C (b) F = k (Δq ) /r = k (480 C)2 /(5.0 m)2 = 8.3 × 1013 N, and is repulsive a = F/m = (8.3 × 1013 N)/(1 kg) = 8.3 × 1013 m/s 21.82 EXECUTE: (c) Even the slightest charge imbalance in matter would lead to explosive repulsion! IDENTIFY: The positive sphere will be deflected in the direction of the electric field but the negative sphere will be deflected in the direction opposite to the electric field Since the spheres hang at rest, they are in equilibrium so the forces on them must balance The external forces on each sphere are gravity, the tension in the string, the force due to the uniform electric field and the electric force due to the other sphere SET UP: The electric force on one sphere due to the other is FC = k q2 in the horizontal direction, the r2 force on it due to the uniform electric field is FE = qE in the horizontal direction, the gravitational force is mg vertically downward and the force due to the string is T directed along the string For equilibrium ∑ Fx = and ∑ Fy = EXECUTE: (a) The positive sphere is deflected in the same direction as the electric field, so the one that is deflected to the left is positive (b) The separation between the two spheres is 2(0.530 m)sin 25o = 0.4480 m FC = k q2 r2 = (8.99 × 109 N ⋅ m /C2 )(72.0 × 10−9 C) (0.4480 m)2 = 2.322 × 10−4 N FE = qE ∑ Fy = gives mg ∑ Fx = gives T sin 25o + FC − FE = mg tan 25o + FC = qE cos 25o Combining the equations and solving for E gives mg tan 25o + FC (6.80 × 10−6 kg)(9.8 m/s ) tan 25o + 2.322 × 10−4 N E= = = 3.66 × 103 N/C q 72.0 × 10−9 C EVALUATE: Since the charges have opposite signs, they attract each other, which tends to reduce the angle between the strings Therefore if their charges were negligibly small, the angle between the strings would be greater than 50° IDENTIFY: The only external force acting on the electron is the electrical attraction of the proton, and its acceleration is toward the center of its circular path (that is, toward the proton) Newton’s second law applies to the proton and Coulomb’s law gives the electrical force on it due to the proton v2 e2 v2 SET UP: Newton’s second law gives FC = m Using the electrical force for FC gives k = m r r r T cos 25o − mg = and T = 21.83 EXECUTE: Solving for v gives v = 21.84 ke2 (8.99 × 109 N ⋅ m /C2 )(1.60 × 10−19 C) = = 2.19 × 106 m/s mr (9.109 × 10−31 kg)(5.29 × 10−11 m) EVALUATE: This speed is less than 1% the speed of light, so it is reasonably safe to use Newtonian physics IDENTIFY: Since we can ignore gravity, the only external force acting on the moving sphere is the electrical attraction of the stationary sphere, and its acceleration is toward the center of its circular path (that is, toward the stationary sphere) Newton’s second law applies to the moving sphere and Coulomb’s law gives the electrical force on it due to the stationary sphere qq v2 v2 SET UP: Newton’s second law gives FC = m Using the electrical force for FC gives k 22 = m r r r © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field EXECUTE: Solving for r gives r = 21.85 k q1q2 = 21-41 (8.99 × 109 N ⋅ m /C2 )(4.3 × 10−6 C)(7.5 × 10−6 C) = 0.925 m mv (9.00 × 10−9 kg)(5.9 × 103 m/s) EVALUATE: We can safely ignore gravity in most cases because it is normally much weaker than the electric force IDENTIFY and SET UP: Use the density of copper to calculate the number of moles and then the number of atoms Calculate the net charge and then use Coulomb’s law to calculate the force ⎛4 ⎞ ⎛4 ⎞ EXECUTE: (a) m = ρV = ρ ⎜ π r ⎟ = (8.9 × 103 kg/m3 ) ⎜ π ⎟ (1.00 × 10−3 m)3 = 3.728 × 10−5 kg ⎝ ⎠ ⎝3 ⎠ n = m/M = (3.728 × 10−5 kg)/(63.546 × 10−3 kg/mol) = 5.867 × 10−4 mol N = nN A = 3.5 × 1020 atoms (b) N e = (29)(3.5 × 1020 ) = 1.015 × 1022 electrons and protons qnet = eN e − ( 0.99900)eN e = (0.100 × 10−2 )(1.602 × 10−19 C)(1.015 × 1022 ) = 1.6 C 21.86 q2 (1.6 C)2 = 2.3 × 1010 N (1.00 m) EVALUATE: The amount of positive and negative charge in even small objects is immense If the charge of an electron and a proton weren’t exactly equal, objects would have large net charges IDENTIFY: Apply constant acceleration equations to a drop to find the acceleration Then use F = ma to find the force and F = q E to find q F =k r =k SET UP: Let D = 2.0 cm be the horizontal distance the drop travels and d = 0.30 mm be its vertical displacement Let + x be horizontal and in the direction from the nozzle toward the paper and let + y be vertical, in the direction of the deflection of the drop a x = and a y = a EXECUTE: Find the time of flight: t = D/v = (0.020 m)/(20 m/s) = 0.00100 s d = at a= 2d t2 = 2(3.00 × 10−4 m) (0.001 s) = 600 m/s 21.87 (1.4 × 10−11 kg)(600 m/s ) = 1.05 × 10−13 C 8.00 × 104 N/C EVALUATE: Since q is positive the vertical deflection is in the direction of the electric field IDENTIFY: Eq (21.3) gives the force exerted by the electric field This force is constant since the electric field is uniform and gives the proton a constant acceleration Apply the constant acceleration equations for the x- and y-components of the motion, just as for projectile motion (a) SET UP: The electric field is upward so the electric force on the positively charged proton is upward G G and has magnitude F = eE Use coordinates where positive y is downward Then applying ∑ F = ma to the proton gives that a x = and a y = −eE/m In these coordinates the initial velocity has components Then a = F/m = qE/m gives q = ma/E = vx = +v0 cos α and v y = + v0 sin α , as shown in Figure 21.87a Figure 21.87a © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-42 Chapter 21 EXECUTE: Finding hmax : At y = hmax the y-component of the velocity is zero v y = 0, v0 y = v0 sin α , a y = −eE/m, y − y0 = hmax = ? v 2y = v02 y + 2a y ( y − y0 ) y − y0 = hmax = v 2y − v02 y 2a y −v02 sin α mv02 sin α = 2( −eE/m) 2eE (b) Use the vertical motion to find the time t: y − y0 = 0, v0 y = v0 sin α , a y = −eE/m, t = ? y − y0 = v0 yt + 12 a yt With y − y0 = this gives t = − 2v0 y ay =− 2(v0 sin α ) 2mv0 sin α = eE −eE/m Then use the x-component motion to find d: a x = 0, v0 x = v0 cos α , t = 2mv0 sin α /eE , x − x0 = d = ? 2 ⎛ 2mv0 sin α ⎞ mv0 2sin α cos α mv0 sin 2α x − x0 = v0 xt + 12 a xt gives d = v0 cos α ⎜ = ⎟= eE eE eE ⎝ ⎠ (c) The trajectory of the proton is sketched in Figure 21.87b Figure 21.87b (d) Use the expression in part (a): hmax = Use the expression in part (b): d = [(4.00 × 105 m/s)(sin 30.0°)]2 (1.673 × 10−27 kg) 2(1.602 × 10−19 C)(500 N/C) (1.673 × 10−27 kg)(4.00 × 105 m/s) sin 60.0° (1.602 × 10−19 C)(500 N/C) = 0.418 m = 2.89 m EVALUATE: In part (a), a y = −eE/m = −4.8 × 1010 m/s This is much larger in magnitude than g, the acceleration due to gravity, so it is reasonable to ignore gravity The motion is just like projectile motion, except that the acceleration is upward rather than downward and has a much different magnitude hmax and d increase when α or v0 increase and decrease when E increases 21.88 IDENTIFY: E x = E1x + E2 x Use Eq (21.7) for the electric field due to each point charge G SET UP: E is directed away from positive charges and toward negative charges −9 q1 C 2 4.00 × 10 = ( 99 × 10 N ⋅ m /C ) = +99.9 N/C EXECUTE: (a) E x = +50.0 N/C E1x = 2 4π ⑀0 r1 (0.60 m) E x = E1x + E2 x , so E2 x = Ex − E1x = +50.0 N/C − 99.9 N/C = 249.9 N/C Since E2 x is negative, q2 must be negative q2 = E2 x r22 (1/4π ⑀0 ) = (49.9 N/C)(1.20 m)2 8.99 × 10 N ⋅ m /C = 7.99 × 10−9 C q2 = −7.99 × 10−9 C (b) E x = −50.0 N/C E1x = +99.9 N/C, as in part (a) E2 x = Ex − E1x = −149.9 N/C q2 is negative q2 = E2 x r22 (1/4π ⑀0 ) = (149.9 N/C)(1.20 m)2 8.99 × 109 N ⋅ m /C2 = 2.40 × 10−8 C q2 = −2.40 × 10−8 C EVALUATE: q2 would be positive if E2 x were positive © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field 21.89 21-43 IDENTIFY: Divide the charge distribution into infinitesimal segments of length dx Calculate E x and E y due to a segment and integrate to find the total field SET UP: The charge dQ of a segment of length dx is dQ = (Q/a )dx The distance between a segment at x and the charge q is a + r − x (1 − y ) −1 ≈ + y when y EXECUTE: (a) dEx = a + r = x, so E x = G G (b) F = qE = a Qdx Q⎛1 ⎞ dQ so E x = = ⎜ − ⎟ 4π ⑀0 ∫0 a ( a + r − x) 4π ⑀0 a ⎝ r a + r ⎠ 4π ⑀0 (a + r − x )2 Q⎛ 1⎞ − ⎟ E y = ⎜ 4π ⑀0 a ⎝ x − a x ⎠ qQ ⎛ 1 ⎞ˆ ⎜ − ⎟ i 4π ⑀0 a ⎝ r a + r ⎠ kqQ kqQ kqQ qQ ((1 − a/x) −1 − 1) = (1 + a/x + ⋅⋅⋅ − 1) ≈ ≈ (Note π ⑀0 r ax ax x that for x a, r = x − a ≈ x.) The charge distribution looks like a point charge from far away, so the force EVALUATE: (c) For x a, F = 21.90 takes the form of the force between a pair of point charges IDENTIFY: Use Eq (21.7) to calculate the electric field due to a small slice of the line of charge and G integrate as in Example 21.10 Use Eq (21.3) to calculate F SET UP: The electric field due to an infinitesimal segment of the line of charge is sketched in Figure 21.90 sin θ = y x + y2 x cosθ = x + y2 Figure 21.90 Slice the charge distribution up into small pieces of length dy The charge dQ in each slice is dQ = Q (dy/a ) The electric field this produces at a distance x along the x-axis is dE Calculate the G components of dE and then integrate over the charge distribution to find the components of the total field ⎛ dQ ⎞ Q ⎛ dy ⎞ EXECUTE: dE = ⎜⎜ ⎟⎟ = ⎜ ⎟ 4π ⑀0 ⎝ x + y ⎠ 4π ⑀0 a ⎜⎝ x + y ⎟⎠ ⎞ Qx ⎛ dy dE x = dE cosθ = ⎜⎜ ⎟ 3/2 4π ⑀0 a ⎝ ( x + y ) ⎟⎠ ⎞ Q ⎛ ydy dE y = − dE sin θ = − ⎜⎜ ⎟ 3/2 4π ⑀0 a ⎝ ( x + y ) ⎟⎠ E x = ∫ dE x = − Qx a dy 4π ⑀0 a Ñ0 ( x + y )3/2 Qx ⎡ ⎢ = 4π ⑀0a ⎢ x ⎣ a ⎤ Q ⎥ = 2⎥ π ⑀0 x x + y ⎦0 y a E y = ∫ dEy = − Q a ydy 4π ⑀ 0a Ñ0 ( x + y )3/2 x + a2 ⎡ ⎤ Q ⎛1 ⎢− ⎥ =− = − ⎜ − ⎜ 2⎥ π ⑀ 4π ⑀0a ⎢ a x ⎝ x + y ⎦0 x + a2 ⎣ Q ⎞ ⎟ ⎟ ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-44 Chapter 21 G G (b) F = q0 E Fx = −qE x = qQ ⎛ 1 − qQ ; Fy = − qE y = ⎜ − ⎜ 2 4π ⑀0 x x + a 4π ⑀0a ⎝ x x + a2 −1/2 ⎛ a2 ⎞ 1⎛ a2 ⎞ a2 = ⎜1 + ⎟ = ⎜1 − ⎟ = − ⎜ ⎟ ⎜ x ⎝ x ⎟⎠ x x x ⎠ x2 + a x ⎝ qQ qQ ⎛ 1 a ⎞ qQa Fx ≈ − , Fy ≈ ⎜ − + ⎟= 4π ⑀0a ⎜⎝ x x x3 ⎟⎠ 8π ⑀ x3 4π ⑀0 x G qQ EVALUATE: For x a, Fy Fx and F ≈ Fx = and F is in the − x-direction For x a the 4π P0 x charge distribution Q acts like a point charge IDENTIFY: Apply Eq (21.9) from Example 21.10 G SET UP: a = 2.50 cm Replace Q by Q Since Q is negative, E is toward the line of charge and (c) For x a, 21.91 ⎞ ⎟ ⎟ ⎠ G Q E=− iˆ 4π ⑀0 x x + a G Q 1 7.00 × 10−9 C EXECUTE: E = − iˆ = − iˆ = (−6110 N/C) iˆ 4π ⑀0 x x + a 4π ⑀0 (0.100 m) (0.100 m)2 + (0.025 m) 21.92 (b) The electric field is less than that at the same distance from a point charge (6300 N/C) For large x, ⎞ 1⎛ a2 ⎞ Q⎛ a2 − + ⋅⋅⋅ ⎟ The first ( x + a ) −1/2 = (1 + a /x ) −1/2 ≈ ⎜1 − ⎟ , which gives E x →∞ = ⎜ 2 ⎜ ⎟ ⎜ ⎟ x x ⎝ 2x ⎠ 4π ⑀0 x ⎝ x ⎠ correction term to the point charge result is negative (c) For a 1% difference, we need the first term in the expansion beyond the point charge result to be less a2 ≈ 0.010 ⇒ x ≈ a 1/(2(0.010)) = (0.025 m) 1/0.020 ⇒ x ≈ 0.177 m than 0.010: 2x2 EVALUATE: At x = 10.0 cm (part b), the exact result for the line of charge is 3.1% smaller than for a point charge It is sensible, therefore, that the difference is 1.0% at a somewhat larger distance, 17.7 cm G kQ G IDENTIFY: The electrical force has magnitude F = and is attractive Apply ∑ F = ma to the earth r SET UP: For a circular orbit, a = v2 2π r The mass of the earth is mE = 5.97 × 1024 kg, The period T is r v the orbit radius of the earth is 1.50 × 1011 m and its orbital period is 3.146 × 107 s EXECUTE: F = ma gives Q= 21.93 mE 4π 2r kT = kQ r = mE 4π 2r v2 , so v2 = r T2 (5.97 × 1024 kg)(4)(π )(1.50 × 1011 m)3 (8.99 × 109 N ⋅ m /C2 )(3.146 × 107 s)2 = 2.99 × 1017 C EVALUATE: A very large net charge would be required IDENTIFY: Apply Eq (21.11) SET UP: σ = Q/A = Q/π R (1 + y )−1/2 ≈ − y /2, when y EXECUTE: (a) E = E= σ 2⑀ [1 − ( R /x + 1) −1/2 ] ⎡ ⎞ 7.00 pC/π (0.025 m) ⎢ ⎛ (0.025 m) 1− ⎜ + 1⎟ ⎟ ⎢ ⎜⎝ (0.200 m) 2⑀0 ⎠ ⎣ −1/2 ⎤ ⎥ = 1.56 N/C, in the + x-direction ⎥ ⎦ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field (b) For x R, E = 21.94 21-45 σ σ R2 σπ R Q [1 − (1 − R /2 x + ⋅⋅⋅)] ≈ = = 2 2⑀0 2⑀0 x 4π ⑀0 x 4π ⑀0 x (c) The electric field of (a) is less than that of the point charge (0.90 N/C) since the first correction term to the point charge result is negative (1.58 − 1.56) (d) For x = 0.200 m, the percent difference is = 0.01 = 1% For x = 0.100 m, 1.56 (6.30 − 6.00) = 0.047 ≈ 5% Edisk = 6.00 N/C and Epoint = 6.30 N/C, so the percent difference is 6.30 EVALUATE: The field of a disk becomes closer to the field of a point charge as the distance from the disk increases At x = 10.0 cm, R/x = 25% and the percent difference between the field of the disk and the field of a point charge is 5% IDENTIFY: When the forces on it balance, the acceleration of a molecule is zero and it moves with constant velocity SET UP: The electrical force is FE = qE and the viscous drag force is FD = KRv q Kv = R E Eq ⎛ Eq ⎞ ⎛ ET T =⎜ (b) The speed is constant and has magnitude v = Therefore x = vt = ⎜ ⎟ KR ⎝ KR ⎠ ⎝ K ET ⎛ ET ⎞ q ⎛q⎞ ⎛q⎞ ⎛q⎞ ⎛q⎞ (c) x = ⎜ is constant ⎜ ⎟ = ⎜ ⎟ and ⎜ ⎟ = 3⎜ ⎟ ⎟ , where K ⎝ K ⎠R ⎝ R ⎠2 ⎝ R ⎠1 ⎝ R ⎠3 ⎝ R ⎠1 EXECUTE: (a) F = FD so qE = KRv and 21.95 ⎞q ⎟ ⎠R ⎛ ET ⎞⎛ q ⎞ ⎛ ET ⎞⎛ q ⎞ ⎛ ET ⎞⎛ q ⎞ ⎛ ET ⎞⎛ q ⎞ x2 = ⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ = x1; x3 = ⎜ ⎟⎜ ⎟ = 3⎜ ⎟⎜ ⎟ = 3x1 K R K R K R ⎝ ⎠⎝ ⎠2 ⎝ ⎠⎝ ⎠1 ⎝ ⎠⎝ ⎠3 ⎝ K ⎠⎝ R ⎠1 EVALUATE: The distance a particle moves is not proportional to its charge, but rather is proportional to the ratio of its charge to its radius (size) G G IDENTIFY: Find the resultant electric field due to the two point charges Then use F = qE to calculate the force on the point charge SET UP: Use the results of Problems 21.90 and 21.89 EXECUTE: (a) The y-components of the electric field cancel, and the x-component from both charges, as ⎞ −2Q ⎛ 1 given in Problem 21.90, is E x = ⎜⎜ − ⎟ Therefore, 1/2 4π ⑀0 a ⎝ y ( y + a ) ⎟⎠ G G ⎞ˆ −2Qq ⎛ 1 −2Qq Qqa ˆ F= i (1 − (1 − a /2 y + ⋅⋅⋅)) iˆ = − ⎜⎜ − ⎟⎟ i If y a, F ≈ 1/2 4π ⑀0 a ⎝ y ( y + a ) ⎠ 4π ⑀ ay 4π ⑀0 y (b) If the point charge is now on the x-axis the two halves of the charge distribution provide different G G Qq ⎛ 1⎞ forces, though still along the x-axis, as given in Problem 21.89: F+ = qE+ = − ⎟ iˆ ⎜ 4π ⑀0 a ⎝ x − a x ⎠ G G G G G Qq ⎛ 1 ⎞ˆ Qq ⎛ ⎞ˆ and F− = qE− = − − + ⎜ − ⎟ i Therefore, F = F+ + F− = ⎜ ⎟ i For 4π ⑀0 a ⎝ x x + a ⎠ 4π ⑀0 a ⎝ x − a x x + a ⎠ G x a, F ≈ ⎞ ⎛ a a2 ⎞⎞ Qq ⎛ ⎛ a a 2Qqa ˆ ⎜ ⎜1 + + + ⎟ − + ⎜ − + − ⎟ ⎟ iˆ = i ⎜ ⎟ ⎜ ⎟ ⎟ 4π ⑀0 ax ⎜⎝ ⎝ π ⑀0 x3 x x x x ⎠ ⎝ ⎠⎠ EVALUATE: If the charge distributed along the x-axis were all positive or all negative, the force would be proportional to 1/y in part (a) and to 1/x in part (b), when y or x is very large 21.96 IDENTIFY: Apply ∑ Fx = and ∑ Fy = to the sphere, with x horizontal and y vertical G SET UP: The free-body diagram for the sphere is given in Figure 21.96 The electric field E of the sheet is directed away from the sheet and has magnitude E = σ 2⑀0 (Eq 21.12) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-46 Chapter 21 EXECUTE: ∑ Fy = gives T cos α = mg and T = T= mg qσ ∑ Fx = gives T sin α = and cos α 2⑀0 qσ mg qσ qσ Combining these two equations we have = and tan α = Therefore, 2⑀0 sin α cos α 2⑀0 sin α 2⑀0mg ⎛ qσ ⎞ ⎟ ⎝ 2⑀0mg ⎠ α = arctan ⎜ EVALUATE: The electric field of the sheet, and hence the force it exerts on the sphere, is independent of the distance of the sphere from the sheet Figure 21.96 21.97 IDENTIFY: Divide the charge distribution into small segments, use the point charge formula for the electric field due to each small segment and integrate over the charge distribution to find the x and y components of the total field SET UP: Consider the small segment shown in Figure 21.97a EXECUTE: A small segment that subtends angle dθ has length a dθ and contains charge ⎛ adθ ⎞ 2Q dQ = ⎜ dθ ( 12 π a is the total ⎟Q = ⎜ πa ⎟ π ⎝2 ⎠ length of the charge distribution.) Figure 21.97a The charge is negative, so the field at the origin is directed toward the small segment The small segment is located at angle θ as shown in the sketch The electric field due to dQ is shown in Figure 21.97b, along with its components dE = dE = dQ 4π ⑀0 a Q 2π ⑀0a dθ Figure 21.97b © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field 21-47 dE x = dE cosθ = (Q/2π 2⑀0a )cosθ dθ E x = ∫ dE x = π /2 Q 2π 2⑀0a Ñ0 cosθ dθ = Q (sin θ 2π 2⑀0a π /2 )= Q 2π 2⑀0a dE y = dE sin θ = (Q/2π 2⑀0a )sin θ dθ E y = ∫ dE y = Q 2π ⑀0 Ñ π /2 a2 sin θ dθ = Q 2π 2⑀0a (− cosθ π /2 )= Q 2π 2⑀ 0a EVALUATE: Note that Ex = E y , as expected from symmetry 21.98 IDENTIFY: We must add the electric field components of the positive half and the negative half SET UP: From Problem 21.97, the electric field due to the quarter-circle section of positive charge has Q Q components E x = + 2 , E y = − 2 The field due to the quarter-circle section of negative 2π ⑀0a 2π ⑀0a Q Q , Ey = + 2 2π 2⑀0a 2π ⑀0a EXECUTE: The components of the resultant field is the sum of the x- and y-components of the fields due Q to each half of the semicircle The y-components cancel, but the x-components add, giving E x = + 2 , π ⑀0a in the + x-direction EVALUATE: Even though the net charge on the semicircle is zero, the field it produces is not zero because of the way the charge is arranged IDENTIFY: Each wire produces an electric field at P due to a finite wire These fields add by vector addition Q The field due to the negative wire points to the left, SET UP: Each field has magnitude 4π ⑀0 x x + a charge has components E x = + 21.99 while the field due to the positive wire points downward, making the two fields perpendicular to each other and of equal magnitude The net field is the vector sum of these two, which is Q cos 45° In part (b), the electrical force on an electron at P is eE Enet = 2E1 cos 45° = 4π ⑀0 x x + a EXECUTE: (a) The net field is Enet = Enet = Q cos 45° 4π ⑀0 x x + a 2(9.00 × 109 N ⋅ m /C2 )(2.50 × 10−6 C)cos 45° (0.600 m) (0.600 m) + (0.600 m) = 6.25 × 104 N/C The direction is 225° counterclockwise from an axis pointing to the right at point P (b) F = eE = (1.60 × 10−19 C)(6.25 × 104 N/C) = 1.00 × 10−14 N, opposite to the direction of the electric 21.100 field, since the electron has negative charge EVALUATE: Since the electric fields due to the two wires have equal magnitudes and are perpendicular to each other, we only have to calculate one of them in the solution IDENTIFY: Each sheet produces an electric field that is independent of the distance from the sheet The net field is the vector sum of the two fields SET UP: The formula for each field is E = σ /2⑀0 , and the net field is the vector sum of these, Enet = σB σ A σB ±σ A , where we use the + or − sign depending on whether the fields are in the ± = 2⑀0 2⑀0 2⑀ same or opposite directions and σ B and σ A are the magnitudes of the surface charges EXECUTE: (a) The two fields oppose and the field of B is stronger than that of A, so Enet = σ B σ A σ B − σ A 11.6 μ C/m − 9.50 μ C/m − = = = 1.19 × 105 N/C, to the right −12 2 2⑀0 2⑀0 2⑀0 2(8.85 × 10 C /N ⋅ m ) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-48 Chapter 21 (b) The fields are now in the same direction, so their magnitudes add Enet = (11.6 μ C/m + 9.50 μ C/m )/2⑀0 = 1.19 × 106 N/C, to the right (c) The fields add but now point to the left, so Enet = 1.19 × 106 N/C, to the left 21.101 EVALUATE: We can simplify the calculations by sketching the fields and doing an algebraic solution first IDENTIFY: Each sheet produces an electric field that is independent of the distance from the sheet The net field is the vector sum of the two fields SET UP: The formula for each field is E = σ /2⑀0 , and the net field is the vector sum of these, Enet = σB 2⑀0 ± σA 2⑀0 = σB ±σ A , where we use the + or − sign depending on whether the fields are in the 2⑀ same or opposite directions and σ B and σ A are the magnitudes of the surface charges EXECUTE: (a) The fields add and point to the left, giving Enet = 1.19 × 106 N/C (b) The fields oppose and point to the left, so Enet = 1.19 × 105 N/C (c) The fields oppose but now point to the right, giving Enet = 1.19 × 105 N/C 21.102 EVALUATE: We can simplify the calculations by sketching the fields and doing an algebraic solution first IDENTIFY: The sheets produce an electric field in the region between them which is the vector sum of the fields from the two sheets SET UP: The force on the negative oil droplet must be upward to balance gravity The net electric field between the sheets is E = σ /⑀0 , and the electrical force on the droplet must balance gravity, so qE = mg EXECUTE: (a) The electrical force on the drop must be upward, so the field should point downward since the drop is negative (b) The charge of the drop is 5e, so qE = mg (5e)(σ /⑀0 ) = mg and σ= 21.103 mg⑀0 (324 × 10−9 kg)(9.80 m/s )(8.85 × 10−12 C2 /N ⋅ m ) = = 35.1 C/m 5e 5(1.60 × 10−19 C) EVALUATE: Balancing oil droplets between plates was the basis of the Milliken Oil-Drop Experiment which produced the first measurement of the mass of an electron IDENTIFY and SET UP: Example 21.11 gives the electric field due to one infinite sheet Add the two fields as vectors G EXECUTE: The electric field due to the first sheet, which is in the xy-plane, is E1 = (σ /2⑀0 ) kˆ for z > and G G E1 = − (σ /2⑀ )kˆ for z < We can write this as E1 = (σ /2⑀0 )( z/ z )kˆ , since z/ z = +1 for z > and z/ z = − z/z = −1 for z < Similarly, we can write the electric field due to the second sheet as G E2 = − (σ /2⑀ )( x/ x )iˆ, since its charge density is −σ The net field is G G G E = E1 + E2 = (σ /2⑀0 )( −( x/ x )iˆ + ( z/ z )kˆ ) EVALUATE: The electric field is independent of the y-component of the field point since displacement in the ± y -direction is parallel to both planes The field depends on which side of each plane the field is located 21.104 IDENTIFY: Apply Eq (21.11) for the electric field of a disk The hole can be described by adding a disk of charge density −σ and radius R1 to a solid disk of charge density +σ and radius R2 SET UP: The area of the annulus is π ( R22 − R12 )σ The electric field of a disk, Eq (21.11) is E= σ ⎡ − 1/ ( R/x) + ⎤ ⎥⎦ 2⑀ ⎢⎣ EXECUTE: (a) Q = Aσ = π ( R22 − R12 )σ G σ ⎛⎡ 2 ⎤ ⎡ ⎤⎞ x (b) E ( x) = ⎜ ⎢1 − 1/ ( R2 /x) + ⎥ − ⎢1 − 1/ ( R1/x) + ⎥ ⎟ iˆ ⎦ ⎣ ⎦⎠ x 2⑀0 ⎝ ⎣ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field ( 21-49 ) G x ˆ σ 1/ ( R1/x) + − 1/ ( R2 /x) + E ( x) = i The electric field is in the + x-direction at points above 2⑀0 x the disk and in the −x-direction at points below the disk, and the factor (c) Note that 1/ ( R1/x ) + = x ˆ i specifies these directions x x x (1 + ( x/R1 ) ) −1/2 ≈ This gives R1 R1 G σ ⎛1 ⎞x ˆ σ ⎛1 ⎞ ˆ E ( x) = i= ⎜ − ⎟ ⎜ − ⎟ x i Sufficiently close means that ( x/R1) 2⑀0 ⎝ R1 R2 ⎠ x 2⑀0 ⎝ R1 R2 ⎠ (d) Fx = − qEx = k= qσ ⎛ 1 ⎞ ⎜ − ⎟ x The force is in the form of Hooke’s law: Fx = −kx, with 2⑀0 ⎝ R1 R2 ⎠ qσ ⎛ 1 ⎞ ⎜ − ⎟ 2⑀0 ⎝ R1 R2 ⎠ f = 2π k = m 2π qσ ⎛ 1 ⎞ ⎜ − ⎟ 2⑀0m ⎝ R1 R2 ⎠ EVALUATE: The frequency is independent of the initial position of the particle, so long as this position is sufficiently close to the center of the annulus for ( x/R1 ) to be small 21.105 IDENTIFY: Apply Coulomb’s law to calculate the forces that q1 and q2 exert on q3 , and add these force vectors to get the net force SET UP: Like charges repel and unlike charges attract Let + x be to the right and + y be toward the top of the page EXECUTE: (a) The four possible force diagrams are sketched in Figure 21.105a Only the last picture can result in a net force in the − x -direction (b) q1 = −2.00 μ C, q3 = +4.00 μ C, and q2 > G G (c) The forces F1 and F2 and their components are sketched in Figure 21.105b Fy = = − q2 = q1 q3 q2 q3 1 sin θ1 + sin θ This gives 4π ⑀0 (0.0400 m) 4π ⑀0 (0.0300 m) sin θ1 3/5 27 q1 = q1 = q1 = 0.843 μ C 16 sin θ 16 4/5 64 (d) Fx = F1x + F2 x and Fy = 0, so F = q3 q1 q2 ⎛ + ⎜⎜ 4π ⑀0 ⎝ (0.0400 m) (0.0300 m) 3⎞ ⎟ = 56.2 N ⎟⎠ G EVALUATE: The net force F on q3 is in the same direction as the resultant electric field at the location of q3 due to q1 and q2 Figure 21.105 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 21-50 21.106 Chapter 21 IDENTIFY: Calculate the electric field at P due to each charge and add these field vectors to get the net field SET UP: The electric field of a point charge is directed away from a positive charge and toward a negative charge Let + x be to the right and let + y be toward the top of the page EXECUTE: (a) The four possible diagrams are sketched in Figure 21.106a The first diagram is the only one in which the electric field must point in the negative y-direction (b) q1 = −3.00 μ C, and q2 < G G (c) The electric fields E1 and E2 and their components are sketched in Figure 24.106b cos θ1 = , 13 k q1 k q2 12 12 12 This gives + sin θ1 = , cos θ = and sin θ = E x = = − 13 13 13 (0.050 m) 13 (0.120 m) 13 k q2 (0.120 m) k q1 Solving for q2 gives q2 = 7.2 μ C, so q2 = −7.2 μ C Then 12 (0.050 m) = k q1 kq2 12 − = −1.17 × 107 N/C E = 1.17 × 107 N/C (0.050 m) 13 (0.120 m) 13 G EVALUATE: With q1 known, specifying the direction of E determines both q2 and E Ey = − Figure 21.106 21.107 IDENTIFY: To find the electric field due to the second rod, divide that rod into infinitesimal segments of length dx, calculate the field dE due to each segment and integrate over the length of the rod to find the G G total field due to the rod Use dF = dq E to find the force the electric field of the second rod exerts on each infinitesimal segment of the first rod SET UP: An infinitesimal segment of the second rod is sketched in Figure 21.107 dQ = (Q/L )dx′ EXECUTE: (a) dE = L E x = Ñ dE x = k dQ ( x + a/2 + L − x′) = kQ dx′ L ( x + a/2 + L − x′)2 L kQ L dx′ kQ ⎡ kQ ⎛ 1 ⎤ ⎞ = = − ⎜ ⎟ L Ñ0 ( x + a/2 + L − x′) L ⎢⎣ x + a/2 + L − x′ ⎥⎦ L ⎝ x + a/2 x + a/2 + L ⎠ 2kQ ⎛ 1 ⎞ − ⎜ ⎟ L ⎝ x + a 2L + x + a ⎠ (b) Now consider the force that the field of the second rod exerts on an infinitesimal segment dq of the first rod This force is in the + x -direction dF = dq E Ex = F = ∫ E dq = Ñ L + a/2 EQ a/ L dx = ( 2kQ 2 L L + a/2 ⎛ Ña/2 1 ⎞ − ⎜ ⎟ dx 2 + + + x a L x a ⎝ ⎠ ) 2kQ kQ ⎛ a + L + a ⎞⎛ L + 2a ⎞ ⎞ F= [ln (a + x)]aL/2+ a/2 − [ln(2 L + x + a)]aL/2+ a/2 = 1n ⎜ ⎜⎛ ⎟⎜ ⎟ ⎟ 2a L L ⎠⎝ L + 2a ⎠ ⎠ ⎝⎝ F= ⎛ (a + L)2 ⎞ 1n ⎜⎜ ⎟⎟ L2 ⎝ a (a + L) ⎠ kQ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electric Charge and Electric Field (c) For a L, F = 21-51 ⎛ a (1 + L/a ) ⎞ kQ 1n ⎜⎜ ⎟⎟ = (2 1n (1 + L/a ) − ln(1 + L/a )) L2 ⎝ a (1 + L/a ) ⎠ L kQ z2 Therefore, for a L, ⎞ ⎛ L L2 ⎞ ⎞ kQ kQ ⎛ ⎛ L L2 − + ⋅⋅⋅ ⎟ ⎟ ≈ F ≈ ⎜ ⎜ − + ⋅⋅⋅ ⎟ − ⎜ ⎟ ⎜ a ⎟⎟ a L ⎜⎝ ⎜⎝ a 2a a ⎠ ⎝ ⎠⎠ EVALUATE: The distance between adjacent ends of the rods is a When a L the distance between the rods is much greater than their lengths and they interact as point charges For small z, ln(1 + z ) ≈ z − Figure 21.107 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... (9 × 109 N ⋅ m /C2 )(3.00 × 10−6 C)(8.00 × 10−6 C) r132 We need Fx = F21 − F31 = −7.00 N, so 3.375 N − r13 = = 0.216 N ⋅ m 0.216 N ⋅ m r132 r132 , in the −x-direction = −7.00 N 0.216 N ⋅ m = 0.144... publisher Electric Charge and Electric Field 21 -13 v y = 8.00 × 105 m/s Then v = vx2 + v 2y = 1.79 × 106 m/s EVALUATE: v y = v0 y + a yt gives a y = 6.4 × 1 013 m/s The electric field between the plates... E1 sin θ , E2 y = + E2 sin θ E y = E1 y + E2 y = E1 y = E1 sin θ = 2(862.8 N/C)(0.800) = 138 0 N/C E = 138 0 N/C, in the + y -direction EVALUATE: Point a is symmetrically placed between identical