Soluções Halliday 5ª edição volume 3
Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane
The solutions here are somewhat brief, as they are designed for the instructor, not for the student. Check with the publishers before electronically posting any part of these solutions; website, ftp, or I have been somewhat casual about subscripts whenever it is obvious that a problem is one dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution. Although this does not change the validity of the answer, it will sometimes obfuscate the approach There are some traditional formula, such as x 0x
which are not used in the text. The worked solutions use only material from the text, so there may be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an easier approach existed. But if it was not in the text, I did not use it here. I also tried to avoid reinventing the wheel. There are some exercises and problems in the text which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer I adopt a di erent approach for rounding of signi cant gures than previous authors; in partic- ular, I usually round intermediate answers. As such, some of my answers will di er from those in Exercises and Problems which are enclosed in a box also appear in the Student’s Solution Manual with considerably more detail and, when appropriate, include discussion on any physical implications of the answer. These student solutions carefully discuss the steps required for solving problems, point out the relevant equation numbers, or even specify where in the text additional information can be found. When two almost equivalent methods of solution exist, often both are presented. You are encouraged to refer students to the Student’s Solution Manual for these exercises and problems. However, the material from the Student’s Solution Manual must not be copied.
E25-1 The charge transferred is Q = (2:5 104 C=s)(20 106 s) = 5:0 101 C:
E25-2 Use Eq. 25-4: s (8:99109Nm2=C2)(26:3106C)(47:1106C) r = = 1:40 m (5:66 N)
E25-3 Use Eq. 25-4: (8:99109Nm2=C2)(3:12106C)(1:48106C) F = = 2:74 N: (0:123 m)2
E25-4 (a) The forces are equal, so m1a1 = m2a2, or m2 = (6:31107kg)(7:22 m=s2)=(9:16 m=s2) = 4:97107kg: (b) Use Eq. 25-4: s (6:31107kg)(7:22 m=s2)(3:20103m)2 q = = 7:201011C (8:99109Nm2=C2) E25-5 (a) Use Eq. 25-4, 1 q1q2 1 (21:3 C)(21:3 C) F = = = 1:77 N 40 r2 4(8:851012 C2=N m2) (1:52 m)2 12
(b) In part (a) we found F12; to solve part (b) we need to rst nd F13. Since q3 = q2 and We must assess the direction of the force of q3 on q1; it will be directed along the line which connects the two charges, and will be directed away from q3. The diagram below shows the directions.
F 23 F 12 F 23 F net F 12 From this diagram we want to nd the magnitude of the net force on q1. The cosine law is appropriate here: net 12 13 12 13 = 9:40 N2;
E25-6 Originally F0 = CQ2 = 0:088 N, where C is a constant. When sphere 3 touches 1 the 0 charge on both becomes Q0=2. When sphere 3 the touches sphere 2 the charge on each becomes (Q0 + Q0=2)=2 = 3Q0=4. The force between sphere 1 and 2 is then F = C(Q0=2)(3Q0=4) = (3=8)CQ2 = (3=8)F0 = 0:033 N: 0
The forces on q3 are F~ 31 and F~ 32. These forces are given by the vector form of Coulomb’s Law, Eq. 25-5, E25-7
1 q3q1 1 q3q1 40 r2 40 (2d)2 31 1 q3q2 1 q3q2 F~ 32 = ^r32 = ^r32: 40 r2 40 (d)2 32
These two forces are the only forces which act on q3, so in order to have q3 in equilibrium the forces must be equal in magnitude, but opposite in direction. In short, 1 q3q1 1 q3q2 40 (2d)2 40 (d)2 q1 q2 ^r31 = ^r32: 41 Note that ^r31 and ^r32 both point in the same direction and are both of unit length. We then get
q1 = 4q2: E25-8 The horizontal and vertical contributions from the upper left charge and lower right charge are straightforward to nd. The contributions from the upper left charge require slightly more work. pp The diagonal distance is 2a; the components will be weighted by cos 45 = 2=2. The diagonal charge will contribute pp 1 (q)(2q) 2 2 q2 40 ( 2a)2 2 80 a2 pp 1 (q)(2q) 2 2 q2 Fy = p ^j = ^j: 40 ( 2a)2 2 80 a2 (a) The horizontal component of the net force is then p 1 (2q)(2q) 2 q2 4 a2 8 a2 0p 0 4 + 2=2 q2 40 a2 = (4:707)(8:99109N m2=C2)(1:13106C)2=(0:152 m)2^i = 2:34 N^i: (b) The vertical component of the net force is then p 1 (q)(2q) 2 q2 40 a2 80 a 2 p 2 + 2=2 q2 80 a2 = (1:293)(8:99109Nm2=C2)(1:13106C)2=(0:152m)2^j=0:642N^j:
E25-9 The magnitude of the force on the negative charge from each positive charge is F = (8:99109N m2=C2)(4:18106C)(6:36106C)=(0:13 m)2 = 14:1 N: The force from each positive charge is directed along the side of the triangle; but from symmetry only the component along the bisector is of interest. This means that we need to weight the above answer by a factor of 2 cos(30) = 1:73. The net force is then 24:5 N.
E25-10 Let the charge on one sphere be q, then the charge on the other sphere is Q = (52:6 106C) q. Then 1 qQ 40 r2 (8:99109Nm2=C2)q(52:6106C q) = (1:19 N)(1:94 m)2:
E25-11 This problem is similar to Ex. 25-7. There are some additional issues, however. It is easy enough to write expressions for the forces on the third charge
qq 40 r2 31 1 q3q2 F~ 32 = ^r32: 40 r2 32 Then 1 q3q1 1 q3q2 40 r2 40 r2 31 32 q1 q2 ^r31 = 2 ^r32: r2 r 31 32
The only way to satisfy the vector nature of the above expression is to have ^r31 = ^r32; this means that q3 must be collinear with q1 and q2. q3 could be between q1 and q2, or it could be on either side. Let’s resolve this issue now by putting the values for q1 and q2 into the expression:
(1:07 C) (3:28 C) r2 r2 31 32 r2 ^r31 = (3:07)r2 ^r32: 32 31
Since squared quantities are positive, we can only get this to work if ^r31 = ^r32, so q3 is not between q1 and q2. We are then left with 32 31 so that q3 is closer to q1 than it is to q2. Then r32 = r31 + r12 = r31 + 0:618 m, and if we take the square root of both sides of the above expression, p p (0:618 m) = (3:07)r31 r31;
E25-12 The magnitude of the magnetic force between any two charges is kq2=a2, where a = 0:153 m. The force between each charge is directed along the side of the triangle; but from symmetry only the component along the bisector is of interest. This means that we need to weight the above answer by a factor of 2 cos(30) = 1:73. The net force on any charge is then 1:73kq2=a2. The distance from any charge to the center of the equilateral triangle is x, given by x2 = (a=2)2 + (d x)2. Then x = a2=8d + d=2 = 0:644a: The angle between the strings and the plane of the charges is , given by sin = x=(1:17 m) = (0:644)(0:153 m)=(1:17 m) = 0:0842;
The force of gravity on each ball is directed vertically and the electric force is directed horizontally. The two must then be related by tan = F E=F G;
so or q = 1:29107C: On any corner charge there are seven forces; one from each of the other seven charges. The net force will be the sum. Since all eight charges are the same all of the forces will be repulsive. We need to sketch a diagram to show how the charges are labeled.
E25-13 2 6 8 4 7 5 The magnitude of the force of charge 2 on charge 1 is 1 q2 40 r2 12 1
3 where r12 = a, the length of a side. Since both charges are the same we wrote q2. By symmetry we expect that the magnitudes of F12, F13, and F14 will all be the same and they will all be at right angles to each other directed along the edges of the cube. Written in terms of vectors the forces
would be 1 q2 40 a2 1 q2 40 a2 1 q2 F~ 14 = k^: 40 a2 The force from charge 5 is 1 q2 15 4 r2 0 15 and is directed along the side diagonal away from charge 5. The distance r15 is also the side diagonal distance, and can be found from 15 then 1 q2 F15 = : 40 2a2 By symmetry we expect that the magnitudes of F15, F16, and F17 will all be the same and they will all be directed along the diagonals of the faces of the cube. In terms of components we would have 1 q2 p p 40 2a2 1 q2 p p 40 2a2 1 q2 p p F~ 17 = ^i= 2 + ^j= 2 : 40 2a2 The last force is the force from charge 8 on charge 1, and is given by 1 q2 40 r2 18 and is directed along the cube diagonal away from charge 8. The distance r18 is also the cube diagonal distance, and can be found from 18 then in term of components 1 q2 p p p F~ 18 = ^i= 3 + ^j= 3 + k^= 3 : 40 3a2 We can add the components together. By symmetry we expect the same answer for each com- ponents, so we’ll just do one. How about ^i. This component has contributions from charge 2, 6, 7, and 8: 1 q2 121 40 a2 1 2 2 3 3 or 1 q2 (1:90) 40 a2 p The three components add according to Pythagoras to pick up a nal factor of 3, so q2 F net = (0:262) : 0a2
E25-14 (a) Yes. Changing the sign of y will change the sign of Fy; since this is equivalent to putting the charge q0 on the \other” side, we would expect the force to also push in the \other” (b) The equation should look Eq. 25-15, except all y’s should be replaced by x’s. Then 1 q0 q Fx = p : 40 x x2 + L2=4 (c) Setting the particle a distance d away should give a force with the same magnitude as 1 q0 q F= p : 40 d d2 + L2=4 This force is directed along the 45 line, so Fx = F cos 45 and Fy = F sin 45. p (d) Let the distance be d = x2 + y2, and then use the fact that Fx=F = cos = x=d. Then x 1 x q0 q Fx = F = : d 40 (x2 + y2 + L2=4)3=2 and y 1 y q0 q Fy = F = : d 40 (x2 + y2 + L2=4)3=2 E25-15 (a) The equation is valid for both positive and negative z, so in vector form it would read 1 q0 q z F~ = Fzk^ = k^: 40 (z2 + R2)3=2 (b) The equation is not valid for both positive and negative z. Reversing the sign of z should p reverse the sign of Fz, and one way to x this is to write 1 = z= z2. Then
1 2q0 qz 1 1 F~ = Fzk^ = p p k^: 40 R2 z2 z2
E25-16 Divide the rod into small di erential lengths dr, each with charge dQ = (Q=L)dr. Each di erential length contributes a di erential force 1 q dQ 1 qQ dF = = dr: 40 r2 40 r2L Integrate: Z Z x+L 1 qQ x 40 r2L 1 qQ 1 1 = 40 L x x + L E25-17 You must solve Ex. 16 before solving this problem! q0 refers to the charge that had been called q in that problem. In either case the distance from q0 will be the same regardless of the sign of q; if q = Q then q will be on the right, while if q = Q then q will be on the left. Setting the forces equal to each other one gets
1 qQ 1 1 1 qQ 40 L x x + L 40 r2 or p r = x(x + L):
E25-18 You must solve Ex. 16 and Ex. 17 before solving this problem. If all charges are positive then moving q0 o axis will result in a net force away from the axis. If q = Q then both q and Q are on the same side of q0. Moving q0 closer to q will result in the attractive force growing faster than the repulsive force, so q0 will move away from equilibrium.
E25-19 We can start with the work that was done for us on Page 577, except since we are concerned with sin = z=r we would have 1 q0 dz z dFx = dF sin = : 2 2p 40 (y + z ) y2 + z2 We will need to take into consideration that changes sign for the two halves of the rod. Then ! q0 Z 0 z dz L=2 Z +z dz Fx = + ; 40 L=2 (y2 + z2)3=2 0 (y2 + z )3=2 2
Z L=2 q0 z dz 20 0 (y2 + z2)3=2 L=2 q0 1 20 y2 + z2 0 ! q0 1 1 = p : 20 y y2 + (L=2)2
E25-20 Use Eq. 25-15 to nd the magnitude of the force from any one rod, but write it as 1 qQ 4 r r2 + L2=4 0
where r2 = z2 + L2=4. The component of this along the z axis is Fz = F z=r. Since there are 4 rods, we have 1 qQz 1 qQz 0 r r + L =4 0 (z + L =4) z + L =2 2222222
Equating the electric force with the force of gravity and solving for Q, 0mg p qz putting in the numbers, (8:851012C2=Nm2)(3:46107kg)(9:8m=s2) p ((0:214m)2+(0:25m)2=4) (0:214m)2+(0:25m)2=2 (2:451012C)(0:214 m) so Q = 3:07106C:
In each case we conserve charge by making sure that the total number of protons is the same on both sides of the expression. We also need to conserve the number of neutrons. (a) Hydrogen has one proton, Beryllium has four, so X must have ve protons. Then X must be (b) Carbon has six protons, Hydrogen has one, so X must have seven. Then X is Nitrogen, N. (c) Nitrogen has seven protons, Hydrogen has one, but Helium has two, so X has 7 + 1 2 = 6 protons. This means X is Carbon, C.
E25-22 (a) Use Eq. 25-4: (8:99109Nm2=C2)(2)(90)(1:601019C)2 F = = 290 N: (121015m)2 (b) a = (290 N)=(4)(1:661027kg) = 4:41028m=s2.
E25-23 Use Eq. 25-4: (8:99109Nm2=C2)(1:601019C)2 F = = 2:89109N: (2821012m)2 E25-24 (a) Use Eq. 25-4: s (3:7109N)(5:01010m)2 q = = 3:201019C: (8:99109Nm2=C2) (b) N = (3:201019C)=(1:601019C) = 2.
E25-25 Use Eq. 25-4, 1 q1q2 ( 1 1:6 1019 C)( 1 1:6 1019 C) F = = 3 3 = 3:8 N: 40 r2 4(8:85 1012 C2=N m2)(2:6 1015 m)2 12
(b) The penny has enough electrons to make a total charge of 1:37105C. The fraction is then (1:15107C)=(1:37105C) = 8:401013:
E25-27 Equate the magnitudes of the forces: 1 q2 40 r2 so s (8:99109Nm2=C2)(1:601019C)2 r = = 5:07 m (9:111031kg)(9:81 m=s2)
E25-29 The mass of water is (250 cm3)(1:00 g/cm3) = 250 g. The number of moles of water is (250 g)=(18:0 g/mol) = 13:9 mol. The number of water molecules is (13:9 mol)(6:021023mol1) = 8:371024. Each molecule has ten protons, so the total positive charge is Q = (8:371024)(10)(1:601019C) = 1:34107C:
E25-30 The total positive charge in 0:250 kg of water is 1:34107C. Mary’s imbalance is then q1 = (52:0)(4)(1:34107C)(0:0001) = 2:79105C;
while John’s imbalance is The electrostatic force of attraction is then 1 q1q2 (2:79105)(4:86105) F = = (8:99109N m2=C2) = 1:61018N: 40 r2 (28:0 m) 2
(a) The gravitational force of attraction between the Moon and the Earth is GMEMM R2 where R is the distance between them. If both the Earth and the moon are provided a charge q, then the electrostatic repulsion would be E25-31
1 q2 FE = : 40 R2 Setting these two expression equal to each other, q2 40
which has solution p q = 5:71 1013 C: (b) We need (5:71 1013 C)=(1:60 1019 C) = 3:57 1032 protons on each body. The mass of protons needed is then (3:57 1032)(1:67 1027 kg) = 5:97 1065 kg: Ignoring the mass of the electron (why not?) we can assume that hydrogen is all protons, so we need that much hydrogen.
Assume that the spheres initially have charges q1 and q2. The force of attraction between them is 1 q1q2 40 r2 12 where r12 = 0:500 m. The net charge is q1 + q2, and after the conducting wire is connected each sphere will get half of the total. The spheres will have the same charge, and repel with a force of 1 (q + q ) 1 (q + q ) 1212212 F2 = = 0:0360 N: 40 r2 12
Since we know the separation of the spheres we can nd q1 + q2 quickly, P25-1
q q1 + q2 = 2 40r2 (0:0360 N) = 2:00 C 12 1 (2:00 C q2)q2 40 r2 12 22 0 = q2 + (2:00 C)q2 + (1:73 C) : 2 2
P25-2 The electrostatic force on Q from each q has magnitude qQ=40a2, where a is the length of thepside of the square. The magnitude of the vertical (horizontal) component of the force of Q on 0 (a) In order to have a zero net force on Q the magnitudes of the two contributions must balance, so p 2Q2 qQ 16 a2 4 a2 00 p (b) No.
P25-3 (a) The third charge, q3, will be between the rst two. The net force on the third charge will be zero if 1 q q3 1 4q q3 40 r2 40 r2 31 32 which will occur if 12 = r31 r32 The total distance is L, so r31 + r32 = L, or r31 = L=3 and r32 = 2L=3. Now that we have found the position of the third charge we need to nd the magnitude. The second and third charges both exert a force on the rst charge; we want this net force on the rst charge to be zero, so 1 q q3 1 q 4q 4 r2 4 r2 0 13 0 12 or q3 4q (L=3)2 L2 which has solution q3 = 4q=9. The negative sign is because the force between the rst and second charge must be in the opposite direction to the force between the rst and third charge. (b) Consider what happens to the net force on the middle charge if is is displaced a small distance z. If the charge 3 is moved toward charge 1 then the force of attraction with charge 1 will increase. But moving charge 3 closer to charge 1 means moving charge 3 away from charge 2, so the force of attraction between charge 3 and charge 2 will decrease. So charge 3 experiences more attraction to ward the charge that it moves toward, and less attraction to the charge it moves away from. Sounds unstable to me.
P25-4 (a) The electrostatic force on the charge on the right has magnitude q2 40×2 The weight of the ball is W = mg, and the two forces are related by F=W=tansin=x=2L: Combining, 2Lq2 = 40mgx3, so 2 1=3 qL x= : 20 (b) Rearrange and solve for q, s 2(8:851012C2=N m2)(0:0112 kg)(9:81 m=s2)(4:70102m)3 q = = 2:28108C: (1:22 m)
P25-5 (a) Originally the balls would not repel, so they would move together and touch; after touching the balls would \split” the charge ending up with q=2 each. They would then repel again. (b) The new equilibrium separation is (q=2)2 1=3 1=3 L1 x0 = = x = 2:96 cm: 20mg 4 P25-6 Take the time derivative of the expression in Problem 25-4. Then dx 2 x dq 2 (4:70102m) = = (1:20109C=s) = 1:65103m=s: dt 3 q dt 3 (2:28108C) P25-7 The force between the two charges is 1 (Q q)q F= : 40 r2 12
We want to maximize this force with respect to variation in q, this means nding dF =dq and setting it equal to 0. Then dF d 1 (Q q)q 1 Q 2q ==: dq dq 4 r2 4 r2 0 12 0 12 2
P25-8 Displace the charge q a distance y. The net restoring force on q will be approximately qQ 1 y qQ 16 F 2 = y: 40 (d=2)2 (d=2) 40 d3
Since F =y is e ectively a force constant, the period of oscillation is r 3 3 1=2 m 0m d T = 2 = : k qQ P25-9 Displace the charge q a distance x toward one of the positive charges Q. The net restoring force on q will be
qQ 1 1 40 (d=2 x)2 (d=2 + x)2 qQ 32 x: 40 d3
Since F =x is e ectively a force constant, the period of oscillation is r 3 3 1=2 m 0m d T = 2 = : k 2qQ
(b) Removing a positive Cesium ion is equivalent to adding a singly charged negative ion at that same location. The net force is then 0
where r is the distance between the Chloride ion and the newly placed negative ion, or p r = 3(0:20109m)2 The force is then (1:61019C)2 F = = 1:92109N: 4(8:851012C2=N m2)3(0:20109m)2
We can pretend that this problem is in a single plane containing all three charges. The magnitude of the force on the test charge q0 from the charge q on the left is 1 q q0 Fl = : 4 (a2 + R2) 0 P25-11
A force of identical magnitude exists from the charge on the right. we need to add these two forces as vectors. Only the components along R will survive, and each force will contribute an amount
R R2 + a2 so the net force on the test particle will be 2 q q0 R p: 40 (a2 + R2) R2 + a2
We want to nd the maximum value as a function of R. This means take the derivative, and set it equal to zero. The derivative is 2 2q q0 1 3R 40 (a2 + R2)3=2 (a2 + R2)5=2 which will vanish when a2 + R2 = 3R2;
E26-1 E = F=q = ma=q. Then E = (9:111031kg)(1:84109m=s2)=(1:601019C) = 1:05102N=C:
E26-2 The answers to (a) and (b) are the same! E26-3 F = W , or Eq = mg, so mg (6:64 1027 kg)(9:81 m=s2) E = = = 2:03 107 N=C: q 2(1:60 1019 C) The alpha particle has a positive charge, this means that it will experience an electric force which is in the same direction as the electric eld. Since the gravitational force is down, the electric force, and consequently the electric eld, must be directed up.
(b) F = Eq = (1:5103N=C)(1:601019C) = 2:41016N: (c) F = mg = (1:671027kg)(9:81 m=s2) = 1:61026N: (d) (2:41016N)=(1:61026N) = 1:51010:
E26-5 Rearrange E = q=40r2, q = 4(8:851012C2=N m2)(0:750 m)2(2:30 N=C) = 1:441010C:
E26-7 Use Eq. 26-12 for points along the perpendicular bisector. Then 1 p (3:56 1029 C m) E = = (8:99 109N m2=C2) = 1:95 104N=C: 4 x3 (25:4 109 m)3 0
E26-8 If the charges on the line x = a where +q and q instead of +2q and 2q then at the center of the square E = 0 by symmetry. This simpli es the problem into nding E for a charge +q at (a; 0) and q at (a; a). This is a dipole, and the eld is given by Eq. 26-11. For this exercise we have x = a=2 and d = a, so 1 qa 40 [2(a=2)2]3=2 or, putting in the numbers, E = 1:11105N=C.
E26-9 The charges at 1 and 7 are opposite and can be e ectively replaced with a single charge of 6q at 7. The same is true for 2 and 8, 3 and 9, on up to 6 and 12. By symmetry we expect the eld to point along a line so that three charges are above and three below. That would mean 9:30.
E26-10 If both charges are positive then Eq. 26-10 would read E = 2E+ sin , and Eq. 26-11 would look like 1qx 40 x2 + (d=2)2 x2 + (d=2)2 p
Treat the two charges on the left as one dipole and treat the two charges on the right as a second dipole. Point P is on the perpendicular bisector of both dipoles, so we can use Eq. 26-12 For the dipole on the left p = 2aq and the electric eld due to this dipole at P has magnitude 1 2aq El = 40 (x + a)3 E26-11
For the dipole on the right p = 2aq and the electric eld due to this dipole at P has magnitude 1 2aq Er = 40 (x a)3 The net electric eld at P is the sum of these two elds, but since the two component elds point in opposite directions we must actually subtract these values, E = Er El;
2aq 1 1 40 (x a)3 (x + a)3 aq 1 1 1 =: 20 x3 (1 a=x)3 (1 + a=x)3 We can use the binomial expansion on the terms containing 1 a=x, aq 1 20 x3 aq 1 20 x3 3(2qa2) =: 20×4
E26-12 Do a series expansion on the part in the parentheses 1 1 R2 2 R 1 1 1 = : p 22 1 + R2=z2 2 z 2z Substitute this in, R2 Q Ez = : 20 2z2 40z2
E26-13 At the surface z = 0 and Ez = =20. Half of this value occurs when z is given by 1z 2 z2 + R2 p which can be written as z2 + R2 = (2z)2. Solve this, and z = R= 3.
E26-14 Look at Eq. 26-18. The electric eld will be a maximum when z=(z2 + R2)3=2 is a maximum. Take the derivative of this with respect to z, and get 1 3 2z2 z2 + R2 3z2 =: (z2 + R2)3=2 2 (z2 + R2)5=2 (z2 + R2)5=2 p This will vanish when the numerator vanishes, or when z = R= 2.
E26-15 (a) The electric eld strength just above the center surface of a charged disk is given by Eq. 26-19, but with z = 0, E= 20 The surface charge density is = q=A = q=(R2). Combining, q = 2 R2E = 2(8:85 1012 C2=N m2)(2:5 102m)2(3 106 N=C) = 1:04 107C: 0
Notice we used an electric eld strength of E = 3 106 N=C, which is the eld at air breaks down (b) We want to nd out how many atoms are on the surface; if a is the cross sectional area of one atom, and N the number of atoms, then A = N a is the surface area of the disk. The number of atoms is A (0:0250 m)2 N = = = 1:31 1017 a (0:015 1018 m2) (c) The total charge on the disk is 1:04 107C, this corresponds to (1:04 107C)=(1:6 1019C) = 6:5 1011 electrons. (We are ignoring the sign of the charge here.) If each surface atom can have at most one excess electron, then the fraction of atoms which are charged is (6:5 1011)=(1:31 1017) = 4:96 106;
E26-16 Imagine switching the positive and negative charges. The electric eld would also need to switch directions. By symmetry, then, the electric eld can only point vertically down. Keeping only that component, Z =2 1 d 0 40 r2 2 =: 40 r2
E26-17 We want to t the data to Eq. 26-19, z Ez = 1 p : 20 z2 + R2 We can nd very easily if we assume that the measurements have no error because then at the surface (where z = 0), the expression for the electric eld simpli es to
E= : 20 Then = 2 E = 2(8:854 1012 C2=N m2)(2:043 107 N=C) = 3:618 104 C=m2. 0 Finding the radius will take a little more work. We can choose one point, and make that the reference point, and then solve for R. Starting with
and then rearranging, 20Ez z z2 + R2 20Ez 1 1 + (R=z)2 1 20Ez 1 + (R=z)2 1 (1 20Ez=)2 s R1 = 1: z (1 20Ez=)2
Using z = 0:03 m and Ez = 1:187 107 N=C, along with our value of = 3:618 104 C=m , we 2 nd s R1 = 1; z (1 2(8:8541012C2=Nm2)(1:187107N=C)=(3:618104C=m2))2 R = 2:167(0:03 m) = 0:065 m: (b) And now nd the charge from the charge density and the radius, q = R2 = (0:065 m)2(3:618 104 C=m2) = 4:80 C:
(b) Integrate: Z L+a 1 a 40 11 40 a L + a q1 40 a(L + a) (c) If a L then L can be replaced with 0 in the above expression.
E26-24 (a) The electric eld is zero nearer to the smaller charge; since the charges have opposite signs it must be to the right of the +2q charge. Equating the magnitudes of the two elds, 2q 5q 40×2 40(x + a)2 or p p 5x = 2(x + a);
which has solution p 2a x = p p = 2:72a: 5 2
E x d E26-26 (a) At point A, 1 q 2q 1 q 40 d2 (2d)2 40 2d2
At point B, 1 q 2q 1 6q 40 (d=2)2 (d=2)2 40 d2 At point C, 1 q 2q 1 7q 4 (2d)2 d2 4 4d2 00
(a) The electric eld does (negative) work on the electron. The magnitude of this work is W = F d, where F = Eq is the magnitude of the electric force on the electron and d is the distance through which the electron moves. Combining, E26-27
which gives the work done by the electric eld on the electron. The electron originally possessed a kinetic energy of K = 1 mv2, since we want to bring the electron to a rest, the work done must be 2 negative. The charge q of the electron is negative, so E~ and ~d are pointing in the same direction, By the work energy theorem, 1 W = K = 0 mv2: 2 We put all of this together and nd d, W mv2 (9:111031kg)(4:86 106 m=s)2 d = = = = 0:0653 m: qE 2qE 2(1:601019C)(1030 N=C) (b) Eq = ma gives the magnitude of the acceleration, and vf = vi + at gives the time. But vf = 0. Combining these expressions,
mv (9:111031kg)(4:86 106 m=s) t = i = = 2:69108 s: Eq (1030 N=C)(1:601019C) (c) We will apply the work energy theorem again, except now we don’t assume the nal kinetic energy is zero. Instead, W = K = Kf Ki;
and dividing through by the initial kinetic energy to get the fraction lost, W Kf Ki Ki Ki
But Ki = 1mv2, and W = qEd, so the fractional change is 2 W qEd (1:601019C)(1030 N=C)(7:88103m) = = = 12:1%: Ki 1mv2 1(9:111031kg)(4:86106m=s)2 22
E26-28 (a) a = Eq=m = (2:16104N=C)(1:601019C)=(1:671027kg) = 2:071012m=s2. pp (b) v = 2ax = 2(2:071012m=s2)(1:22102m) = 2:25105m=s:
E26-29 (a) E = 2q=40r2, or (1:88107C) E = = 5:85105N=C: 2(8:851012C2=N m2)(0:152 m=2)2
E26-30 (a) The average speed between the plates is (1:95102m)=(14:7109s) = 1:33106m=s. The speed with which the electron hits the plate is twice this, or 2:65106m=s. (b) The acceleration is a = (2:65106m=s)=(14:7109s) = 1:801014m=s2. The electric eld then has magnitude E = ma=q, or E = (9:111031kg)(1:801014m=s2)=(1:601019C) = 1:03103N=C: E26-31 The drop is balanced if the electric force is equal to the force of gravity, or Eq = mg. The mass of the drop is given in terms of the density by
4 m = V = r3: 3 Combining, mg 4r3g 4(851 kg=m3)(1:64106m)3(9:81 m=s2) q = = = = 8:111019C: E 3E 3(1:92105N=C) We want the charge in terms of e, so we divide, and get q (8:111019C) = = 5:07 5: e (1:601019C)
E26-32 (b) F = (8:99109N m2=C2)(2:16106C)(85:3109C)=(0:117m)2 = 0:121 N: (a) E2 = F=q = (0:121 N)=(2:16106C) = 5:60104N=C: 1 E1 = F=q2 = (0:121 N)=(85:3109C) = 1:42106N=C:
If each value of q measured by Millikan was a multiple of e, then the di erence between any two values of q must also be a multiple of q. The smallest di erence would be the smallest multiple, and this multiple might be unity. The di erences are 1.641, 1.63, 1.60, 1.63, 3.30, 3.35, 3.18, 3.24, all times 1019 C. This is a pretty clear indication that the fundamental charge is on the order of 1:6 1019 C. If so, the likely number of fundamental charges on each of the drops is shown below in a table arranged like the one in the book: E26-33
4 8 12 5 10 14 7 11 16 The total number of charges is 87, while the total charge is 142:69 1019 C, so the average charge per quanta is 1:64 1019 C.
E26-34 Because of the electric eld the acceleration toward the ground of a charged particle is not g, but g Eq=m, where the sign depends on the direction of the electric eld. (a)Ifthelowerplateispositivelychargedthena=gEq=m.Replaceginthependulumperiod expression by this, and then s L T = 2 : g Eq=m (b) If the lower plate is negatively charged then a = g + Eq=m. Replace g in the pendulum period expression by this, and then s L T = 2 : g + Eq=m
E26-35 The ink drop travels an additional time t0 = d=vx, where d is the additional horizontal distance between the plates and the paper. During this time it travels an additional vertical distance y0 = v t0, where v = at = 2y=t = 2yv =L. Combining, yyx
2yv t0 2yd 2(6:4104m)(6:8103m) y0 = x = = = 5:44104m; L L (1:6102m) so the total de ection is y + y0 = 1:18103m.
E26-36 (a) p = (1:48109C)(6:23106m) = 9:221015C m: Use = pE sin , where is the angle between ~p and E~ . For this dipole p = qd = 2ed or p = 2(1:6 1019 C)(0:78 109 m) = 2:5 1028 C m. For all three cases pE = (2:5 1028 C m)(3:4 106N=C) = 8:5 1022 N m: E26-37
(b) For the perpendicular case = 90, so sin = 1, and = 8:5 1022 N m:. (c) For the anti-parallel case = 180, so sin = 0, and = 0.
(d) Use Eq. 26-12 and F = Eq. Then 40x3F q 4(8:851012C2=N m2)(0:285m)3(5:221016N) (3:16106C) = 4:251022C m: E26-39 The point-like nucleus contributes an electric eld 1 Ze 40 r2 while the uniform sphere of negatively charged electron cloud of radius R contributes an electric eld given by Eq. 26-24, 1 Zer E = : 40 R3
The net electric eld is just the sum, Ze 1 r E= 40 r2 R3
E26-40 The shell theorem rst described for gravitation in chapter 14 is applicable here since both electric forces and gravitational forces fall o as 1=r2. The net positive charge inside the sphere of The net force on either electron will be zero when e2 eQ 4e2 d3 e2d d2 (d=2)2 d2 4R3 R3
P26-1 (a) Let the positive charge be located closer to the point in question, then the electric eld from the positive charge is 1q E+ = 40 (x d=2)2 The negative charge is located farther from the point in question, so 1q E = 40 (x + d=2)2 The net electric eld is the sum of these two elds, but since the two component elds point in opposite direction we must actually subtract these values, 1q1q 40 (z d=2)2 40 (z + d=2)2 1q 1 1 = 40 z2 (1 d=2z)2 (1 + d=2z)2 We can use the binomial expansion on the terms containing 1 d=2z, 1q 40 z2 1 qd = 20 z3
(b) The electric eld is directed away from the positive charge when you are closer to the positive charge; the electric eld is directed toward the negative charge when you are closer to the negative charge. In short, along the axis the electric eld is directed in the same direction as the dipole moment.
P26-2 The key to this problem will be the expansion of 1 1 3 zd 1 : (x2 + (z d=2)2)3=2 (x2 + z2)3=2 2 x2 + z2
p for d x2 + z2. Far from the charges the electric eld of the positive charge has magnitude 1q 4 x2 + (z d=2)2 0
the components of this are 1q x Ex;+ = ; x2 2 p 40 + z x2 + (z d=2)2 1 q (z d=2) Ez;+ = : x2 2 p 40 + z x2 + (z d=2)2 Expand both according to the rst sentence, then
1 xq 3 zd 40 (x2 + z2)3=2 2 x2 + z2 1 (z d=2)q 3 zd Ez;+ = 1 + : 40 (x2 + z2)3=2 2 x2 + z2 Similar expression exist for the negative charge, except we must replace q with q and the + in the parentheses with a , and z d=2 with z + d=2 in the Ez expression. All that is left is to add the expressions. Then
1 xq 3 zd 1 xq 3 zd Ex = 1 + + 1 ; 40 (x2 + z2)3=2 2 x2 + z2 40 (x2 + z2)3=2 2 x2 + z2 1 3xqzd 40 (x2 + z2)5=2 1 (z d=2)q 3 zd 1 (z + d=2)q 3 zd Ez = 1 + + 1 ; 40 (x2 + z2)3=2 2 x2 + z2 40 (x2 + z2)3=2 2 x2 + z2 1 3z2dq 1 dq 40 (x2 + z2)5=2 40 (x2 + z2)3=2 1 (2z2 x2)dq =: 40 (x2 + z2)5=2 p P26-3 (a) Each point on the ring is a distance z2 + R2 from the point on the axis in question. Since all points are equal distant and subtend the same angle from the axis then the top half of the ring contributes q1 z 4 (x2 + R2) z2 + R2 0
while the bottom half contributes a similar expression. Add, and q1 + q2 z q z Ez = = ; 4 (z2 + R2)3=2 4 (z2 + R2)3=2 00 (b) The perpendicular component would be zero if q1 = q2. It isn’t, so it must be the di erence q1 q2 which is of interest. Assume this charge di erence is evenly distributed on the top half of the ring. If it is a positive di erence, then E? must point down. We are only interested then in the vertical component as we integrate around the top half of the ring. Then Z q 1 (q1 2)= 4 z2 + R2 00 q1 q2 1 =: 22 z2 + R2 0
Add the contributions: 1 q 2q q 40 (z + d)2 z2 (z d)2 2 2 q 2d 3d 2d 3d 40z2 z z2 z z2 q 6d2 3Q 40z2 z2 40z4 where Q = 2qd2.
A monopole eld falls o as 1=r2. A dipole eld falls o as 1=r3, and consists of two oppositely charge monopoles close together. A quadrupole eld (see Exercise 11 above or read Problem 4) falls o as 1=r4 and (can) consist of two otherwise identical dipoles arranged with anti- parallel dipole moments. Just taking a leap of faith it seems as if we can construct a 1=r6 eld First we need an octopole which is constructed from a quadrupole. We want to keep things as simple as possible, so the construction steps are P26-5
2. The dipole is a charge +q at x = 0 and a charge q at x = a. We’ll call this a dipole at x = a=2 3. The quadrupole is the dipole at x = a=2, and a second dipole pointing the other way at x = a=2. The charges are then q at x = a, +2q at x = 0, and q at x = a.
4. The octopole will be two stacked, o set quadrupoles. There will be q at x = a, +3q at x = 0, 3q at x = a, and +q at x = 2a.
5. Finally, our distribution with a far eld behavior of 1=r6. There will be +q at x = 2a, 4q at x = a, +6q at x = 0, 4q at x = a, and +q at x = 2a.
P26-6 The vertical component of E~ is simply half of Eq. 26-17. The horizontal component is given by a variation of the work required to derive Eq. 26-16, 1 dz z 40 y2 + z2 y2 + z2 p
which integrates to zero if the limits are 1 to +1, but in this case, Z1 1 Ez = dEz = : 0 40 z
P26-7 (a) Swap all positive and negative charges in the problem and the electric eld must reverse direction. But this is the same as ipping the problem over; consequently, the electric eld must point parallel to the rod. This only holds true at point P , because point P doesn’t move when you ip the rod.
(b) We are only interested in the vertical component of the eld as contributed from each point on the rod. We can integrate only half of the rod and double the answer, so we want to evaluate
Ez (c) The previous expression expansion to P26-8 Evaluate 1 dz z 40 y2 + z2 p y2 + z2 0 p 2 (L=2)2 + y2 y =p: 40 y (L=2)2 + y 2
is exact. If y L, then the expression simpli es with a Taylor L2 40 y3 Z L=2
R 1 z dq Z 0 40 (z2 + r2)3=2 where r is the radius of the ring, z the distance to the plane of the ring, and dq the di erential charge on the ring. But r2 + z2 = R2, and dq = (2r dr), where = q=2R2. Then p Z R 2 r2 q R r dr 0 40 R5 q1 =: 40 3R2
The key statement is the second italicized paragraph on page 595; the number of eld lines through a unit cross-sectional area is proportional to the electric eld strength. If the exponent is n, then the electric eld strength a distance r from a point charge is P26-9
kq rn and the total cross sectional area at a distance r is the area of a spherical shell, 4r2. Then the number of eld lines through the shell is proportional to kq EA = 4r2 = 4kqr2n: rn Note that the number of eld lines varies with r if n 6= 2. This means that as we go farther from the point charge we need more and more eld lines (or fewer and fewer). But the eld lines can only start on charges, and we don’t have any except for the point charge. We have a problem; we really do need n = 2 if we want workable eld lines.
P26-10 The distance traveled by the electron will be d1 = a1t2=2; the distance traveled by the proton will be d2 = a2t2=2. a1 and a2 are related by m1a1 = m2a2, since the electric force is the same (same charge magnitude). Then d1 + d2 = (a1 + a2)t2=2 is the 5.00 cm distance. Divide by the proton distance, and then d1 + d2 a1 + a2 m2 = = + 1: d2 a2 m1
P26-11 This is merely a fancy projectile motion problem. vx = v0 cos while vy;0 = v0 sin . The x and y positions are x = vxt and 1 ax2 y = at2 + vy;0t = + x tan : 2 2v2 cos2 0
The acceleration of the electron is vertically down and has a magnitude of F Eq (1870 N=C)(1:61019C) a = = = = 3:2841014m=s2: m m (9:111031kg) We want to nd out how the vertical velocity of the electron at the location of the top plate. If we get an imaginary answer, then the electron doesn’t get as high as the top plate.
q p = 7:226105m=s: This is a real answer, so this means the electron either hits the top plate, or it misses both plates. The time taken to reach the height of the top plate is vy (7:226105m=s) (5:83106m=s) sin(39) t = = = 8:972109s: a (3:2841014m=s2)
In this time the electron has moved a horizontal distance of x = (5:83106m=s) cos(39)(8:972109s) = 4:065102m: This is clearly on the upper plate.
P26-12 Near the center of the ring z R, so a Taylor expansion yields z E= : 20 R2
The force on the electron is F = Ee, so the e ective \spring” constant is k = e=20R2. This means r r r k e eq != = = : m 20mR2 40mR3
P26-13 U = pE cos , so the work required to ip the dipole is W = pE [cos(0 + ) cos 0] = 2pE cos 0:
P26-14 If the torque on a system is given by j j = , where is a constant, then the frequency p of oscillation of the system is f = =I=2. In this case = pE sin pE, so p f = pE=I=2:
Use the a variation of the exact result from Problem 26-1. The two charge are positive, but since we will eventually focus on the area between the charges we must subtract the two eld contributions, since they point in opposite directions. Then P26-15
q11 Ez = 40 (z a=2)2 (z + a=2)2 and then take the derivative,
dEz q 1 1 = : dz 20 (z a=2)3 (z + a=2)3 Applying the binomial expansion for points z a,
dEz 8q 1 1 1 dz 20 a3 (2z=a 1)3 (2z=a + 1)3 8q 1 20 a3 8q 1 =: 0 a3 There were some fancy sign ips in the second line, so review those steps carefully! (b) The electrostatic force on a dipole is the di erence in the magnitudes of the electrostatic forces on the two charges that make up the dipole. Near the center of the above charge arrangement the electric eld behaves as dEz dz z=0
The net force on a dipole is dEz dEz F+ F = q(E+ E) = q Ez(0) + z+ Ez(0) z dz dz z=0 z=0
where the \+” and \-” subscripts refer to the locations of the positive and negative charges. This last line can be simpli ed to yield dEz dEz q (z+ z) = qd : dz dz z=0 z=0
(b) E = (2:0 m2)^j (2 N=C)^j = 4N m2=C: (c) E = (2:0 m2)^j [(3 N=C)^i + (4 N=C)k^] = 0: (d) In each case the eld is uniform so we can simply evaluate E = E~ A~ , where A~ has six parts, one for every face. The faces, however, have the same size but are organized in pairs with opposite directions. These will cancel, so the total ux is zero in all three cases.
E27-3 (a) The at base is easy enough, since according to Eq. 27-7, Z E = E~ dA~ :
There are two important facts to consider in order to integrate this expression. E~ is parallel to the axis of the hemisphere, E~ points inward while d~A points outward on the at base. E~ is uniform, so it is everywhere the same on the at base. Since E~ is anti-parallel to dA~ , E~ dA~ = E dA, then ZZ E = E~ dA~ = E dA:
Since E~ is uniform we can simplify this as ZZ E = E dA = E dA = EA = R2E:
The last steps are just substituting the area of a circle for the at side of the hemisphere. (b) We must rst sort out the dot product
E dA R We can simplify the vector part of the problem with E~ dA~ = cos E dA, so ZZ E = E~ dA~ = cos E dA
Once again, E~ is uniform, so we can take it out of the integral, ZZ E = cos E dA = E cos dA
We’ll integrate around the axis, from 0 to 2. We’ll integrate from the axis to the equator, from 0 to =2. Then Z Z 2 Z =2 E = E cos dA = E R2 cos sin d d : 00
Pulling out the constants, doing the integration, and then writing 2 cos sin as sin(2), Z =2 Z =2 E 00
Change variables and let = 2, then we have 1 Z E = R2E sin d = R2E: 02
E27-4 Through S1, E = q=0. Through S2, E = q=0. Through S3, E = q=0. Through S4, E = 0. Through S5, E = q=0.
E27-5 By Eq. 27-8, q (1:84 C) = = = 2:08105 N m2=C: E 0 (8:851012 C2=N m2) E27-6 The total ux through the sphere is E = (1 + 2 3 + 4 5 + 6)(103N m2=C) = 3103N m2=C: The charge inside the die is (8:851012C2=N m2)(3103N m2=C) = 2:66108C:
E27-7 The total ux through a cube would be q=0. Since the charge is in the center of the cube we expect that the ux through any side would be the same, or 1=6 of the total ux. Hence the ux through the square surface is q=60.
E27-8 If the electric eld is uniform then there are no free charges near (or inside) the net. The ux through the netting must be equal to, but opposite in sign, from the ux through the opening. The ux through the opening is Ea2, so the ux through the netting is Ea2.
E27-9 There is no ux through the sides of the cube. The ux through the top of the cube is (58 N=C)(100 m)2 = 5:8105N m2=C. The ux through the bottom of the cube is (110 N=C)(100 m)2 = 1:1106N m2=C: The total ux is the sum, so the charge contained in the cube is q = (8:851012C2=N m2)(5:2105N m2=C) = 4:60106C:
E27-10 (a) There is only a ux through the right and left faces. Through the right face R = (2:0 m )^j (3 N=C m)(1:4 m)^j = 8:4 N m2=C: 2
There are eight cubes which can be \wrapped” around the charge. Each cube has three external faces that are indistinguishable for a total of twenty-four faces, each with the same ux E. The total ux is q=0, so the ux through one face is E = q=240. Note that this is the ux through faces opposite the charge; for faces which touch the charge the electric eld is parallel to the surface, so the ux would be zero.
E27-11 E27-12 Use Eq. 27-11, = 20rE = 2(8:851012C2=N m2)(1:96 m)(4:52104N=C) = 4:93106C=m: E27-13 (a) q = A = (2:0106C=m2)(0:12 m)(0:42 m) = 3:17107C: (b) The charge density will be the same! q = A = (2:0 106C=m2)(0:08 m)(0:28 m) = 1:41107C:
E27-14 The electric eld from the sheet on the left is of magnitude El = =20, and points directly away from the sheet. The magnitude of the electric eld from the sheet on the right is the same, (a) To the left of the sheets the two elds add since they point in the same direction. This means (c) To the right of the sheets the two elds add since they point in the same direction. This means that the electric eld is E~ = (=0)^i.
E27-15 The electric eld from the plate on the left is of magnitude El = =20, and points directly toward the plate. The magnitude of the electric eld from the plate on the right is the same, but it (a) To the left of the plates the two elds cancel since they point in the opposite directions. This (b) Between the plates the two electric elds add since they point in the same direction. This (c) To the right of the plates the two elds cancel since they point in the opposite directions. This means that the electric eld is E~ = 0.
E27-16 The magnitude of the electric eld is E = mg=q. The surface charge density on the plates is = 0E = 0mg=q, or (8:851012C2=N m2)(9:111031kg)(9:81 m=s2) = = 4:941022C=m2: (1:601019C)
We don’t really need to write an integral, we just need the charge per unit length in the cylinder to be equal to zero. This means that the positive charge in cylinder must be +3:60nC=m. This positive charge is uniformly distributed in a circle of radius R = 1:50 cm, so E27-17
3:60nC=m 3:60nC=m = = = 5:09C=m3: R2 (0:0150 m)2
E27-18 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. The E~ eld will be perpendicular to the surface, so Gauss’ law will simplify to III q =0 = E~ dA~ = E dA = E dA = 4r2E: enc
(a) For point P1 the charge enclosed is qenc = 1:26107C, so (1:26107C) E = = 3:38106N=C: 4(8:851012C2=N m2)(1:83102m)2 (b) Inside a conductor E = 0.
The proton orbits with a speed v, so the centripetal force on the proton is FC = mv2=r. This centripetal force is from the electrostatic attraction with the sphere; so long as the proton is outside the sphere the electric eld is equivalent to that of a point charge Q (Eq. 27-15), E27-19
1Q E= : 40 r2 If q is the charge on the proton we can write F = Eq, or mv2 1 Q =q r 40 r2
Solving for Q, 40mv2r q 4(8:851012 C2=N m2)(1:671027kg)(294103m=s)2(0:0113 m) (1:601019C) = 1:13109C:
E27-20 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. The E~ eld will be perpendicular to the surface, so Gauss’ law will simplify to III q =0 = E~ dA~ = E dA = E dA = 4r2E: enc
(a) At r = 0:120 m qenc = 4:06108C. Then (4:06108C) E = = 2:54104N=C: 4(8:851012C2=N m2)(1:20101m)2 (b) At r = 0:220 m qenc = 5:99108C. Then
(5:99108C) E = = 1:11104N=C: 4(8:851012C2=N m2)(2:20101m)2 (c) At r = 0:0818 m qenc = 0 C. Then E = 0.
E27-21 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical shell. The E~ eld will be perpendicular to the curved surface and parallel to the end surfaces, so Gauss’ law will simplify to IZZ qenc=0 = E~ dA~ = E dA = E dA = 2rLE;
where L is the length of the cylinder. Note that = q=2rL represents a surface charge density. (a) r = 0:0410 m is between the two cylinders. Then (24:1106C=m2)(0:0322 m) E = = 2:14106N=C: (8:851012C2=N m2)(0:0410 m) (b) r = 0:0820 m is outside the two cylinders. Then (24:1106C=m2)(0:0322 m) + (18:0106C=m2)(0:0618 m) E = = 4:64105N=C: (8:851012C2=N m2)(0:0820 m) The negative sign is because it is pointing inward.
E27-22 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical shell. The E~ eld will be perpendicular to the curved surface and parallel to the end surfaces, so Gauss’ law will simplify to IZZ qenc=0 = E~ dA~ = E dA = E dA = 2rLE;
where L is the length of the cylinder. The charge enclosed is Z 2 2 qenc = dV = L r R
The electric eld is given by 2 2 2 2 L r R r R E= = : 20rL 20r At the surface, (2R)2 2 R 3R Es = = : 202R 40 Solve for r when E is half of this: 3R r2 R2 8 2r 0 = 4r2 3rR 4R2:
E27-23 The electric eld must do work on the electron to stop it. The electric eld is given by E = =20. The work done is W = F d = Eqd. d is the distance in question, so 20K 2(8:851012C2=N m2)(1:15105 eV) d = = = 0:979 m q (2:08106C=m2)e
E27-24 Let the spherical Gaussian surface have a radius of R and be centered on the origin. Choose the orientation of the axis so that the in nite line of charge is along the z axis. The electric eld is then directed radially outward from the z axis with magnitude E = =20, where is the perpendicular distance from the z axis. Now we want to evaluate I E = E~ dA~ ;
over the surface of the sphere. In spherical coordinates, dA = R2 sin d d , = R sin , and E~ dA~ = EA sin . Then I 2R E = sin R d d = : 20 0
(a) The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical shell. The E~ eld will be perpendicular to the curved surface and parallel to the end surfaces, so Gauss’ law will simplify to E27-25
IZZ where L is the length of the cylinder. Now for the qenc part. If the (uniform) volume charge density is , then the charge enclosed in the Gaussian cylinder is ZZ qenc = dV = dV = V = r2L:
Combining, r2L=0 = E2rL or E = r=20: (b) Outside the charged cylinder the charge enclosed in the Gaussian surface is just the charge in the cylinder. Then ZZ qenc = dV = dV = V = R2L:
and and then nally R2 E= : 20r (b) E = q=0 = (1:52104C)=(8:851012C2=N m2) = 1:72107N m2=C: (c) E = =0 = (8:13106C=m2)=(8:851012C2=N m2) = 9:19105N=C
E27-27 (a) = (2:4106C)=4(0:65 m)2 = 4:52107C=m2: (b) E = = = (4:52107C=m2)=(8:851012C2=N m2) = 5:11104N=C: 0
E27-29 (a) The near eld is given by Eq. 27-12, E = =20, so (6:0106C)=(8:0102 m)2 E = 5:3107N=C: 2(8:851012 C2=N m2) (b) Very far from any object a point charge approximation is valid. Then 1 q 1 (6:0106C) E = = = 60N=C: 40 r2 4(8:851012 C2=N m2) (30 m)2
P27-1 For a spherically symmetric mass distribution choose a spherical Gaussian shell. Then III ~g dA~ = g dA = g dA = 4r2g:
Then g gr2 4G G or Gm g= : r2 P27-2 (a) The ux through all surfaces except the right and left faces will be zero. Through the left face, p l = EyA = b aa2: Through the right face, p r = EyA = b 2aa2: The net ux is then pp = ba5=2( 2 1) = (8830 N=C m1=2)(0:130 m)5=2( 2 1) = 22:3 N m2=C: (b) The charge enclosed is q = (8:851012C2=N m2)(22:3 N m2=C) = 1:971010C.
The net force on the small sphere is zero; this force is the vector sum of the force of gravity W , the electric force FE , and the tension T .
P27-3 T F E W These forces are related by Eq = mg tan : We also have E = =20, so 20mg tan q 2(8:851012 C2=N m2)(1:12106kg)(9:81 m=s2) tan(27:4) (19:7109C) = 5:11109C=m2:
P27-4 The materials are conducting, so all charge will reside on the surfaces. The electric eld inside any conductor is zero. The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. The E~ eld will be perpendicular to the surface, so Gauss’ law will simplify to III q =0 = E~ dA~ = E dA = E dA = 4r2E: enc
(e) Since E = 0 inside the shell, qenc = 0, this requires that q reside on the inside surface. This is no charge on the outside surface.
P27-5 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical shell. The E~ eld will be perpendicular to the curved surface and parallel to the end surfaces, so Gauss’ law will simplify to IZZ qenc=0 = E~ dA~ = E dA = E dA = 2rLE;
(a) Outside the conducting shell qenc = +q 2q = q. Then E = q=20rL. The negative sign indicates that the eld is pointing inward toward the axis of the cylinder. (b) Since E = 0 inside the conducting shell, qenc = 0, which means a charge of q is on the inside surface of the shell. The remaining q must reside on the outside surface of the shell. (c) In the region between the cylinders qenc = +q. Then E = +q=20rL. The positive sign indicates that the eld is pointing outward from the axis of the cylinder.
P27-6 Subtract Eq. 26-19 from Eq. 26-20. Then z E= p : 20 z2 + R2
This problem is closely related to Ex. 27-25, except for the part concerning qenc. We’ll set up the problem the same way: the Gaussian surface will be a (imaginary) cylinder centered on the axis of the physical cylinder. For Gaussian surfaces of radius r < R, there is no charge enclosed We've already worked out the integral P27-7 Z tube for the cylinder, and then from Gauss' law, Z qenc = 0 E~ dA~ = 20rlE: tube (a) When r < R there is no enclosed charge, so the left hand vanishes and consequently E = 0 (b) When r > R there is a charge l enclosed, so
P27-8 This problem is closely related to Ex. 27-25, except for the part concerning qenc. We’ll set up the problem the same way: the Gaussian surface will be a (imaginary) cylinder centered on the axis of the physical cylinders. For Gaussian surfaces of radius r < a, there is no charge enclosed We've already worked out the integral Z E~ dA~ = 2rlE; tube for the cylinder, and then from Gauss' law, Z qenc = 0 E~ dA~ = 20rlE: tube (a) When r < a there is no enclosed charge, so the left hand vanishes and consequently E = 0 (b) When b > r > a there is a charge l enclosed, so
E= : 20r P27-9 Uniform circular orbits require a constant net force towards the center, so F = Eq = q=20r. The speed of the positron is given by F = mv2=r; the kinetic energy is K = mv2=2 = F r=2. Combining, q 40 (30109C=m)(1:61019C) 4((8:85 1012 C2=N m2) = 4:311017 J = 270 eV:
P27-10 = 20rE, so q = 2(8:851012C2=N m2)(0:014 m)(0:16 m)(2:9104N=C) = 3:6109C: P27-11 (a) Put a spherical Gaussian surface inside the shell centered on the point charge. Gauss’ law states I q E~ dA~ = enc : 0 Since there is spherical symmetry the electric eld is normal to the spherical Gaussian surface, and it is everywhere the same on this surface. The dot product simpli es to E~ dA~ = E dA, while since E is a constant on the surface we can pull it out of the integral, and we end up with I q 0 H2 where q is the point charge in the center. Now dA = 4r , where r is the radius of the Gaussian surface, so q E= : 40r2 (b) Repeat the above steps, except put the Gaussian surface outside the conducting shell. Keep it centered on the charge. Two things are di erent from the above derivation: (1) r is bigger, and
(2) there is an uncharged spherical conducting shell inside the Gaussian surface. Neither change will a ect the surface integral or qenc, so the electric eld outside the shell is still q 40r2 (c) This is a subtle question. With all the symmetry here it appears as if the shell has no e ect; the eld just looks like a point charge eld. If, however, the charge were moved o center the eld inside the shell would become distorted, and we wouldn’t be able to use Gauss’ law to nd it. So Outside the shell, however, we can’t tell what is going on inside the shell. So the electric eld outside the shell looks like a point charge eld originating from the center of the shell regardless of where inside the shell the point charge is placed! (d) Yes, q induces surface charges on the shell. There will be a charge q on the inside surface (f) No, as the electric eld from the outside charge won’t make it through a conducting shell. (g) No, this is not a contradiction, because the outside charge never experienced any electrostatic attraction or repulsion from the inside charge. The force is between the shell and the outside charge.
P27-12 The repulsive electrostatic forces must exactly balance the attractive gravitational forces. Then 1 q2 m2 40 r2 r2 p or m = q= 40G: Numerically, (1:601019C) m = q = 1:86109kg: 4(8:851012C2=N m2)(6:671011N m2=kg2)
P27-13 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. The E~ eld will be perpendicular to the surface, so Gauss’ law will simplify to III q =0 = E~ dA~ = E dA = E dA = 4r2E: enc
qenc = q + 4 r2dr, or R Zr qenc = q + 4 Ar dr = q + 2A(r2 a2): a
P27-14 (a) The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. The E~ eld will be perpendicular to the surface, so Gauss’ law will simplify to III qenc=0 = E~ dA~ = E dA = E dA = 4r2E:
(b) The electric eld in the hole is given by E~ h = E~ E~ b, where E~ is the eld from part (a) and E~ b is the eld that would be produced by the matter that would have been in the hole had the hole not been there. Then E~ b = ~b=30;
where ~b is a vector pointing from the center of the hole. Then ~r ~b E~ h = = (~r ~b): 30 30 30 But ~r ~b = ~a, so E~ h = ~a=30.
If a point is an equilibrium point then the electric eld at that point should be zero. If it is a stable point then moving the test charge (assumed positive) a small distance from the equilibrium point should result in a restoring force directed back toward the equilibrium point. In other words, there will be a point where the electric eld is zero, and around this point there will be an electric eld pointing inward. Applying Gauss’ law to a small surface surrounding our point P , we have a net inward ux, so there must be a negative charge inside the surface. But there should be nothing inside the surface except an empty point P , so we have a contradiction.
P27-15 P27-16 (a) Follow the example on Page 618. By symmetry E = 0 along the median plane. The charge enclosed between the median plane and a surface a distance x from the plane is q = Ax. Then E = Ax=0A = A=0: (b) Outside the slab the charge enclosed between the median plane and a surface a distance x from the plane is is q = Ad=2, regardless of x. The E = Ad=2=0A = d=20: P27-17 (a) The total charge is the volume integral over the whole sphere, Z Q = dV:
This is actually a three dimensional integral, and dV = A dr, where A = 4r2. Then Z Q = dV;
1 R4 = SR3: ZR r S2 0R 4S (b) Put a spherical Gaussian surface inside the sphere centered on the center. We can use Gauss’ law here because there is spherical symmetry in the entire problem, both inside and outside the Gaussian surface. Gauss’ law states I E~ dA~ = qenc : 0
Since there is spherical symmetry the electric eld is normal to the spherical Gaussian surface, and it is everywhere the same on this surface. The dot product simpli es to E~ dA~ = E dA, while since E is a constant on the surface we can pull it out of the integral, and we end up with I qenc 0 Now dA = 4r2, where r is the radius of the Gaussian surface, so H
qenc E= : 40r2 We aren’t done yet, because the charge enclosed depends on the radius of the Gaussian surface. We need to do part (a) again, except this time we don’t want to do the whole volume of the sphere, we only want to go out as far as the Gaussian surface. Then Z qenc = dV;
Zr r S2 0R 4S 1 4 R4 r4 = S : R Combine these last two results and S r4 40r2 R S r2 40 R Q r2 =: 40 R4 In the last line we used the results of part (a) to eliminate S from the expression.
P27-18 (a) Inside the conductor E = 0, so a Gaussian surface which is embedded in the conductor but containing the hole must have a net enclosed charge of zero. The cavity wall must then have a (b) The net charge on the conductor is +10:0 C; the charge on the outer surface must then be +13:0 C.
P27-19 (a) Inside the shell E = 0, so the net charge inside a Gaussian surface embedded in the (d) Yes.
Throughout this chapter we will use the convention that V (1) = 0 unless explicitly stated otherwise. Then the potential in the vicinity of a point charge will be given by Eq. 28-18, V = q=40r.
E28-1 (a) Let U12 be the potential energy of the interaction between the two \up” quarks. Then (2=3)2e(1:601019C) U12 = (8:99109N m2=C2) = 4:84105eV: (1:321015m) (b) Let U13 be the potential energy of the interaction between an \up” quark and a \down” quark. Then (1=3)(2=3)e(1:601019C) U13 = (8:99109N m2=C2) = 2:42105eV (1:321015m) Note that U13 = U23. The total electric potential energy is the sum of these three terms, or zero.
E28-2 There are six interaction terms, one for every charge pair. Number the charges clockwise from the upper left hand corner. Then p p U24 = q2=40( 2a): Add these terms and get 2 q2 q2
U = p 4 = 0:206 2 40a 0a The amount of work required is W = U .
(a) We build the electron one part at a time; each part has a charge q = e=3. Moving the rst part from in nity to the location where we want to construct the electron is easy and takes no work at all. Moving the second part in requires work to change the potential energy to 1 q1q2 40 r Bringing in the third part requires work against the force of repulsion between the third charge and both of the other two charges. Potential energy then exists in the form U13 and U23, where all three charges are the same, and all three separations are the same. Then U12 = U13 = U12, so the total potential energy of the system is 1 (e=3)2 3 (1:601019 C=3)2 U = 3 = = 2:721014 J 4 r 4(8:851012 C2=N m2) (2:821015 m) 0
(b) Dividing our answer by the speed of light squared to nd the mass, E28-3
2:72 1014 J m = = 3:02 1031 kg: (3:00 108 m=s)2
E28-4 There are three interaction terms, one for every charge pair. Number the charges from the left; let a = 0:146 m. Then (25:5109C)(17:2109C) 40a (25:5109C)(19:2109C) 40(a + x) (17:2109C)(19:2109C) U23 = : 40x Add these and set it equal to zero. Then (25:5)(17:2) (25:5)(19:2) (17:2)(19:2) a a+x x which has solution x = 1:405a = 0:205 m.
E28-5 The volume of the nuclear material is 4a3=3, where a = 8:01015m. Upon dividing in p half each part will have a radius r where 4r3=3 = 4a3=6. Consequently, r = a= 3 2 = 6:351015m. (a) The force of repulsion is (46)2(1:601019C)2 F = = 3000 N 4(8:851012C2=N m2)[2(6:351015m)]2
(b) The potential energy is (46)2e(1:601019C) U = = 2:4108eV 4(8:851012C2=N m2)2(6:351015m)
E28-6 This is a work/kinetic energy problem: 1 mv2 = qV . Then 20 s 2(1:601019C)(10:3103V) v0 = = 6:0107m=s: (9:111031kg)
(a) The energy released is equal to the charges times the potential through which the charge was moved. Then E28-7
U = qV = (30 C)(1:0 109 V) = 3:0 1010 J: (b) Although the problem mentions acceleration, we want to focus on energy. The energy will change the kinetic energy of the car from 0 to Kf = 3:0 1010 J. The speed of the car is then s r 2K 2(3:0 1010 J) v = = = 7100 m=s: m (1200 kg) (c) The energy required to melt ice is given by Q = mL, where L is the latent heat of fusion. Then Q (3:0 1010 J) m = = = 90; 100kg: L (3:33105J=kg)
E28-8 (a) U = (1:601019C)(1:23109V) = 1:971010J: E28-9 This is an energy conservation problem: 1 mv2 = qV ; V = q=40(1=r1 1=r2). Com- 2 bining, s q2 1 1
20m r1 r2 s (3:1106C)2 1 1
= ; 2(8:851012C2=N m2)(18106kg) (0:90103m) (2:5103m) = 2600 m=s:
E28-10 This is an energy conservation problem: 1 q2 1 m(2v)2 = mv2: 2 40r 2 Rearrange, q2 60mv2 (1:601019C)2 = = 1:42109m: 6(8:851012C2=N m2)(9:111031kg)(3:44105m=s)2)
E28-11 (a) V = (1:601019C)=4(8:851012C2=N m2)(5:291011m) = 27:2 V. (c) For uniform circular orbits F = mv2=r; the force is electrical, or F = e2=40r2. Kinetic energy is K = mv2=2 = F r=2, so e2 (1:601019C) K = = = 13:6 eV: 80r 8(8:851012C2=N m2)(5:291011m)
(d) The ionization energy is (K + U ), or Eion = [(13:6 eV) + (27:2 eV)] = 13:6 eV:
E28-12 (a) The electric potential at A is (5:0106 6 1 q1 q2 C) (2:010 C) V = + = (8:99109N m2=C) + = 6:0104V: A 40 r1 r2 (0:15 m) (0:05 m)
The electric potential at B is 1 q q (5:0106C) (2:0106C)
VB = 1 + 2 = (8:99109N m2=C) + = 7:8105V: 40 r2 r1 (0:05 m) (0:15 m)
E28-13 (a) The magnitude of the electric eld would be found from F (3:90 1015 N) E = = = 2:44 104 N=C: q (1:60 1019 C) (b) The potential di erence between the plates is found by evaluating Eq. 28-15,
Zb V = E~ d~s: a The electric eld between two parallel plates is uniform and perpendicular to the plates. Then E~ d~s = E ds along this path, and since E is uniform, Zb Zb Zb aaa
E28-14 V = Ex, so 20 2(8:851012C2=N m2) x = V = (48 V) = 7:1103m (0:12106C=m2)
E28-15 The electric eld around an in nitely long straight wire is given by E = =20r. The potential di erence between the inner wire and the outer cylinder is given by Zb V = (=20r) dr = (=20) ln(a=b): a
The electric eld near the surface of the wire is then given by V (855 V) E = = = = 1:32108V=m: 20a a ln(a=b) (6:70107m) ln(6:70107 m=1:05102 m) The electric eld near the surface of the cylinder is then given by V (855 V) E = = = = 8:43103V=m: 20a a ln(a=b) (1:05102m) ln(6:70107 m=1:05102 m) E28-16 V = Ex = (1:92105N=C)(1:50102m) = 2:88103V:
E28-17 (a) This is an energy conservation problem: 1 (2)(79)e2 (2)(79)e(1:601019C) K = = (8:99109N m2=C) = 3:2107 eV 40 r (7:01015m) (b) The alpha particles used by Rutherford never came close to hitting the gold nuclei.
E28-18 This is an energy conservation problem: mv2=2 = eq=40r, or s (1:601019C)(1:761015C) v = = 2:13104m=s 2(8:851012C2=N m2)(1:22102m)(9:111031kg)
E28-19 1 q 1 (1:16C) VA = = = 5060 V; 40 r 4(8:85 1012 C2=N m2) (2:06 m) and 1 q 1 (1:16C) VB = = = 8910 V; 40 r 4(8:85 1012 C2=N m2) (1:17 m) (b) The answer is the same, since when concerning ourselves with electric potential we only care about distances, and not directions.
E28-20 The number of \excess” electrons on each grain is 40rV 4(8:851012C2=N m)(1:0106m)(400 V) n = = = 2:8105 e (1:601019C) E28-21 The excess charge on the shuttle is q = 40rV = 4(8:851012C2=N m)(10 m)(1:0 V) = 1:1109C
E28-22 q = 1:37105C, so (1:37105C) V = (8:99109N m2=C2) = 1:93108V: (6:37106m) E28-23 The ratio of the electric potential to the electric eld strength is
V 1q 1q = = = r: E 40 r 40 r2
In this problem r is the radius of the Earth, so at the surface of the Earth the potential is V = Er = (100 V=m)(6:38106m) = 6:38108 V:
E28-24 Use Eq. 28-22: (1:47)(3:341030C m) V = (8:99109N m2=C2) = 1:63105V: (52:0109m)2 E28-25 (a) When nding VA we need to consider the contribution from both the positive and the negative charge, so 1 q VA = qa + 40 a + d There will be a similar expression for VB ,
1q VB = qa + : 40 a + d
1 q 1 q VA VB = qa + qa + ; 40 a + d 40 a + d q11 20 a a + d q a+d a 20 a(a + d) a(a + d) qd =: 20 a(a + d) (b) Does it do what we expect when d = 0? I expect it the di erence to go to zero as the two points A and B get closer together. The numerator will go to zero as d gets smaller. The denominator, however, stays nite, which is a good thing. So yes, Va VB ! 0 as d ! 0.
E28-26 (a) Since both charges are positive the electric potential from both charges will be positive. There will be no nite points where V = 0, since two positives can’t add to zero. (b) Between the charges the electric eld from each charge points toward the other, so E~ will pp vanish when q=x2 = 2q=(d x)2. This happens when d x = 2x, or x = d=(1 + 2).
p (a) V at C is 2(2:13106C) V = (8:99109N m2=C2) = 2:76106V (1:39102m) (c) Don’t forget about the potential energy of the original two charges! (2:13106C)2 U = (8:99109N m2=C2) = 2:08 J 0 (1:96102m) Add this to the answer from part (b) to get 7:35 J.
E28-28 The potential is given by Eq. 28-32; at the surface V s = R=20, half of this occurs when p 3R=4 = z:
We can nd the linear charge density by dividing the charge by the circumference,
Q 2R where Q refers to the charge on the ring. The work done to move a charge q from a point x to the origin will be given by E28-29
1Q1Q 4 R2 4 R2 + x2 00 qQ 1 1 = p : 4 R R2 + x2 0
Putting in the numbers, ! (5:931012C)(9:12109C) 1 1 = 1:861010J: 4(8:851012C2=N m2) 1:48m (1:48m)2 + (3:07m)2 p
E28-30 (a) The electric eld strength is greatest where the gradient of V is greatest. That is (b) The least absolute value occurs where the gradient is zero, which is between b and c and again between e and f .
E28-31 The potential on the positive plate is 2(5:52 V) = 11:0 V; the electric eld between the plates is E = (11:0 V)=(1:48102m) = 743 V=m.
E28-33 The radial potential gradient is just the magnitude of the radial component of the electric eld, @V Er = @r Then @V 1 q @r 40 r2 1 79(1:60 1019C) 4(8:85 1012 C2=N m2) (7:0 1015m)2 = 2:321021 V=m:
E28-34 Evaluate @V =@r, and Ze 1 r E= +2 : 40 r2 2R3
E28-35 Ex=@V=@x=2(1530V=m2)x.Atthepointinquestion,E=2(1530V=m2)(1:28 102m) = 39:2 V=m.
E28-36 Draw the wires so that they are perpendicular to the plane of the page; they will then \come out of” the page. The equipotential surfaces are then lines where they intersect the page, and they look like
E28-37 (a) jVB VAj = jW=qj = j(3:94 1019 J)=(1:60 1019 C)j = 2:46 V. The electric eld did work on the electron, so the electron was moving from a region of low potential to a region of (b) VC is at the same potential as VB (both points are on the same equipotential line), so (c) VC is at the same potential as VB (both points are on the same equipotential line), so VC VB = 0 V.
E28-38 (a) For point charges r = q=40V , so r = (8:99109N m2=C2)(1:5108C)=(30 V) = 4:5 m (b) No, since V / 1=r.
E28-39 The dotted lines are equipotential lines, the solid arrows are electric eld lines. Note that there are twice as many electric eld lines from the larger charge!
E28-41 This can easily be done with a spreadsheet. The following is a sketch; the electric eld is the bold curve, the potential is the thin curve.
sphere radius r
E28-42 Originally V = q=40r, where r is the radius of the smaller sphere. (a) Connecting the spheres will bring them to the same potential, or V1 = V2. (b) q1 + q2 = q; V1 = q1=40r and V2 = q2=402r; combining all of the above q2 = 2q1 and q1 = q=3 and q2 = 2q=3.
E28-43 (a) q = 4R2, so V = q=40R = R=0, or V = (1:601019C=m2)(6:37106m)=(8:851012C2=N m2) = 0:115 V (b) Pretend the Earth is a conductor, then E = =epsilon0, so
E = (1:601019C=m2)=(8:851012C2=N m2) = 1:81108V=m: E28-44 V = q=40R, so V = (8:99109N m2=C2)(15109C)=(0:16 m) = 850 V:
E28-45 (a) q = 40RV = 4(8:851012C2=N m2)(0:152 m)(215 V) = 3:63109C (b) = q=4R2 = (3:63109C)=4(0:152 m)2 = 1:25108C=m2.
(a) The total charge (Q = 57:2nC) will be divided up between the two spheres so that they are at the same potential. If q1 is the charge on one sphere, then q2 = Q q1 is the charge on the other. Consequently 1 q1 1 Q q1 40 r1 40 r2 Qr2 q1 = : r2 + r1 Putting in the numbers, we nd Qr1 (57:2 nC)(12:2 cm) q1 = = = 38:6 nC; r2 + r1 (5:88 cm) + (12:2 cm) (b) The potential on each sphere should be the same, so we only need to solve one. Then 1 q1 1 (38:6 nC) = = 2850 V: 4 r 4(8:85 1012 C2=N m2) (12:2 cm) 01 E28-47
(b) V = q=40r, so r = q=40V , and then r = (8:99109N m2=C2)(31:5109C)=(1:20103V) = 0:236 m: That is (0:236 m) (0:162 m) = 0:074 m above the surface.
E28-49 (a) Apply the point charge formula, but solve for the charge. Then 1q 40 r q = 4(8:85 1012 C2=N m2)(1 m)(106 V) = 0:11 mC:
Now that’s a fairly small charge. But if the radius were decreased by a factor of 100, so would the charge (1:10 C). Consequently, smaller metal balls can be raised to higher potentials with less (b) The electric eld near the surface of the ball is a function of the surface charge density, E = =0. But surface charge density depends on the area, and varies as r2. For a given potential, the electric eld near the surface would then be given by qV E= = = : 0 40r2 r Note that the electric eld grows as the ball gets smaller. This means that the break down eld is more likely to be exceeded with a low voltage small ball; you’ll get sparking.
E28-50 A \Volt” is a Joule per Coulomb. The power required by the drive belt is the product (3:41106V)(2:83103C=s) = 9650 W.
P28-1 (a) According to Newtonian mechanics we want K = 1 mv2 to be equal to W = qV 2 which means mv2 (0:511 MeV) V = = = 256 kV: 2q 2e (b) Let’s do some rearranging rst.
“# 1 1 2 K1 mc2 1 2 p K1 mc2 1 2 p 1p K +1 mc2 1 K 2 +1 mc2 and nally, s 1 = 1 K + 12 mc2 Putting in the numbers, v u1 t (256 keV) 2 +1 (511 keV) so v = 0:746c.
P28-2 (a) The potential of the hollow sphere is V = q=40r. The work required to increase the charge by an amount dq is dW = V =; dq. Integrating, e q e2 Z W = dq = : 0 40r 80r This corresponds to an electric potential energy of e(1:601019C) W = = 2:55105 eV = 4:081014J: 8(8:851012C2=N m2)(2:821015m) (b) This would be a mass of m = (4:081014J)=(3:00108m=s)2 = 4:531031kg.
P28-3 The negative charge is held in orbit by electrostatic attraction, or mv2 qQ =: r 40r2 The kinetic energy of the charge is 1 qQ K = mv2 = : 2 80r The electrostatic potential energy is qQ 40r so the total energy is qQ E= : 80r The work required to change orbit is then
qQ 1 1 W= : 80 r1 r2 R P28-4 (a) V = E dr, so r qr qr2 Z V = dr = : 4 R3 8 R3 00 0
(c) If instead of V = 0 at r = 0 as was done in part (a) we take V = 0 at r = 1, then V = q=40R on the surface of the sphere. The new expression for the potential inside the sphere will look like V = V 0 + V , where V 0 is the answer from part (a) and V is a constant so that the ss surface potential is correct. Then q qR2 3qR2 4 R 8 R3 8 R3 000
and then qr2 3qR2 q(3R2 r2) V= + = : 8 R3 8 R3 8 R3 000
P28-5 The total electric potential energy of the system is the sum of the three interaction pairs. One of these pairs does not change during the process, so it can be ignored when nding the change in potential energy. The change in electrical potential energy is then 2 2 2 qqq11 U = 2 2 = : 40rf 40ri 20 rf ri
In this case ri = 1:72 m, while rf = 0:86 m. The change in potential energy is then
11 U = 2(8:99109N m2=C2)(0:122 C)2 = 1:56108J (0:86 m) (1:72 m) The time required is t = (1:56108)=(831 W) = 1:87105s = 2:17 days:
P28-6 (a) Apply conservation of energy: qQ qQ 40d 40K (b) Apply conservation of energy: qQ 1 40(2d) 2 p so, combining with the results in part (a), v = K=m.
P28-7 (a) First apply Eq. 28-18, but solve for r. Then q (32:0 1012 C) r = = = 562 m: 40V 4(8:85 1012 C2=N m2)(512 V) (b) If two such drops join together the charge doubles, and the volume of water doubles, but the p radius of the new drop only increases by a factor of 3 2 = 1:26 because volume is proportional to The potential on the surface of the new drop will be 1 qnew 40 rnew 1 2qold 40 3 2 rold 1 qold = (2)2=3 = (2)2=3V old: 40 rold
P28-9 (a) The potential at any point will be the sum of the contribution from each charge, 1 q1 1 q2 40 r1 40 r2 where r1 is the distance the point in question from q1 and r2 is the distance the point in question from q2. Pick a point, call it (x; y). Since q1 is at the origin, p r1 = x2 + y2:
Since q2 is at (d; 0), where d = 9:60 nm, p r2 = (x d)2 + y2:
De ne the \Stanley Number” as S = 40V . Equipotential surfaces are also equi-Stanley surfaces. In particular, when V = 0, so does S. We can then write the potential expression in a sightly simpli ed form q1 q2 S= + : r1 r2 If S = 0 we can rearrange and square this expression.
q1 q2 r1 r2 r2 r2 q2 q2 12 x2 + y2 (x d)2 + y2 q2 q2 12
Let = q2=q1, then we can write x +y = ( 2 1)x2 + 2xd + ( 2 1)y2 = d2:
We complete the square for the ( 2 1)x2 + 2xd term by adding d2=( 2 1) to both sides of the equation. Then “# 2 d1 ( 2 1) x + + y2 = d2 1 + : 21 21 The center of the circle is at d (9:60 nm) = = 5:4 nm: 2 1 (10=6)2 1 (b) The radius of the circle is v u u 1+ 1 t 21 21 which can be simpli ed to j(10=6)j d = (9:6 nm) = 9:00 nm: 2 1 (10=6)2 1 (c) No.
P28-10 An annulus is composed of di erential rings of varying radii r and width dr; the charge on any ring is the product of the area of the ring, dA = 2r dr, and the surface charge density, or k 2k dq = dA = 2r dr = dr: r3 r2 The potential at the center can be found by adding up the contributions from each ring. Since we are at the center, the contributions will each be dV = dq=40r. Then b 1 k 2 a2 Z k dr k 1 b V= = := : 20 r3 40 a2 b2 40 b2a2 a
The total charge on the annulus is b 11 Z ba 2k Q = dr = 2k = 2k : a r2 a b ba
Combining, Q a+b V= : 80 ab q 1 1 1 40 r + d r r d q 1 1 40r 1 + d=r 1 d=r qdd 40r r r q 2d 1+ : 40r r P28-12 (a) Add the contributions from each di erential charge: dq = dy. Then y+L Z y+L V = dy = ln : y 40y 40 y
(b) Take the derivative: @V y L L Ey = = = : @y 40 y + L y2 40 y(y + L) (c) By symmetry it must be zero, since the system is invariant under rotations about the axis of the rod. Note that we can’t determine E? from derivatives because we don’t have the functional form of V for points o -axis! P28-13 (a) We follow the work done in Section 28-6 for a uniform line of charge, starting with Eq. 28-26, 1 dx 40 x2 + y2 ZL 1 kx dx 40 0 x2 + y2
kp L 40 0 kp = L2 + y2 y : 40 (b) The y component of the electric eld can be found from @V @y which (using a computer-aided math program) is ! ky Ey = 1 p : 40 L2 + y 2 (c) We could nd Ex if we knew the x variation of V . But we don’t; we only found the values of (d) We want to nd y such that the ratio
k k p L2 + y2 y = (L) 40 40 p is one-half. Simplifying, L2 + y2 y = L=2; which can be written as L2 + y2 = L2=4 + Ly + y2;
P28-14 The spheres are small compared to the separation distance. Assuming only one sphere at a potential of 1500 V, the charge would be q = 4 rV = 4(8:851012C2=N m)(0:150 m)(1500 V) = 2:50108C: 0
The potential from the sphere at a distance of 10.0 m would be (0:150 m) V = (1500 V) = 22:5 V: (10:0 m) This is small compared to 1500 V, so we will treat it as a perturbation. This means that we can assume that the spheres have charges of q = 4 rV = 4(8:851012C2=N m)(0:150 m)(1500 V + 22:5 V) = 2:54108C: 0
Calculating the fraction of excess electrons is the same as calculating the fraction of excess charge, so we’ll skip counting the electrons. This problem is e ectively the same as Exercise 28-47; we have a total charge that is divided between two unequal size spheres which are at the same potential on the surface. Using the result from that exercise we have Qr1 r2 + r1 where Q = 6:2 nC is the total charge available, and q1 is the charge left on the sphere. r1 is the radius of the small ball, r2 is the radius of Earth. Since the fraction of charge remaining is q1=Q, we can write q1 r1 r1 = = 2:0 108: Q r2 + r1 r2 P28-15
P28-16 The positive charge on the sphere would be q = 40rV = 4(8:851012C2=N m2)(1:08102m)(1000 V) = 1:20109C: The number of decays required to build up this charge is n = 2(1:20109C)=(1:601019C) = 1:501010:
The extra factor of two is because only half of the decays result in an increase in charge. The time required is t = (1:501010)=(3:70108s1) = 40:6 s:
P28-18 (a) Outside of an isolated charged spherical object E = q=40r2 and V = q=40r. Then E = V=r. Consequently, the sphere must have a radius larger than r = (9:15106V)=(100 (c) wv = (320106C=s), so (320106C=s) = = 2:00105C=m2: (0:485 m)(33:0 m=s)
(a) The charge which ows through a cross sectional surface area in a time t is given by q = it; where i is the current. For this exercise we have q = (4:82 A)(4:60 60 s) = 1330 C E29-1
E29-3 (a) j = nqv = (2:101014=m3)2(1:601019C)(1:40105m=s) = 9:41 A=m2: Since the ions have positive charge then the current density is in the same direction as the velocity. (b) We need an area to calculate the current.
(b) v = j=ne = (2:59105A=m2)=(8:491028=m3)(1:601019C) = 1:911015m=s: d
The current rating of a fuse of cross sectional area A would be imax = (440 A=cm2)A;
and if the fuse wire is cylindrical A = d2=4. Then E29-5 s 4 (0:552 A) d = = 4:00102 cm: (440 A=m2)
E29-6 Current density is current divided by cross section of wire, so the graph would look like:
4 I (A/mil^2 x10^-3) 3 2 1 50 100 150 200 d(mils)
E29-7 The current is in the direction of the motion of the positive charges. The magnitude of the current is i = (3:11018=s + 1:11018=s)(1:601019C) = 0:672 A:
E29-8 (a) The total current is i = (3:501015=s + 2:251015=s)(1:601019C) = 9:20104A: (b) The current density is j = (9:20104A)=(0:165103m)2 = 1:08104A=m2:
E29-9 (a) j = (8:70106=m3)(1:601019C)(470103m=s) = 6:54107A=m2: (b) i = (6:54107A=m2)(6:37106m)2 = 8:34107A:
E29-10 i = wv, so = (95:0106A)=(0:520 m)(28:0 m=s) = 6:52106C=m2:
The drift velocity is given by Eq. 29-6, j i (115 A) vd = = = = 2:71104m=s: ne Ane (31:2106m2)(8:491028=m3)(1:601019C) The time it takes for the electrons to get to the starter motor is x (0:855 m) t = = = 3:26103s: v (2:71104m=s) That’s about 54 minutes.
E29-11 E29-12 V = iR = (50103A)(1800 ) = 90 V: The resistance of an object with constant cross section is given by Eq. 29-13, L (11; 000 m) R = = (3:0 107 m) = 0:59 : A (0:0056 m2)
E29-14 The slope is approximately [(8:2 1:7)=1000] cm=C, so
1 = 6:5103 cm=C 4103=C 1:7 cm E29-15 (a) i = V =R = (23 V)=(15103 ) = 1500 A: (b) j = i=A = (1500 A)=(3:0103m)2 = 5:3107A=m2: (c) = RA=L = (15103 )(3:0103m)2=(4:0 m) = 1:1107 m. The material is possibly E29-13
E29-16 Use the equation from Exercise 29-17. R = 8 ; then T = (8 )=(50 )(4:3103=C) = 37 C: The nal temperature is then 57C.
E29-17 Start with Eq. 29-16, and multiply through by L=A, LL AA to get R R0 = R0 av(T T0): E29-18 The wire has a length L = (250)2(0:122 m) = 192 m. The diameter is 0.129 inches; the cross sectional area is then A = (0:129 0:0254 m)2=4 = 8:43106m2:
The resistance is R = L=A = (1:69108 m)(192 m)=(8:43106m2) = 0:385 : E29-19 If the length of each conductor is L and has resistivity , then L 4L RA = = D2=4 D2 and L 4L RB = = : (4D2=4 D2=4) 3D2 The ratio of the resistances is then RA = 3: RB
E29-20 R = R, so L =(d1=2)2 = L =(d =2)2. Simplifying, =d2 = =d2. Then 11 222 1122
d2 = (1:19103m)p(9:68108 m)=(1:69108 m) = 2:85103m: (b) V = iR = (6:00103A)(2:65106 ) = 1:59108V: (c) R = V=i = (1:59108V)=(750103A) = 2:12108 :
E29-22 Since V = iR, then if V and i are the same, then R must be the same. (a) Since R = R, 1L1=r2 = 2L2=r2, or =r2 = =r2. Then 1 21122 p riron=rcopper = (9:68108 m)(1:6910 m) = 2:39: 8
(b) Start with the de nition of current density: i V V j= = = : A RA L Since V and L is the same, but is di erent, then the current densities will be di erent.
Conductivity is given by Eq. 29-8, ~j = E~ . If the wire is long and thin, then the magnitude of the electric eld in the wire will be given by E V =L = (115 V)=(9:66 m) = 11:9 V=m: E29-23
We can now nd the conductivity, j (1:42104A=m2) = = = 1:19103( m)1: E (11:9 V=m)
E29-24 (a) vd = j=en = E=en: Then v = (2:701014= m)(120 V=m)=(1:601019C)(620106=m3 + 550106=m3) = 1:73102m=s: d
(b) j = E = (2:701014= m)(120 V=m) = 3:241014A=m2: E29-25 (a) R=L = =A, so j = i=A = (R=L)i=. For copper, j = (0:152103 =m)(62:3 A)=(1:69108 m) = 5:60105A=m2;
for aluminum, j = (0:152103 =m)(62:3 A)=(2:75108 m) = 3:44105A=m2: (b) A = L=R; if is density, then m = lA = l=(R=L). For copper, m = (1:0 m)(8960 kg=m3)(1:69108 m)=(0:152103 =m) = 0:996 kg;
for aluminum, m = (1:0 m)(2700 kg=m3)(2:75108 m)=(0:152103 =m) = 0:488 kg:
10 8 R (Kilo-ohms) 6 4 2 1 2 3 4 V(Volts)
E29-27 (a) The resistance is de ned as V (3:55 106 V=A2)i2 R = = = (3:55 106 V=A2)i: ii When i = 2:40 mA the resistance would be R = (3:55 106 V=A2)(2:40 103A) = 8:52 k : (b) Invert the above expression, and i = R=(3:55 106 V=A2) = (16:0 )=(3:55 106 V=A2) = 4:51 A:
E29-28 First, n = 3(6:021023)(2700 kg=m3)(27:0103kg) = 1:811029=m3. Then m (9:111031kg) = = = 7:151015s: ne2 (1:811029=m3)(1:601019C)2(2:75108 m)
E29-29 (a) E = E0=e = q=40eR2, so (1:00106C) E= = 4(8:851012C2=N m2)(4:7)(0:10 m)2
(b) E = E0 = q=40R2, so (1:00106C) E= = 4(8:851012C2=N m2)(0:10 m)2 (c) ind = 0(E0 E) = q(1 1=e)=4R2. Then (1:00106C) 1 ind = 1 = 6:23106C=m2: 4(0:10 m)2 (4:7)
E29-30 Midway between the charges E = q=0d, so q = (8:851012C2=N m2)(0:10 m)(3106V=m) = 8:3106C:
(a) At the surface of a conductor of radius R with charge Q the magnitude of the electric eld is given by 1 40 while the potential (assuming V = 0 at in nity) is given by
1 V = QR: 40 E29-31 The potential on the sphere that would result in \sparking” is V = ER = (3106N=C)R: (b) It is \easier” to get a spark o of a sphere with a smaller radius, because any potential on (c) The points of a lighting rod are like small hemispheres; the electric eld will be large near these points so that this will be the likely place for sparks to form and lightning bolts to strike.
If there is more current owing into the sphere than is owing out then there must be a change in the net charge on the sphere. The net current is the di erence, or 2 A. The potential on the surface of the sphere will be given by the point-charge expression, 1q 40 r and the charge will be related to the current by q = it. Combining, 1 it 40 r or 4 V r 4(8:85 1012 C2=N m2)(980 V)(0:13 m) 0 t = = = 7:1 ms: i (2 A) P29-1
P29-2 The net current density is in the direction of the positive charges, which is to the east. There are two electrons for every alpha particle, and each alpha particle has a charge equal in magnitude to two electrons. The current density is then = 1:0105C=m2:
P29-3 (a) The resistance of the segment of the wire is R = L=A = (1:69108 m)(4:0102m)=(2:6103m)2 = 3:18105 : The potential di erence across the segment is V = iR = (12 A)(3:18105 ) = 3:8104V: (c) The drift speed is v = j=en = i=Aen, so v = (12 A)=(2:6103m)2(1:61019C)(8:491028=m3) = 4:16105m=s: The electrons will move 1 cm in (1:0102m)=(4:16105m=s) = 240 s.
p (b) The speed of the particles in the beam is given by v = 2K=m, so p v = 2(22:4 MeV)=4(932 MeV=c2) = 0:110c: It takes (0:180 m)=(0:110)(3:00108m=s) = 5:45109s for the beam to travel 18.0 cm. The number of charges is then N = it=q = (250109A)(5:45109s)=(3:21019C) = 4260: (c) W = qV , so V = (22:4 MeV)=2e = 11:2 MV:
P29-5 (a) The time it takes to complete one turn is t = (250 m)=c. The total charge is q = it = (30:0 A)(950 m)=(3:00108m=s) = 9:50105C: (b) The number of charges is N = q=e, the total energy absorbed by the block is then U = (28:0109 eV)(9:50105C)=e = 2:66106J: This will raise the temperature of the block by T = U=mC = (2:66106J)=(43:5 kg)(385J=kgC) = 159 C: RR P29-6 (a) i = j dA = 2 jr dr;
Z i = 2 0Rj0(1 r=R)r dr = 2j0(R2=2 R3=3R) = j0R2=6: (b) Integrate, again: Z i = 2 0Rj0(r=R)r dr = 2j0(R3=3R) = j0R2=3:
P29-7 (a) Solve 20 = 0[1 + (T 20C)], or T = 20C + 1=(4:3103=C) = 250C: (b) Yes, ignoring changes in the physical dimensions of the resistor.
P29-8 The resistance when on is (2:90 V)=(0:310 A) = 9:35 . The temperature is given by T = 20C + (9:35 1:12 )=(1:12 )(4:5103=C) = 1650C:
Originally we have a resistance R1 made out of a wire of length l1 and cross sectional area A1. The volume of this wire is V1 = A1l1. When the wire is drawn out to the new length we have l2 = 3l1, but the volume of the wire should be constant so A2 = A1=3: The original resistance is l1 R1 = : A1 The new resistance is l2 3l1 A2 A1=3 or R2 = 54 .
P29-10 (a) i = (35:8 V)=(935 ) = 3:83102A: (d) E = (35:8 V)=(0:158 m) = 227 V=m: P29-9
P29-11 (a) = (1:09103 )(5:5103m)2=4(1:6 m) = 1:62108 m. This is possibly silver. (b) R = (1:62108 m)(1:35103m)4=(2:14102m)2 = 6:08108 :
P29-12 (a) L=L = 1:7105 for a temperature change of 1:0 C. Area changes are twice this, Take the di erential of RA = L: R dA+A dR = dL+L d, or dR = dL=A+L d=AR dA=A. For nite changes this can be written as R L A =+: RLA = = 4:3103. Since this term is so much larger than the other two it is the only signi cant e ect.
P29-13 We will use the results of Exercise 29-17, R R0 = R0 av(T T0):
To save on subscripts we will drop the \av” notation, and just specify whether it is carbon \c” or The disks will be e ectively in series, so we will add the resistances to get the total. Looking only at one disk pair, we have = R0;c + R0;i + (R0;c c + R0;i i) (T T0):
This last equation will only be constant if the coecient for the term (T T0) vanishes. Then
but R = L=A, and the disks have the same cross sectional area, so Lcc c + Lii i = 0;
or Lc i i (9:68108 m)(6:5103=C) = = = 0:036: Li c c (3500108 m)(0:50103=C)
P29-14 The current entering the cone is i. The current density as a function of distance x from the left end is then i j= : [a + x(b a)=L]2 The electric eld is given by E = j. The potential di erence between the ends is then ZL ZL i iL V = E dx = dx = 0 0 [a + x(b a)=L]2 ab
P29-15 The current is found from Eq. 29-5, Z where the region of integration is over a spherical shell concentric with the two conducting shells but between them. The current density is given by Eq. 29-10, ~j = E~ =;
and we will have an electric eld which is perpendicular to the spherical shell. Consequently, ZZ 11 i = E~ dA~ = E dA
By symmetry we expect the electric eld to have the same magnitude anywhere on a spherical shell which is concentric with the two conducting shells, so we can bring it out of the integral sign, and then Z2 1 4r E i = E dA = ;
where E is the magnitude of the electric eld on the shell, which has radius r such that b > r > a. The above expression can be inverted to give the electric eld as a function of radial distance, since the current is a constant in the above expression. Then E = i=4r2 The potential is given by Za V = E~ d~s;
b we will integrate along a radial line, which is parallel to the electric eld, so Za V = E dr;
b a i Z b 4r2 i a dr Z 4 b r i 1 1 = : 4 a b We divide this expression by the current to get the resistance. Then
11 R= 4 a b P29-16 Since =p=vd, p / vd. For an ideal gas the kinetic energy is proportional to the temperature, so / K / T .
We apply Eq. 30-1, E30-2 (a)C=V=q=(73:01012C)=(19:2V)=3:801012F: (b) The capacitance doesn’t change! (c) V = q=C = (2101012C)=(3:801012F) = 55:3 V: E30-3 q = CV = (26:0106F)(125 V) = 3:25103C: E30-4 (a) C = 0A=d = (8:851012F=m)(8:22102m)2=(1:3110 m) = 1:4310 F: 3 10 (b) q = CV = (1:431010F)(116 V) = 1:66108C.
Eq. 30-11 gives the capacitance of a cylinder, L (0:0238 m) C = 20 = 2(8:851012 F=m) = 5:4610 F: 13 ln(b=a) ln((9:15mm)=(0:81mm))
E30-6 (a) A = Cd=0 = (9:701012F)(1:20103m)=(8:851012F=m) = 1:32103m2: 0 (c) V = q0=C = [V ]0C0=C = [V ]0d=d0. Using this formula, the new potential di erence would be [V ]0 = (13:0 V)(1:10103m)=(1:20103m) = 11:9 V: The potential energy has changed by (11:9 V) (30:0 V) = 1:1 V.
E30-7 (a) From Eq. 30-8, (0:040 m)(0:038 m) C = 4(8:851012F=m) = 8:451011F: (0:040 m) (0:038 m) (b) A = Cd= = (8:451011F)(2:00103m)=(8:851012F=m) = 1:91102m2: 0
E30-8 Let a = b + d, where d is the small separation between the shells. Then ab (b + d)b ab d b2 40 = 0A=d: d
E30-1 E30-5 E30-9 The charge on each of the capacitors is then q = CV = (1:00 106 F)(110 V) = 1:10 104 C: If there are N capacitors, then the total charge will be N q, and we want this total charge to be 1:00 C. Then (1:00 C) (1:00 C) N = = = 9090: q (1:10 104 C)
E30-10 First nd the equivalent capacitance of the parallel part: C = C + C = (10:3106F) + (4:80106F) = 15:1106F: eq 1 2 Then nd the equivalent capacitance of the series part: 111 = + = 3:23105F1: C (15:1106F) (3:90106F) eq Then the equivalent capacitance of the entire arrangement is 3:10106F.
E30-11 First nd the equivalent capacitance of the series part: 111 = + = 3:05105F1: C (10:3106F) (4:80106F) eq The equivalent capacitance is 3:28106F. Then nd the equivalent capacitance of the parallel part: Ceq = C1 + C2 = (3:28106F) + (3:90106F) = 7:18106F: This is the equivalent capacitance for the entire arrangement.
E30-12 For one capacitor q = CV = (25:0106F)(4200 V) = 0:105 C. There are three capaci- tors, so the total charge to pass through the ammeter is 0:315 C.
E30-13 (a) The equivalent capacitance is given by Eq. 30-21, 111115 =+= + = Ceq C1 C2 (4:0F) (6:0F) (12:0F) (b) The charge on the equivalent capacitor is q = CV = (2:40F)(200 V) = 0:480 mC. For series capacitors, the charge on the equivalent capacitor is the same as the charge on each of the capacitors. This statement is wrong in the Student Solutions! (c) The potential di erence across the equivalent capacitor is not the same as the potential di erence across each of the individual capacitors. We need to apply q = CV to each capacitor using the charge from part (b). Then for the 4:0F capacitor, q (0:480 mC) C (4:0F) and for the 6:0F capacitor, q (0:480 mC) V = = = 80 V: C (6:0F) Note that the sum of the potential di erences across each of the capacitors is equal to the potential di erence across the equivalent capacitor.
E30-14 (a) The equivalent capacitance is Ceq = C1 + C2 = (4:0F) + (6:0F) = (10:0F): (c) For parallel capacitors, the potential di erence across the equivalent capacitor is the same as (b) For the 4:0F capacitor, q = CV = (4:0F)(200 V) = 8:0104 C;
and for the 6:0F capacitor, q = CV = (6:0F)(200 V) = 12:0104 C:
0A 0A d deq = = = : Ceq 3C 3
0A 0A deq = = = 3d: Ceq C=3 E30-16 (a) The maximum potential across any individual capacitor is 200 V; so there must be at least (1000 V)=(200 V) = 5 series capacitors in any parallel branch. This branch would have an equivalent capacitance of C = C=5 = (2:0106F)=5 = 0:40106F: eq (b) For parallel branches we add, which means we need (1:2106F)=(0:40106F) = 3 parallel branches of the combination found in part (a).
E30-17 Look back at the solution to Ex. 30-10. If C3 breaks down electrically then the circuit is (a) q = (10:3106F)(115 V) = 1:18103C: 1
E30-18 The 108F capacitor originally has a charge of q = (108106F)(52:4 V) = 5:66103C. After it is connected to the second capacitor the 108F capacitor has a charge of q = (108 106F)(35:8 V) = 3:87103C. The di erence in charge must reside on the second capacitor, so the capacitance is C = (1:79103C)=(35:8 V) = 5:00105F:
Consider any junction other than A or B. Call this junction point 0; label the four nearest junctions to this as points 1, 2, 3, and 4. The charge on the capacitor that links point 0 to point 1 is q1 = CV01; where V01 is the potential di erence across the capacitor, so V01 = V0 V1; where V0 is the potential at the junction 0, and V1 is the potential at the junction 1. Similar expressions For the junction 0 the net charge must be zero; there is no way for charge to cross the plates of the capacitors. Then q1 + q2 + q3 + q4 = 0; and this means E30-19
CV01 + CV02 + CV03 + CV04 = 0 or V01 + V02 + V03 + V04 = 0: Let V0i = V0 Vi, and then rearrange,
or 1 V0 = (V1 + V2 + V3 + V4) : 4 E30-20 U = uV = 0E2V=2, where V is the volume. Then
E30-21 The total capacitance is (2100)(5:0106F) = 1:05102F: The total energy stored is 11 U = C(V )2 = (1:05102F)(55103V)2 = 1:59107 J: 22 The cost is $0:03 (1:59107J) = $0:133: 3600103J
22 (b) (3:05106J)=(3600103J=kW h) = 0:847kW h: E30-23 (a) The capacitance of an air lled parallel-plate capacitor is given by Eq. 30-5, 0A (8:851012F=m)(42:0 104m2) C = = = 2:861011 F: d (1:30 103m) (b) The magnitude of the charge on each plate is given by q = CV = (2:861011 F)(625 V) = 1:79108 C: (c) The stored energy in a capacitor is given by Eq. 30-25, regardless of the type or shape of the capacitor, so 11 U = C(V )2 = (2:861011 F)(625 V)2 = 5:59 J: 22 (d) Assuming a parallel plate arrangement with no fringing e ects, the magnitude of the electric eld between the plates is given by Ed = V , where d is the separation between the plates. Then E=V=d=(625V)=(0:00130m)=4:81105V=m: (e) The energy density is Eq. 30-28, 11 u = 0E2 = ((8:851012F=m))(4:81105 V=m)2 = 1:02 J=m3: 22 E30-24 The equivalent capacitance is given by 1=Ceq = 1=(2:12106F) + 1=(3:88106F) = 1=(1:37106F):
The energy stored is U = 1 (1:37106F)(328 V)2 = 7:37102J: 2 E30-25 V=r = q=40r2 = E, so that if V is the potential of the sphere then E = V=r is the electric eld on the surface. Then the energy density of the electric eld near the surface is 12 2 1 (8:8510 F=m) (8150 V) u = 0E2 = = 7:41102J=m3: 2 2 (0:063 m)
E30-26 The charge on C3 can be found from considering the equivalent capacitance. q3 = (3:10 106F)(112 V) = 3:4710 C: The potential across C3 is given by [V ]3 = (3:4710 C)=(3:90 4 4 The potential across the parallel segment is then (112 V) (89:0 V) = 23:0 V. So [V ]1 = Then q1 = (10:3106F)(23:0 V) = 2:37104C and q2 = (4:80106F)(23:0 V) = 1:10104C:.
There is enough work on this problem without deriving once again the electric eld between charged cylinders. I will instead refer you back to Section 26-4, and state 1q 20 Lr E30-27
where q is the magnitude of the charge on a cylinder and L is the length of the cylinders. The energy density as a function of radial distance is found from Eq. 30-28, 1 1 q2 u = 0E2 = 2 82 L2r2 0
The total energy stored in the electric eld is given by Eq. 30-24, 1 q2 q2 ln(b=a) 2 C 2 20L where we substituted into the last part Eq. 30-11, the capacitance of a cylindrical capacitor. p We want to show that integrating a volume integral from r = a to r = ab over the energy density function will yield U=2. Since we want to do this problem the hard way, we will pretend we don’t know the answer, and integrate from r = a to r = c, and then nd out what c is. Then Z 1 2 c 2 L 1 q2 ZZZ 82 L2r2 a00 0 q2 c 2 L dr ZZZ 82 L2 r 0a00 2 Zc q dr 40L a r q2 c = ln : 40L a Now we equate this to the value for U that we found above, and we solve for c.
1 q2 ln(b=a) q2 c 2 2 20L 40L a p ab = c:
E30-28 (a) d = 0A=C, or d = (8:851012F=m)(0:350 m2)=(51:31012F) = 6:04103m: (b) C = (5:60)(51:31012 F) = 2:871010F.
E30-29 Originally, C1 = 0A=d1. After the changes, C2 = 0A=d2. Dividing C2 by C1 yields C2=C1 = d1=d2, so = d C2=d C1 = (2)(2:571012F)=(1:321012F) = 3:89: 21
E30-30 TherequiredcapacitanceisfoundfromU=1C(V)2,or 2 C = 2(6:61106J)=(630 V)2 = 3:331011F: The dielectric constant required is = (3:331011F)=(7:401012F) = 4:50. Try transformer oil.
Capacitance with dielectric media is given by Eq. 30-31, e0A C= : d The various sheets have di erent dielectric constants and di erent thicknesses, and we want to maximize C, which means maximizing e=d. For mica this ratio is 54 mm1, for glass this ratio is 35 mm1, and for paran this ratio is 0.20 mm1. Mica wins.
E30-31 E30-32 The minimum plate separation is given by d = (4:13103V)=(18:2106V=m) = 2:27104m: The minimum plate area is then dC (2:27104m)(68:4109F) A = = = 0:627 m2: 0 (2:80)(8:851012F=m)
E30-33 The capacitance of a cylindrical capacitor is given by Eq. 30-11, 1:0103m C = 2(8:851012F=m)(2:6) = 8:63108F: ln(0:588=0:11)
E30-34 (a) U = C0(V )2=2, C0 = e0A=d, and V=d is less than or equal to the dielectric strength (which we will call S). Then V = Sd and 1 2 so the volume is given by V = 2U=e0S2:
This quantity is a minimum for mica, so V = 2(250103J)=(5:4)(8:851012F=m)(160106V=m)2 = 0:41 m3: (b) e = 2U=V 0S2, so e = 2(250103J)=(0:087m3)(8:851012F=m)(160106V=m)2 = 25:
(a) The capacitance of a cylindrical capacitor is given by Eq. 30-11,
L C = 20e : ln(b=a) The factor of e is introduced because there is now a dielectric (the Pyrex drinking glass) between the plates. We can look back to Table 29-2 to get the dielectric properties of Pyrex. The capacitance of our \glass” is then E30-35
(0:15 m) C = 2(8:851012F=m)(4:7) = 7:31010 F: ln((3:8 cm)=(3:6 cm) (b) The breakdown potential is (14 kV/mm)(2 mm) = 28 kV.
e (a) Insert the slab so that it is a distance a above the lower plate. Then the distance between the slab and the upper plate is d a b. Inserting the slab has the same e ect as having two capacitors wired in series; the separation of the bottom capacitor is a, while that of the top The bottom capacitor has a capacitance of C1 = 0A=a; while the top capacitor has a capacitance of C2 = 0A=(d a b): Adding these in series, E30-37
111 Ceq C1 C2 a dab 0A 0A db =: 0A So the capacitance of the system after putting the copper slab in is C = 0A=(d b): (b) The energy stored in the system before the slab is inserted is q2 q2 d Ui = = 2Ci 2 0A while the energy stored after the slab is inserted is q2 q2 d b Uf = = 2Cf 2 0A The ratio is U i=U f = d=(d b): (c) Since there was more energy before the slab was inserted, then the slab must have gone in willingly, it was pulled in!. To get the slab back out we will need to do work on the slab equal to q2 d q2 d b q2 b Ui Uf = = : 2 0A 2 0A 2 0A
E30-38 (a) Insert the slab so that it is a distance a above the lower plate. Then the distance between the slab and the upper plate is d a b. Inserting the slab has the same e ect as having two capacitors wired in series; the separation of the bottom capacitor is a, while that of the top The bottom capacitor has a capacitance of C1 = 0A=a; while the top capacitor has a capacitance of C2 = 0A=(d a b): Adding these in series,
111 Ceq C1 C2 a dab 0A 0A db =: 0A So the capacitance of the system after putting the copper slab in is C = 0A=(d b):
(b) The energy stored in the system before the slab is inserted is C (V )2 (V )2 A i0 Ui = = 2 2d while the energy stored after the slab is inserted is C (V )2 (V )2 A f0 Uf = = 2 2 db The ratio is U i=U f = (d b)=d: (c) Since there was more energy after the slab was inserted, then the slab must not have gone in willingly, it was being repelled!. To get the slab in we will need to do work on the slab equal to the energy di erence.
(V )2 A (V )2 A (V )2 Ab 000 Uf Ui = = : 2 d b 2 d 2 d(d b)
(a) E = V=d = CV=e0A, or (1121012F)(55:0 V) E = = 13400 V=m: (5:4)(8:851012F=m)(96:5104m2) (c) Q0 = Q(1 1=e) = (6:16109C)(1 1=(5:4)) = 5:02109C.
E30-40 (a) E = q=e0A, so (890109C) e = = 6:53 (1:40106V=m)(8:851012F=m)(110104m2) (b) q0 = q(1 1= ) = (890109C)(1 1=(6:53)) = 7:54107C: e
P30-1 The capacitance of the cylindrical capacitor is from Eq. 30-11, 20L C= : ln(b=a) If the cylinders are very close together we can write b = a + d, where d, the separation between the cylinders, is a small number, so 20L 20L C= = : ln ((a + d)=a) ln (1 + d=a) Expanding according to the hint, 20L 2a0L C = d=a d Now 2a is the circumference of the cylinder, and L is the length, so 2aL is the area of a cylindrical plate. Hence, for small separation between the cylinders we have 0A d which is the expression for the parallel plates.
P30-2 (a) C = 0A=x; take the derivative and dC 0 dA 0A dx dT x dT x2 dT 1 dA 1 dx =C : A dT x dT (b) Since (1=A)dA=dT = 2 a and (1=x)dx=dT = s, we need
= 2 = 2(23106=C) = 46106=C: sa Insert the slab so that it is a distance d above the lower plate. Then the distance between the slab and the upper plate is abd. Inserting the slab has the same e ect as having two capacitors wired in series; the separation of the bottom capacitor is d, while that of the top capacitor is abd. The bottom capacitor has a capacitance of C1 = 0A=d; while the top capacitor has a capacitance of C2 = 0A=(a b d): Adding these in series, 111 Ceq C1 C2 d abd 0A 0A ab =: 0A So the capacitance of the system after putting the slab in is C = 0A=(a b): P30-3
P30-4 The potential di erence between any two adjacent plates is V . Each interior plate has a charge q on each surface; the exterior plate (one pink, one gray) has a charge of q on the interior The capacitance of one pink/gray plate pair is C = 0A=d. There are n plates, but only n 1 plate pairs, so the total charge is (n 1)q. This means the total capacitance is C = 0(n 1)A=d.
P30-5 As far as point e is concerned point a looks like it is originally positively charged, and point d is originally negatively charged. It is then convenient to de ne the charges on the capacitors in terms of the charges on the top sides, so the original charge on C1 is q1;i = C1V0 while the original charge on C2 is q2;i = C2V0. Note the negative sign re ecting the opposite polarity of C2. (a) Conservation of charge requires q1;i + q2;i = q1;f + q2;f ;
but since q = CV and the two capacitors will be at the same potential after the switches are closed we can write C1 C2 V0 = V: C1 + C2
With numbers, (1:16 F) (3:22 F) V = (96:6 V) = 45:4 V: (1:16 F) + (3:22 F)
The negative sign means that the top sides of both capacitor will be negatively charged after the (b) The charge on C1 is C1V = (1:16 F)(45:4 V) = 52:7C: (c) The charge on C2 is C2V = (3:22 F)(45:4 V) = 146C:
P30-6 C2 and C3 form an e ective capacitor with equivalent capacitance Ca = C2C3=(C2 + C3). The charge on C1 is originally q0 = C1V0. After throwing the switch the potential across C1 is given by q1 = C1V1. The same potential is across Ca; q2 = q3, so q2 = CaV1. Charge is conserved, so q1 + q2 = q0. Combining some of the above, q0 C1 C1 + Ca C1 + Ca and then C2 C2(C2 + C3) q1 = 1 V0 = 1 V0: C1 + Ca C1C2 + C1C3 + C2C3 Similarly, 1 CaC1 1 1 1 q2 = V0 = + + V0: C1 + Ca C1 C2 C3 q3 = q2 because they are in series.
(a) If terminal a is more positive than terminal b then current can ow that will charge the capacitor on the left, the current can ow through the diode on the top, and the current can charge the capacitor on the right. Current will not ow through the diode on the left. The capacitors are Since the capacitors are identical and series capacitors have the same charge, we expect the capacitors to have the same potential di erence across them. But the total potential di erence across both capacitors is equal to 100 V, so the potential di erence across either capacitor is 50 V. The output pins are connected to the capacitor on the right, so the potential di erence across (b) If terminal b is more positive than terminal a the current can ow through the diode on the left. If we assume the diode is resistanceless in this con guration then the potential di erence across it will be zero. The net result is that the potential di erence across the output pins is 0 V. In real life the potential di erence across the diode would not be zero, even if forward biased. It will be somewhere around 0.5 Volts.
P30-7 P30-8 Divide the strip of width a into N segments, each of width x = a=N . The capacitance of each strip is C = 0ax=y. If is small then 111d = 1 x=d): y d + x sin d + x ( Since parallel capacitances add, 0a a a2 Z X 0 a C = C = (1 x=d)dx = 1 : d 0 d 2d
P30-9 (a) When S2 is open the circuit acts as two parallel capacitors. The branch on the left has an e ective capacitance given by 1111 Cl (1:010 F) (3:0106F) 7:510 F 6 7
while the branch on the right has an e ective capacitance given by 1111 =+=: Cl (2:0106F) (4:0106F) 1:3310 F 6
The charge on either capacitor in the branch on the left is while the charge on either capacitor in the branch on the right is q = (1:33106F)(12 V) = 1:6105C: (b) After closing S2 the circuit is e ectively two capacitors in series. The top part has an e ective capacitance of 6
while the e ective capacitance of the bottom part is C = (3:0106F) + (4:0106F) = (7:0106F): b
The e ective capacitance of the series combination is given by 1111 =+=: C (3:0106F) (7:0106F) 2:1106F eq
The charge on each part is q = (2:1106F)(12 V) = 2:52105C. The potential di erence across the top part is t
and then the charge on the top two capacitors is q1 = (1:0 106F)(8:4 V) = 8:4 106C and q = (2:0106F)(8:4 V) = 1:68105C. The potential di erence across the bottom part is 2
t and then the charge on the top two capacitors is q1 = (3:0 106F)(3:6 V) = 1:08 105C and 2
P30-10 Let V = Vxy. By symmetry V2 = 0 and V1 = V4 = V5 = V3 = V=2. Suddenly the problem is very easy. The charges on each capacitor is q1, except for q2 = 0. Then the equivalent capacitance of the circuit is q q1 + q4 Ceq = = = C1 = 4:0106F: V 2V1
P30-11 (a) The charge on the capacitor with stored energy U0 = 4:0 J is q0, where
q2 U0 = 0 : 2C When this capacitor is connected to an identical uncharged capacitor the charge is shared equally, so that the charge on either capacitor is now q = q0=2. The stored energy in one capacitor is then
q2 q2=4 1 U = = 0 = U0: 2C 2C 4 But there are two capacitors, so the total energy stored is 2U = U0=2 = 2:0 J. (b) Good question. Current had to ow through the connecting wires to get the charge from one capacitor to the other. Originally the second capacitor was uncharged, so the potential di erence across that capacitor would have been zero, which means the potential di erence across the con- necting wires would have been equal to that of the rst capacitor, and there would then have been energy dissipation in the wires according to P = i2R: That’s where the missing energy went.
P30-12 R = L=A and C = 0A=L. Combining, R = 0=C, or R = (9:40 m)(8:851012F=m)=(1101012F) = 0:756 :
2 R (b) U = u dV where dV Then Z 1 e2 e2 1 U = 4pi r2 dr = : R 3220r4 80 R
(c) R = e2=80mc2, or (1:601019C)2 R = = 1:401015m: 8(8:851012F=m)(9:111031kg)(3:00108m=s)2
2 P30-15 According to Problem 14, the force on a plate of a parallel plate capacitor is q2 F= : 20A The force per unit area is then F q2 2 A 20A2 20 where = q=A is the surface charge density. But we know that the electric eld near the surface of a conductor is given by E = =0, so F1 = 0E2: A2
P30-16 A small surface area element dA carries a charge dq = q dA=4R2. There are three forces on the elements which balance, so p(V0=V )dA + q dq=40R2 = p dA;
or pR3 + q2=162 R = pR3: 00 This can be rearranged as q2 = 162 pR(R3 R3): 00
P30-17 The magnitude of the electric eld in the cylindrical region is given by E = =20r; where is the linear charge density on the anode. The potential di erence is given by V = (=20) ln(b=a); where a is the radius of the anode b the radius of the cathode. Combining, E = V =r ln(b=a), this will be a maximum when r = a, so V = (0:180103m) ln[(11:0103m)=(0:180103m)](2:20106V=m) = 1630 V:
P30-18 This is e ectively two capacitors in parallel, each with an area of A=2. Then
0A=2 0A=2 0A e1 + e2 Ceq = e1 + e2 = : ddd2
We will treat the system as two capacitors in series by pretending there is an in nitesi- mally thin conductor between them. The slabs are (I assume) the same thickness. The capacitance of one of the slabs is then given by Eq. 30-31, e10A d=2 P30-19
where d=2 is the thickness of the slab. There would be a similar expression for the other slab. The equivalent series capacitance would be given by Eq. 30-21, 111 Ceq C1 C2 d=2 d=2 e10A e20A d e2 + e1 20A e1e2 20A e1e2 Ceq = : d e2 + e1 P30-20 Treat this as three capacitors. Find the equivalent capacitance of the series combination on the right, and then add on the parallel part on the left. The right hand side is
1 d d 2d e2 + e3 =+=: Ceq e20A=2 e30A=2 0A e2e3 Add this to the left hand side, and
e10A=2 0A e2e3 2d 2d e2 + e3 0A e1 e2e3 =+: 2d 2 e2 + e3
00 (c) W = U 0 U = 2U U = U = 0A(V )2=2d: (a) In the air gap we have 0E2V (8:851012F=m)(6:9103V=m)2(1:15102m2)(4:6103m) Ua = = = 1:11108J: 0 22 (b) The remaining 40% is in the slab.
P30-23 (a) C = 0A=d = (8:851012F=m)(0:118 m2)=(1:2210 m) = 8:5610 F. 2 11 (b) Use the results of Problem 30-24.