Instructor Solutions Manual for
Physics by
Halliday, Resnick, and Krane
Paul Stanley Beloit College
Volume 2
A Note To The Instructor...
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 server access must be restricted to your students.
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 if viewed by a novice.
There are some traditional formula, such as
v2x= v0x2 + 2axx,
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 you to the previous solution.
I adopt a different approach for rounding of significant figures than previous authors; in partic- ular, I usually round intermediate answers. As such, some of my answers will differ from those in the back of the book.
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.
Paul Stanley Beloit College [email protected]
E25-1 The charge transferred is
Q = (2.5× 104C/s)(20× 10−6s) = 5.0× 10−1C.
E25-2 Use Eq. 25-4:
r = s
(8.99×109N·m2/C2)(26.3×10−6C)(47.1×10−6C)
(5.66 N) = 1.40 m
E25-3 Use Eq. 25-4:
F = (8.99×109N·m2/C2)(3.12×10−6C)(1.48×10−6C)
(0.123 m)2 = 2.74 N.
E25-4 (a) The forces are equal, so m1a1= m2a2, or
m2= (6.31×10−7kg)(7.22 m/s2)/(9.16 m/s2) = 4.97×10−7kg.
(b) Use Eq. 25-4:
q = s
(6.31×10−7kg)(7.22 m/s2)(3.20×10−3m)2
(8.99×109N·m2/C2) = 7.20×10−11C
E25-5 (a) Use Eq. 25-4, F = 1
4π0 q1q2
r122 = 1
4π(8.85×10−12C2/N· m2)
(21.3 µC)(21.3 µC)
(1.52 m)2 = 1.77 N
(b) In part (a) we found F12; to solve part (b) we need to first find F13. Since q3 = q2 and r13= r12, we can immediately conclude that F13= F12.
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 12 F23
F23
F 12 θ
F net
From this diagram we want to find the magnitude of the net force on q1. The cosine law is appropriate here:
Fnet2 = F122 + F132 − 2F12F13cos θ,
= (1.77 N)2+ (1.77 N)2− 2(1.77 N)(1.77 N) cos(120◦),
= 9.40 N2, Fnet = 3.07 N.
E25-6 Originally F0 = CQ20 = 0.088 N, where C is a constant. When sphere 3 touches 1 the 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)CQ20= (3/8)F0= 0.033 N.
E25-7 The forces on q3are ~F31and ~F32. These forces are given by the vector form of Coulomb’s Law, Eq. 25-5,
~F31 = 1 4π0
q3q1
r231 ˆr31= 1 4π0
q3q1 (2d)2ˆr31,
~F32 = 1 4π0
q3q2
r232 ˆr32= 1 4π0
q3q2 (d)2ˆr32.
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,
F~31 = −~F32, 1
4π0
q3q1
(2d)2ˆr31 = − 1 4π0
q3q2
(d)2ˆr32, q1
4ˆr31 = −q2
1ˆr32.
Note that ˆr31and ˆr32 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 find. The contributions from the upper left charge require slightly more work.
The diagonal distance is√
2a; the components will be weighted by cos 45◦ =√
2/2. The diagonal charge will contribute
Fx = 1
4π0
(q)(2q) (√
2a)2
√2 2 ˆi =
√2 8π0
q2 a2ˆi,
Fy = 1
4π0 (q)(2q) (√
2a)2
√2 2 ˆj =
√2 8π0
q2 a2ˆj.
(a) The horizontal component of the net force is then
Fx = 1
4π0
(2q)(2q) a2 ˆi +
√2 8π0
q2 a2ˆi,
= 4 +√ 2/2 4π0
q2 a2ˆi,
= (4.707)(8.99×109N· m2/C2)(1.13×10−6C)2/(0.152 m)2ˆi = 2.34 Nˆi.
(b) The vertical component of the net force is then Fy = − 1
4π0
(q)(2q) a2 ˆj +
√2 8π0
q2 a2ˆj,
= −2 +√ 2/2 8π0
q2 a2ˆj,
= (−1.293)(8.99×109N· m2/C2)(1.13×10−6C)2/(0.152 m)2ˆj = −0.642 Nˆj.
E25-9 The magnitude of the force on the negative charge from each positive charge is F = (8.99×109N· m2/C2)(4.18×10−6C)(6.36×10−6C)/(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× 10−6C)− q. Then
1 4π0
r2 = F,
(8.99×109N·m2/C2)q(52.6×10−6C− q) = (1.19 N)(1.94 m)2.
Solve this quadratic expression for q and get answers q1= 4.02×10−5C and q2= 1.24×10−6N.
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
~F31 = 1 4π0
q3q1 r231 ˆr31,
~F32 = 1 4π0
q3q2 r232 ˆr32. Then
~F31 = −~F32, 1
4π0 q3q1
r312 ˆr31 = − 1 4π0
q3q2
r322 ˆr32, q1
r231ˆr31 = −q2
r232ˆr32.
The only way to satisfy the vector nature of the above expression is to have ˆr31=±ˆr32; 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 q1and q2 into the expression:
(1.07 µC)
r231 ˆr31 = −(−3.28 µC) r322 ˆr32, r232ˆr31 = (3.07)r231ˆr32.
Since squared quantities are positive, we can only get this to work if ˆr31= ˆr32, so q3 is not between q1and q2. We are then left with
r322 = (3.07)r231,
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,
r31+ (0.618 m) = p
(3.07)r31, (0.618 m) = p
(3.07)r31− r31, (0.618 m) = 0.752r31,
0.822 m = 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 length of the angle bisector, d, is given by d = a cos(30◦).
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 θ, given by sin θ = x/(1.17 m) = (0.644)(0.153 m)/(1.17 m) = 0.0842, or θ = 4.83◦.
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 θ = FE/FG, so
1.73(8.99×109N· m2/C2)q2/(0.153 m)2= (0.0133 kg)(9.81 m/s2) tan(4.83◦), or q = 1.29×10−7C.
E25-13 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.
2 1
3 4
5 6
7
8
The magnitude of the force of charge 2 on charge 1 is F12= 1
4π0 q2 r212,
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
F~12 = 1 4π0
q2 a2ˆi, F~13 = 1
4π0
q2 a2ˆj, F~14 = 1
4π0 q2 a2
k.ˆ The force from charge 5 is
F15= 1 4π0
q2 r215,
and is directed along the side diagonal away from charge 5. The distance r15is also the side diagonal distance, and can be found from
r152 = a2+ a2= 2a2, then
F15= 1 4π0
q2 2a2.
By symmetry we expect that the magnitudes of F15, F16, and F17will 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
~F15 = 1 4π0
q2 2a2ˆj/√
2 + ˆk/√ 2
,
~F16 = 1 4π0
q2 2a2ˆi/√
2 + ˆk/√ 2
,
~F17 = 1 4π0
q2 2a2ˆi/√
2 + ˆj/√ 2
. The last force is the force from charge 8 on charge 1, and is given by
F18= 1 4π0
q2 r218,
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
r218= a2+ a2+ a2= 3a2, then in term of components
~F18= 1 4π0
q2 3a2ˆi/√
3 + ˆj/√
3 + ˆk/√ 3
.
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 ˆi. This component has contributions from charge 2, 6, 7, and 8:
1 4π0
q2 a2
1 1 + 2
2√ 2+ 1
3√ 3
, or
1 4π0
q2 a2(1.90)
The three components add according to Pythagoras to pick up a final factor of√ 3, so Fnet= (0.262) q2
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”
direction.
(b) The equation should look Eq. 25-15, except all y’s should be replaced by x’s. Then Fx= 1
4π0
q0q xpx2+ L2/4.
(c) Setting the particle a distance d away should give a force with the same magnitude as F = 1
4π0
q0q dpd2+ L2/4.
This force is directed along the 45◦ line, so Fx= F cos 45◦ and Fy= F sin 45◦. (d) Let the distance be d =p
x2+ y2, and then use the fact that Fx/F = cos θ = x/d. Then Fx= Fx
d = 1 4π0
x q0q
(x2+ y2+ L2/4)3/2. and
Fy= Fy d = 1
4π0
y q0q
(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 F = F~ zˆk = 1
4π0
q0q z (z2+ R2)3/2
k.ˆ
(b) The equation is not valid for both positive and negative z. Reversing the sign of z should reverse the sign of Fz, and one way to fix this is to write 1 = z/√
z2. Then F = F~ zk =ˆ 1
4π0 2q0qz
R2
1
√z2 − 1
√z2
k.ˆ
E25-16 Divide the rod into small differential lengths dr, each with charge dQ = (Q/L)dr. Each differential length contributes a differential force
dF = 1 4π0
q dQ r2 = 1
4π0
qQ r2Ldr.
Integrate:
F =
Z dF =
Z x+L x
1 4π0
qQ r2Ldr,
= 1
4π0
qQ L
1 x− 1
x + L
E25-17 You must solve Ex. 16 before solving this problem! q0refers 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
4π0
qQ L
1 x− 1
x + L
= 1
4π0
qQ r2, or
r =p
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 off axis will result in a net force away from the axis.
That’s unstable.
If q =−Q then both q and Q are on the same side of q0. Moving q0closer 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 θ = z/r we would have
dFx= dF sin θ = 1 4π0
q0λ dz (y2+ z2)
z py2+ z2.
We will need to take into consideration that λ changes sign for the two halves of the rod. Then Fx = q0λ
4π0
Z 0
−L/2
−z dz (y2+ z2)3/2 +
Z L/2 0
+z dz (y2+ z2)3/2
! ,
= q0λ 2π0
Z L/2 0
z dz (y2+ z2)3/2,
= q0λ 2π0
−1 py2+ z2
L/2
0
,
= q0λ 2π0
1
y − 1
py2+ (L/2)2
! .
E25-20 Use Eq. 25-15 to find the magnitude of the force from any one rod, but write it as F = 1
4π0
q Q rpr2+ L2/4,
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
F = 1 π0
q Q z
r2pr2+ L2/4, = 1 π0
q Q z
(z2+ L2/4)pz2+ L2/2, Equating the electric force with the force of gravity and solving for Q,
Q = π0mg
qz (z2+ L2/4)p
z2+ L2/2;
putting in the numbers,
π(8.85×10−12C2/N·m2)(3.46×10−7kg)(9.8m/s2)
(2.45×10−12C)(0.214 m) ((0.214m)2+(0.25m)2/4)p
(0.214m)2+(0.25m)2/2 so Q = 3.07×10−6C.
E25-21 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 five protons. Then X must be Boron, B.
(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:
F = (8.99×109N·m2/C2)(2)(90)(1.60×10−19C)2
(12×10−15m)2 = 290 N.
(b) a = (290 N)/(4)(1.66×10−27kg) = 4.4×1028m/s2. E25-23 Use Eq. 25-4:
F = (8.99×109N·m2/C2)(1.60×10−19C)2
(282×10−12m)2 = 2.89×10−9N.
E25-24 (a) Use Eq. 25-4:
q = s
(3.7×10−9N)(5.0×10−10m)2
(8.99×109N·m2/C2) = 3.20×10−19C.
(b) N = (3.20×10−19C)/(1.60×10−19C) = 2.
E25-25 Use Eq. 25-4, F = 1
4π0 q1q2
r212 = (131.6× 10−19C)(131.6× 10−19C)
4π(8.85× 10−12C2/N· m2)(2.6× 10−15m)2 = 3.8 N.
E25-26 (a) N = (1.15×10−7C)/(1.60×10−19C) = 7.19×1011.
(b) The penny has enough electrons to make a total charge of−1.37×105C. The fraction is then (1.15×10−7C)/(1.37×105C) = 8.40×10−13.
E25-27 Equate the magnitudes of the forces:
1 4π0
q2 r2 = mg, so
r = s
(8.99×109N·m2/C2)(1.60×10−19C)2
(9.11×10−31kg)(9.81 m/s2) = 5.07 m E25-28 Q = (75.0 kg)(−1.60×10−19C)/(9.11×10−31kg) =−1.3×1013C.
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)(6.02×1023mol−1) = 8.37×1024. Each molecule has ten protons, so the total positive charge is
Q = (8.37×1024)(10)(1.60×10−19C) = 1.34×107C.
E25-30 The total positive charge in 0.250 kg of water is 1.34×107C. Mary’s imbalance is then q1= (52.0)(4)(1.34×107C)(0.0001) = 2.79×105C,
while John’s imbalance is
q2= (90.7)(4)(1.34×107C)(0.0001) = 4.86×105C, The electrostatic force of attraction is then
F = 1 4π0
q1q2
r2 = (8.99×109N· m2/C2)(2.79×105)(4.86×105)
(28.0 m)2 = 1.6×1018N.
E25-31 (a) The gravitational force of attraction between the Moon and the Earth is FG= 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
FE= 1 4π0
q2 R2. Setting these two expression equal to each other,
q2
4π0 = GMEMM, which has solution
q = p
4π0GMEMM,
= q
4π(8.85×10−12C2/Nm2)(6.67×10−11Nm2/kg2)(5.98×1024kg)(7.36×1022kg),
= 5.71× 1013C.
(b) We need
(5.71× 1013C)/(1.60× 10−19C) = 3.57× 1032 protons on each body. The mass of protons needed is then
(3.57× 1032)(1.67× 10−27kg) = 5.97× 1065kg.
Ignoring the mass of the electron (why not?) we can assume that hydrogen is all protons, so we need that much hydrogen.
P25-1 Assume that the spheres initially have charges q1and q2. The force of attraction between them is
F1= 1 4π0
q1q2
r212 =−0.108 N,
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
F2= 1 4π0
1
2(q1+ q2)12(q1+ q2)
r212 = 0.0360 N.
Since we know the separation of the spheres we can find q1+ q2 quickly, q1+ q2= 2
q
4π0r212(0.0360 N) = 2.00 µC We’ll put this back into the first expression and solve for q2.
−0.108 N = 1 4π0
(2.00 µC− q2)q2
r122 ,
−3.00 × 10−12C2 = (2.00 µC− q2)q2,
0 = −q22+ (2.00 µC)q2+ (1.73 µC)2. The solution is q2= 3.0 µC or q2=−1.0 µC. Then q1=−1.0 µC or q1= 3.0 µC.
P25-2 The electrostatic force on Q from each q has magnitude qQ/4π0a2, where a is the length of the side of the square. The magnitude of the vertical (horizontal) component of the force of Q on Q is√
2Q2/16π0a2.
(a) In order to have a zero net force on Q the magnitudes of the two contributions must balance,
so √
2Q2
16π0a2 = qQ 4π0a2, or q =√
2Q/4. The charges must actually have opposite charge.
(b) No.
P25-3 (a) The third charge, q3, will be between the first two. The net force on the third charge will be zero if
1 4π0
q q3 r231 = 1
4π0
4q q3 r322 , which will occur if
1 r31
= 2 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 find the magnitude. The second and third charges both exert a force on the first charge; we want this net force on the first charge to be zero, so
1 4π0
q q3
r213 = 1 4π0
q 4q r122 ,
or q3
(L/3)2 = 4q L2,
which has solution q3=−4q/9. The negative sign is because the force between the first and second charge must be in the opposite direction to the force between the first 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 F = q2
4π0x2,
The weight of the ball is W = mg, and the two forces are related by F/W = tan θ≈ sin θ = x/2L.
Combining, 2Lq2= 4π0mgx3, so
x = q2L 2π0
1/3 . (b) Rearrange and solve for q,
q = s
2π(8.85×10−12C2/N· m2)(0.0112 kg)(9.81 m/s2)(4.70×10−2m)3
(1.22 m) = 2.28×10−8C.
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
x0= (q/2)2L 2π0mg
1/3
= 1 4
1/3
x = 2.96 cm.
P25-6 Take the time derivative of the expression in Problem 25-4. Then dx
dt = 2 3
x q
dq dt =2
3
(4.70×10−2m)
(2.28×10−8C)(−1.20×10−9C/s) = 1.65×10−3m/s.
P25-7 The force between the two charges is F = 1
4π0
(Q− q)q r122 .
We want to maximize this force with respect to variation in q, this means finding dF/dq and setting it equal to 0. Then
dF dq = d
dq
1 4π0
(Q− q)q r212
= 1
4π0
Q− 2q r212 . This will vanish if Q− 2q = 0, or q = 12Q.
P25-8 Displace the charge q a distance y. The net restoring force on q will be approximately F ≈ 2 qQ
4π0 1 (d/2)2
y
(d/2) = qQ 4π0
16 d3y.
Since F/y is effectively a force constant, the period of oscillation is
T = 2πr m
k = 0mπ3d3 qQ
1/2 .
P25-9 Displace the charge q a distance x toward one of the positive charges Q. The net restoring force on q will be
F = qQ
4π0
1
(d/2− x)2 − 1 (d/2 + x)2
,
4π0
32 d3x.
Since F/x is effectively a force constant, the period of oscillation is
T = 2πr m
k = 0mπ3d3 2qQ
1/2 .
P25-10 (a) Zero, by symmetry.
(b) Removing a positive Cesium ion is equivalent to adding a singly charged negative ion at that same location. The net force is then
F = e2/4π0r2,
where r is the distance between the Chloride ion and the newly placed negative ion, or r =p
3(0.20×10−9m)2 The force is then
F = (1.6×10−19C)2
4π(8.85×10−12C2/N· m2)3(0.20×10−9m)2 = 1.92×10−9N.
P25-11 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
Fl= 1 4π0
q q0
(a2+ R2).
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
Flsin θ = Fl R
√R2+ a2, so the net force on the test particle will be
2 4π0
q q0 (a2+ R2)
√ R
R2+ a2.
We want to find the maximum value as a function of R. This means take the derivative, and set it equal to zero. The derivative is
2q q0
4π0
1
(a2+ R2)3/2 − 3R2 (a2+ R2)5/2
, which will vanish when
a2+ R2= 3R2, a simple quadratic equation with solutions R =±a/√
2.
E26-1 E = F/q = ma/q. Then
E = (9.11×10−31kg)(1.84×109m/s2)/(1.60×10−19C) = 1.05×10−2N/C.
E26-2 The answers to (a) and (b) are the same!
F = Eq = (3.0×106N/C)(1.60×10−19C) = 4.8×10−13N.
E26-3 F = W , or Eq = mg, so E =mg
q = (6.64× 10−27kg)(9.81 m/s2)
2(1.60× 10−19C) = 2.03× 10−7N/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 field. Since the gravitational force is down, the electric force, and consequently the electric field, must be directed up.
E26-4 (a) E = F/q = (3.0×10−6N)/(2.0×10−9C) = 1.5×103N/C.
(b) F = Eq = (1.5×103N/C)(1.60×10−19C) = 2.4×10−16N.
(c) F = mg = (1.67×10−27kg)(9.81 m/s2) = 1.6×10−26N.
(d) (2.4×10−16N)/(1.6×10−26N) = 1.5×1010. E26-5 Rearrange E = q/4π0r2,
q = 4π(8.85×10−12C2/N· m2)(0.750 m)2(2.30 N/C) = 1.44×10−10C.
E26-6 p = qd = (1.60×10−19C)(4.30×10−9) = 6.88×10−28C· m.
E26-7 Use Eq. 26-12 for points along the perpendicular bisector. Then E = 1
4π0
p
x3 = (8.99× 109N· m2/C2)(3.56× 10−29C· m)
(25.4× 10−9m)3 = 1.95× 104N/C.
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 simplifies the problem into finding E for a charge +q at (a, 0) and−q at (a, a). This is a dipole, and the field is given by Eq. 26-11. For this exercise we have x = a/2 and d = a, so
E = 1 4π0
qa [2(a/2)2]3/2, or, putting in the numbers, E = 1.11×105N/C.
E26-9 The charges at 1 and 7 are opposite and can be effectively 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 field 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 θ, and Eq. 26-11 would look like
E = 2 1
4π0
q x2+ (d/2)2
x px2+ (d/2)2,
≈ 2 1 4π0
q x2
√x x2 when x d. This can be simplified to E = 2q/4π0x2.
E26-11 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 to find the two fields.
For the dipole on the left p = 2aq and the electric field due to this dipole at P has magnitude El= 1
4π0
2aq (x + a)3 and is directed up.
For the dipole on the right p = 2aq and the electric field due to this dipole at P has magnitude Er= 1
4π0
2aq (x− a)3 and is directed down.
The net electric field at P is the sum of these two fields, but since the two component fields point in opposite directions we must actually subtract these values,
E = Er− El,
= 2aq 4π0
1
(x− a)3 − 1 (x + a)3
,
= aq
2π0
1 x3
1
(1− a/x)3 − 1 (1 + a/x)3
. We can use the binomial expansion on the terms containing 1± a/x,
E ≈ aq
2π0
1
x3((1 + 3a/x)− (1 − 3a/x)) ,
= aq
2π0
1
x3(6a/x) ,
= 3(2qa2) 2π0x4.
E26-12 Do a series expansion on the part in the parentheses
1− 1
p1 + R2/z2 ≈ 1 −
1−1
2 R2
z2
= R2 2z2. Substitute this in,
Ez≈ σ 20
R2 2z2
π
π = Q
4π0z2.
E26-13 At the surface z = 0 and Ez= σ/20. Half of this value occurs when z is given by 1
2 = 1− z
√z2+ R2, which can be written as z2+ R2= (2z)2. Solve this, and z = R/√
3.
E26-14 Look at Eq. 26-18. The electric field 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
(z2+ R2)3/2−3 2
2z2
(z2+ R2)5/2 = z2+ R2− 3z2 (z2+ R2)5/2 . This will vanish when the numerator vanishes, or when z = R/√
2.
E26-15 (a) The electric field strength just above the center surface of a charged disk is given by Eq. 26-19, but with z = 0,
E = σ 20
The surface charge density is σ = q/A = q/(πR2). Combining,
q = 20πR2E = 2(8.85× 10−12C2/N· m2)π(2.5× 10−2m)2(3× 106N/C) = 1.04× 10−7C.
Notice we used an electric field strength of E = 3× 106N/C, which is the field at air breaks down and sparks happen.
(b) We want to find 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
N = A
a = π(0.0250 m)2
(0.015× 10−18m2) = 1.31× 1017 (c) The total charge on the disk is 1.04× 10−7C, this corresponds to
(1.04× 10−7C)/(1.6× 10−19C) = 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× 10−6, which isn’t very many.
E26-16 Imagine switching the positive and negative charges. The electric field would also need to switch directions. By symmetry, then, the electric field can only point vertically down. Keeping only that component,
E = 2
Z π/2 0
1 4π0
λdθ r2 sin θ,
= 2
4π0
λ r2. But λ = q/(π/2), so E = q/π20r2.
E26-17 We want to fit the data to Eq. 26-19, Ez= σ
20
1− z
√z2+ R2
. There are only two variables, R and q, with q = σπR2.
We can find σ very easily if we assume that the measurements have no error because then at the surface (where z = 0), the expression for the electric field simplifies to
E = σ 20.
Then σ = 20E = 2(8.854× 10−12C2/N· m2)(2.043× 107N/C) = 3.618× 10−4C/m2.
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
Ez= σ 20
1− z
√z2+ R2
,
and then rearranging,
20Ez
σ = 1− z
√z2+ R2, 20Ez
σ = 1− 1
p1 + (R/z)2, 1
p1 + (R/z)2 = 1−20Ez
σ ,
1 + (R/z)2 = 1
(1− 20Ez/σ)2, R
z =
s 1
(1− 20Ez/σ)2− 1.
Using z = 0.03 m and Ez = 1.187× 107N/C, along with our value of σ = 3.618× 10−4C/m2, we find
R
z =
s 1
(1− 2(8.854×10−12C2/Nm2)(1.187×107N/C)/(3.618×10−4C/m2))2 − 1, R = 2.167(0.03 m) = 0.065 m.
(b) And now find the charge from the charge density and the radius, q = πR2σ = π(0.065 m)2(3.618× 10−4C/m2) = 4.80 µC.
E26-18 (a) λ =−q/L.
(b) Integrate:
E =
Z L+a a
1 4π0
λ dxx2,
= λ
4π0
1 a− 1
L + a
,
= q
4π0
1 a(L + a), since λ = q/L.
(c) If a L then L can be replaced with 0 in the above expression.
E26-19 A sketch of the field looks like this.
E26-20 (a) F = Eq = (40 N/C)(1.60×10−19C) = 6.4×10−18N
(b) Lines are twice as far apart, so the field is half as large, or E = 20N/C.
E26-21 Consider a view of the disk on edge.
E26-22 A sketch of the field looks like this.
E26-23 To the right.
E26-24 (a) The electric field 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 fields,
2q
4π0x2 = 5q 4π0(x + a)2,
or √
5x =√
2(x + a), which has solution
x =
√2 a
√5−√
2 = 2.72a.
E26-25 This can be done quickly with a spreadsheet.
d
x E
E26-26 (a) At point A,
E = 1 4π0
−q
d2 − −2q (2d)2
= 1
4π0
−q 2d2,
where the negative sign indicates that ~E is directed to the left.
At point B,
E = 1 4π0
q
(d/2)2 − −2q (d/2)2
= 1
4π0
6q d2, where the positive sign indicates that ~E is directed to the right.
At point C,
E = 1 4π0
q
(2d)2 +−2q d2
= 1
4π0
−7q 4d2, where the negative sign indicates that ~E is directed to the left.
E26-27 (a) The electric field 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,
W = ~F· ~d = q~E· ~d,
which gives the work done by the electric field on the electron. The electron originally possessed a kinetic energy of K = 12mv2, since we want to bring the electron to a rest, the work done must be negative. The charge q of the electron is negative, so ~E and ~d are pointing in the same direction, and ~E· ~d = Ed.
By the work energy theorem,
W = ∆K = 0−1 2mv2. We put all of this together and find d,
d = W
qE =−mv2
2qE = −(9.11×10−31kg)(4.86× 106m/s)2
2(−1.60×10−19C)(1030 N/C) = 0.0653 m.
(b) Eq = ma gives the magnitude of the acceleration, and vf = vi+ at gives the time. But vf= 0. Combining these expressions,
t =−mvi
Eq =−(9.11×10−31kg)(4.86× 106m/s)
(1030 N/C)(−1.60×10−19C) = 2.69×10−8s.
(c) We will apply the work energy theorem again, except now we don’t assume the final kinetic energy is zero. Instead,
W = ∆K = Kf− Ki,
and dividing through by the initial kinetic energy to get the fraction lost, W
Ki
= Kf− Ki
Ki
= fractional change of kinetic energy.
But Ki =12mv2, and W = qEd, so the fractional change is W
Ki = qEd
1
2mv2 = (−1.60×10−19C)(1030 N/C)(7.88×10−3m)
1
2(9.11×10−31kg)(4.86× 106m/s)2 =−12.1%.
E26-28 (a) a = Eq/m = (2.16×104N/C)(1.60×10−19C)/(1.67×10−27kg) = 2.07×1012m/s2. (b) v =√
2ax =p2(2.07×1012m/s2)(1.22×10−2m) = 2.25×105m/s.
E26-29 (a) E = 2q/4π0r2, or
E = (1.88×10−7C)
2π(8.85×10−12C2/N· m2)(0.152 m/2)2 = 5.85×105N/C.
(b) F = Eq = (5.85×105N/C)(1.60×10−19C) = 9.36×10−14N.
E26-30 (a) The average speed between the plates is (1.95×10−2m)/(14.7×10−9s) = 1.33×106m/s.
The speed with which the electron hits the plate is twice this, or 2.65×106m/s.
(b) The acceleration is a = (2.65×106m/s)/(14.7×10−9s) = 1.80×1014m/s2. The electric field then has magnitude E = ma/q, or
E = (9.11×10−31kg)(1.80×1014m/s2)/(1.60×10−19C) = 1.03×103N/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 m = ρV = ρ4
3πr3. Combining,
q = mg
E =4πρr3g
3E =4π(851 kg/m3)(1.64×10−6m)3(9.81 m/s2)
3(1.92×105N/C) = 8.11×10−19C.
We want the charge in terms of e, so we divide, and get q
e = (8.11×10−19C)
(1.60×10−19C) = 5.07≈ 5.
E26-32 (b) F = (8.99×109N· m2/C2)(2.16×10−6C)(85.3×10−9C)/(0.117m)2= 0.121 N.
(a) E2= F/q1= (0.121 N)/(2.16×10−6C) = 5.60×104N/C.
E1= F/q2= (0.121 N)/(85.3×10−9C) = 1.42×106N/C.
E26-33 If each value of q measured by Millikan was a multiple of e, then the difference between any two values of q must also be a multiple of q. The smallest difference would be the smallest multiple, and this multiple might be unity. The differences are 1.641, 1.63, 1.60, 1.63, 3.30, 3.35, 3.18, 3.24, all times 10−19C. This is a pretty clear indication that the fundamental charge is on the order of 1.6× 10−19C. 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:
4 8 12
5 10 14 7 11 16
The total number of charges is 87, while the total charge is 142.69× 10−19C, so the average charge per quanta is 1.64× 10−19C.
E26-34 Because of the electric field 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 field.
(a) If the lower plate is positively charged then a = g− Eq/m. Replace g in the pendulum period expression by this, and then
T = 2π s
L 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
T = 2π s
L 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= vyt0, where vy= at = 2y/t = 2yvx/L. Combining,
y0 =2yvxt0 L =2yd
L = 2(6.4×10−4m)(6.8×10−3m)
(1.6×10−2m) = 5.44×10−4m, so the total deflection is y + y0= 1.18×10−3m.
E26-36 (a) p = (1.48×10−9C)(6.23×10−6m) = 9.22×10−15C· m.
(b) ∆U = 2pE = 2(9.22×10−15C· m)(1100 N/C) = 2.03×10−11J.
E26-37 Use τ = pE sin θ, where θ is the angle between ~p and ~E. For this dipole p = qd = 2ed or p = 2(1.6× 10−19C)(0.78× 10−9m) = 2.5× 10−28C· m. For all three cases
pE = (2.5× 10−28C· m)(3.4 × 106N/C) = 8.5× 10−22N· m.
The only thing we care about is the angle.
(a) For the parallel case θ = 0, so sin θ = 0, and τ = 0.
(b) For the perpendicular case θ = 90◦, so sin θ = 1, and τ = 8.5× 10−22N· m..
(c) For the anti-parallel case θ = 180◦, so sin θ = 0, and τ = 0.
E26-38 (c) Equal and opposite, or 5.22×10−16N.
(d) Use Eq. 26-12 and F = Eq. Then p = 4π0x3F
q ,
= 4π(8.85×10−12C2/N· m2)(0.285m)3(5.22×10−16N)
(3.16×10−6C) ,
= 4.25×10−22C· m.
E26-39 The point-like nucleus contributes an electric field E+= 1
4π0
Ze r2,
while the uniform sphere of negatively charged electron cloud of radius R contributes an electric field given by Eq. 26-24,
E−= 1 4π0
−Zer R3 .
The net electric field is just the sum,
E = Ze 4π0
1 r2 − r
R3
E26-40 The shell theorem first described for gravitation in chapter 14 is applicable here since both electric forces and gravitational forces fall off as 1/r2. The net positive charge inside the sphere of radius d/2 is given by Q = 2e(d/2)3/R3= ed3/4R3.
The net force on either electron will be zero when e2
d2 = eQ
(d/2)2 = 4e2 d2
d3
4R3 =e2d R3, which has solution d = R.
P26-1 (a) Let the positive charge be located closer to the point in question, then the electric field from the positive charge is
E+ = 1 4π0
q (x− d/2)2 and is directed away from the dipole.
The negative charge is located farther from the point in question, so E− = 1
4π0 q (x + d/2)2 and is directed toward the dipole.
The net electric field is the sum of these two fields, but since the two component fields point in opposite direction we must actually subtract these values,
E = E+− E−,
= 1
4π0
q
(z− d/2)2 − 1 4π0
q (z + d/2)2,
= 1
4π0
q z2
1
(1− d/2z)2 − 1 (1 + d/2z)2
We can use the binomial expansion on the terms containing 1± d/2z,
E ≈ 1
4π0 q
z2((1 + d/z)− (1 − d/z)) ,
= 1
2π0 qd z3
(b) The electric field is directed away from the positive charge when you are closer to the positive charge; the electric field is directed toward the negative charge when you are closer to the negative charge. In short, along the axis the electric field is directed in the same direction as the dipole moment.
P26-2 The key to this problem will be the expansion of 1
(x2+ (z± d/2)2)3/2 ≈ 1 (x2+ z2)3/2
1∓3
2 zd x2+ z2
.
for d √
x2+ z2. Far from the charges the electric field of the positive charge has magnitude E+= 1
4π0
q
x2+ (z− d/2)2, the components of this are
Ex,+ = 1 4π0
q x2+ z2
x
px2+ (z− d/2)2, Ez,+ = 1
4π0 q x2+ z2
(z− d/2) px2+ (z− d/2)2. Expand both according to the first sentence, then
Ex,+ ≈ 1 4π0
xq (x2+ z2)3/2
1 + 3
2 zd x2+ z2
, Ez,+ = 1
4π0
(z− d/2)q (x2+ z2)3/2
1 + 3
2 zd 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
Ex = 1
4π0 xq (x2+ z2)3/2
1 +3
2 zd x2+ z2
+ 1
4π0
−xq (x2+ z2)3/2
1−3
2 zd x2+ z2
,
= 1
4π0
3xqzd (x2+ z2)5/2,
Ez = 1
4π0
(z− d/2)q (x2+ z2)3/2
1 +3
2 zd x2+ z2
+ 1
4π0
−(z + d/2)q (x2+ z2)3/2
1−3
2 zd x2+ z2
,
= 1
4π0
3z2dq
(x2+ z2)5/2 − 1 4π0
dq (x2+ z2)3/2,
= 1
4π0
(2z2− x2)dq (x2+ z2)5/2.
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
E1z= q1
4π0(x2+ R2)
√ z
z2+ R2, while the bottom half contributes a similar expression. Add, and
Ez=q1+ q2 4π0
z
(z2+ R2)3/2 = q 4π0
z (z2+ R2)3/2, which is identical to Eq. 26-18.
(b) The perpendicular component would be zero if q1= q2. It isn’t, so it must be the difference q1− q2 which is of interest. Assume this charge difference is evenly distributed on the top half of the ring. If it is a positive difference, 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
E⊥ = Z π
0
1 4π0
(q1− q2)/π
z2+ R2 cos θ dθ,
= q1− q2
2π20
1 z2+ R2.
