Physics by Resnick Halliday Krane

333 Pages·1986·1.44 MB·English

Instructor Solutions Manual
for
Physics
by
Halliday, Resnick, and Krane
Paul Stanley
Beloit College
Volume 2
A Note To The Instructor...
Thesolutionsherearesomewhatbrief,astheyaredesignedfortheinstructor,notforthestudent.
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
v2 =v2 +2a x,
x 0x x
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.
ExercisesandProblemswhichareenclosedinaboxalsoappearintheStudent’sSolutionManual
withconsiderablymoredetailand,whenappropriate,includediscussiononanyphysicalimplications
oftheanswer. Thesestudentsolutionscarefullydiscussthestepsrequiredforsolvingproblems,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]
1
E25-1 The charge transferred is
Q=(2.5×104C/s)(20×10−6s)=5.0×10−1C.
E25-2 Use Eq. 25-4:
(8.99×109N·m2/C2)(26.3×10−6C)(47.1×10−6C)
r =(cid:115) =1.40m
(5.66N)
E25-3 Use Eq. 25-4:
(8.99×109N·m2/C2)(3.12×10−6C)(1.48×10−6C)
F = =2.74N.
(0.123m)2
E25-4 (a) The forces are equal, so m a =m a , or
1 1 2 2
m =(6.31×10−7kg)(7.22m/s2)/(9.16m/s2)=4.97×10−7kg.
2
(b) Use Eq. 25-4:
(6.31×10−7kg)(7.22m/s2)(3.20×10−3m)2
q =(cid:115) =7.20×10−11C
(8.99×109N·m2/C2)
E25-5 (a) Use Eq. 25-4,
1 q q 1 (21.3µC)(21.3µC)
F = 1 2 = =1.77N
4π(cid:15) r2 4π(8.85×10−12C2/N·m2) (1.52m)2
0 12
(b) In part (a) we found F ; to solve part (b) we need to first find F . Since q = q and
12 13 3 2
r =r , we can immediately conclude that F =F .
13 12 13 12
We must assess the direction of the force of q on q ; it will be directed along the line which
3 1
connectsthetwocharges,andwillbedirectedawayfromq . Thediagrambelowshowsthedirections.
3
F
23
F q
12
F
F
23 F 12
net
From this diagram we want to find the magnitude of the net force on q . The cosine law is
1
appropriate here:
F 2 = F2 +F2 −2F F cosθ,
net 12 13 12 13
= (1.77N)2+(1.77N)2−2(1.77N)(1.77N)cos(120◦),
= 9.40N2,
F = 3.07N.
net
2
E25-6 Originally F = CQ2 = 0.088N, where C is a constant. When sphere 3 touches 1 the
0 0
charge on both becomes Q /2. When sphere 3 the touches sphere 2 the charge on each becomes
0
(Q +Q /2)/2=3Q /4. The force between sphere 1 and 2 is then
0 0 0
F =C(Q /2)(3Q /4)=(3/8)CQ2 =(3/8)F =0.033N.
0 0 0 0
E25-7 Theforcesonq areF(cid:126) andF(cid:126) . TheseforcesaregivenbythevectorformofCoulomb’s
3 31 32
Law, Eq. 25-5,
1 q q 1 q q
F(cid:126) = 3 1ˆr = 3 1 ˆr ,
31 4π(cid:15) r2 31 4π(cid:15) (2d)2 31
0 31 0
1 q q 1 q q
F(cid:126) = 3 2ˆr = 3 2ˆr .
32 4π(cid:15) r2 32 4π(cid:15) (d)2 32
0 32 0
These two forces are the only forces which act on q , so in order to have q in equilibrium the forces
3 3
must be equal in magnitude, but opposite in direction. In short,
F(cid:126) = −F(cid:126) ,
31 32
1 q q 1 q q
3 1 ˆr = − 3 2ˆr ,
4π(cid:15) (2d)2 31 4π(cid:15) (d)2 32
0 0
q q
1ˆr = − 2ˆr .
4 31 1 32
Note that ˆr and ˆr both point in the same direction and are both of unit length. We then get
31 32
q =−4q .
1 2
E25-8 Thehorizontalandverticalcontributionsfromtheupperleftchargeandlowerrightcharge
arestraightforwardtofind√. Thecontributionsfromtheupperleftchargerequire√slightlymorework.
The diagonal distance is 2a; the components will be weighted by cos45◦ = 2/2. The diagonal
charge will contribute
√ √
1 (q)(2q) 2 2 q2
F = √ ˆi= ˆi,
x 4π(cid:15)0( 2a)2√2 8√π(cid:15)0a2
1 (q)(2q) 2 2 q2
F = √ ˆj= ˆj.
y 4π(cid:15)0( 2a)2 2 8π(cid:15)0a2
(a) The horizontal component of the net force is then
√
1 (2q)(2q) 2 q2
F = ˆi+ ˆi,
x 4π(cid:15) a2 8π(cid:15) a2
0√ 0
4+ 2/2q2
= ˆi,
4π(cid:15) a2
0
= (4.707)(8.99×109N·m2/C2)(1.13×10−6C)2/(0.152m)2ˆi=2.34Nˆi.
(b) The vertical component of the net force is then
√
1 (q)(2q) 2 q2
F = − ˆj+ ˆj,
y 4π(cid:15) a2 8π(cid:15) a2
0√ 0
−2+ 2/2q2
= ˆj,
8π(cid:15) a2
0
= (−1.293)(8.99×109N·m2/C2)(1.13×10−6C)2/(0.152m)2ˆj=−0.642Nˆj.
3
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)/(0.13m)2 =14.1N.
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 2cos(30◦)=1.73. The net force is then 24.5N.
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 qQ
= F,
4π(cid:15) r2
0
(8.99×109N·m2/C2)q(52.6×10−6C−q) = (1.19N)(1.94m)2.
Solve this quadratic expression for q and get answers q =4.02×10−5C and q =1.24×10−6N.
1 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
1 q q
F(cid:126) = 3 1ˆr ,
31 4π(cid:15) r2 31
0 31
1 q q
F(cid:126) = 3 2ˆr .
32 4π(cid:15) r2 32
0 32
Then
F(cid:126) = −F(cid:126) ,
31 32
1 q q 1 q q
3 1ˆr = − 3 2ˆr ,
4π(cid:15) r2 31 4π(cid:15) r2 32
0 31 0 32
q q
1 ˆr = − 2 ˆr .
r2 31 r2 32
31 32
The only way to satisfy the vector nature of the above expression is to have ˆr =±ˆr ; this means
31 32
that q must be collinear with q and q . q could be between q and q , or it could be on either
3 1 2 3 1 2
side. Let’s resolve this issue now by putting the values for q and q into the expression:
1 2
(1.07µC) (−3.28µC)
ˆr = − ˆr ,
r2 31 r2 32
31 32
r2 ˆr = (3.07)r2 ˆr .
32 31 31 32
Since squared quantities are positive, we can only get this to work if ˆr =ˆr , so q is not between
31 32 3
q and q . We are then left with
1 2
r2 =(3.07)r2 ,
32 31
so that q is closer to q than it is to q . Then r = r +r = r +0.618m, and if we take the
3 1 2 32 31 12 31
square root of both sides of the above expression,
r +(0.618m) = (3.07)r ,
31 (cid:112) 31
(0.618m) = (3.07)r −r ,
(cid:112) 31 31
(0.618m) = 0.752r ,
31
0.822m = r
31
4
E25-12 The magnitude of the magnetic force between any two charges is kq2/a2, where a =
0.153m. Theforcebetweeneachchargeisdirectedalongthesideofthetriangle;butfromsymmetry
only the component along the bisector is of interest. This means that we need to weight the above
answer by a factor of 2cos(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=acos(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
sinθ =x/(1.17m)=(0.644)(0.153m)/(1.17m)=0.0842,
or θ =4.83◦.
Theforceofgravityoneachballisdirectedverticallyandtheelectricforceisdirectedhorizontally.
The two must then be related by
tanθ =F /F ,
E G
so
1.73(8.99×109N·m2/C2)q2/(0.153m)2 =(0.0133kg)(9.81m/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.
Thenetforcewillbethesum. Sincealleightchargesarethesamealloftheforceswillberepulsive.
We need to sketch a diagram to show how the charges are labeled.
2
1
4
6
7
3
8
5
The magnitude of the force of charge 2 on charge 1 is
1 q2
F = ,
12 4π(cid:15) r2
0 12
where r =a, the length of a side. Since both charges are the same we wrote q2. By symmetry we
12
expect that the magnitudes of F , F , and F will all be the same and they will all be at right
12 13 14
angles to each other directed along the edges of the cube. Written in terms of vectors the forces
5
would be
1 q2
F(cid:126) = ˆi,
12 4π(cid:15) a2
0
1 q2
F(cid:126) = ˆj,
13 4π(cid:15) a2
0
1 q2
F(cid:126) = kˆ.
14 4π(cid:15) a2
0
The force from charge 5 is
1 q2
F = ,
15 4π(cid:15) r2
0 15
andisdirectedalongthesidediagonalawayfromcharge5. Thedistancer isalsothesidediagonal
15
distance, and can be found from
r2 =a2+a2 =2a2,
15
then
1 q2
F = .
15 4π(cid:15) 2a2
0
By symmetry we expect that the magnitudes of F , F , and F will all be the same and they will
15 16 17
all be directed along the diagonals of the faces of the cube. In terms of components we would have
1 q2 √ √
F(cid:126) = ˆj/ 2+kˆ/ 2 ,
15 4π(cid:15) 2a2 (cid:16) (cid:17)
0
1 q2 √ √
F(cid:126) = ˆi/ 2+kˆ/ 2 ,
16 4π(cid:15) 2a2 (cid:16) (cid:17)
0
1 q2 √ √
F(cid:126) = ˆi/ 2+ˆj/ 2 .
17 4π(cid:15) 2a2 (cid:16) (cid:17)
0
The last force is the force from charge 8 on charge 1, and is given by
1 q2
F = ,
18 4π(cid:15) r2
0 18
and is directed along the cube diagonal away from charge 8. The distance r is also the cube
18
diagonal distance, and can be found from
r2 =a2+a2+a2 =3a2,
18
then in term of components
1 q2 √ √ √
F(cid:126) = ˆi/ 3+ˆj/ 3+kˆ/ 3 .
18 4π(cid:15) 3a2 (cid:16) (cid:17)
0
We can add the components together. By symmetry we expect the same answer for each com-
ponents, so we’ll just do one. How aboutˆi. This component has contributions from charge 2, 6, 7,
and 8:
1 q2 1 2 1
(cid:18) + √ + √ (cid:19),
4π(cid:15)0a2 1 2 2 3 3
or
1 q2
(1.90)
4π(cid:15) a2
0 √
The three components add according to Pythagoras to pick up a final factor of 3, so
q2
F =(0.262) .
net (cid:15) a2
0
6
E25-14 (a) Yes. Changing the sign of y will change the sign of F ; since this is equivalent to
y
putting the charge q on the “other” side, we would expect the force to also push in the “other”
0
direction.
(b) The equation should look Eq. 25-15, except all y’s should be replaced by x’s. Then
1 q q
F = 0 .
x 4π(cid:15)0x x2+L2/4
(cid:112)
(c) Setting the particle a distance d away should give a force with the same magnitude as
1 q q
F = 0 .
4π(cid:15)0d d2+L2/4
(cid:112)
This force is directed along the 45◦ line, so F =F cos45◦ and F =F sin45◦.
x y
(d) Let the distance be d= x2+y2, and then use the fact that F /F =cosθ =x/d. Then
(cid:112) x
x 1 xq q
F =F = 0 .
x d 4π(cid:15) (x2+y2+L2/4)3/2
0
and
y 1 yq q
F =F = 0 .
y d 4π(cid:15) (x2+y2+L2/4)3/2
0
E25-15 (a) The equation is valid for both positive and negative z, so in vector form it would read
1 q qz
F(cid:126) =F kˆ = 0 kˆ.
z 4π(cid:15) (z2+R2)3/2
0
(b) The equation is not valid for both positive and negative√z. Reversing the sign of z should
reverse the sign of F , and one way to fix this is to write 1=z/ z2. Then
z
1 2q qz 1 1
F(cid:126) =Fzkˆ = 4π(cid:15)0 R02 (cid:18)√z2 − √z2(cid:19)kˆ.
E25-16 Divide the rod into small differential lengths dr, each with charge dQ = (Q/L)dr. Each
differential length contributes a differential force
1 qdQ 1 qQ
dF = = dr.
4π(cid:15) r2 4π(cid:15) r2L
0 0
Integrate:
x+L 1 qQ
F = (cid:90) dF =(cid:90) dr,
4π(cid:15) r2L
x 0
1 qQ 1 1
= (cid:18) − (cid:19)
4π(cid:15) L x x+L
0
E25-17 You must solve Ex. 16 before solving this problem! q refers to the charge that had been
0
called q in that problem. In either case the distance from q will be the same regardless of the sign
0
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
(cid:18) − (cid:19)= ,
4π(cid:15) L x x+L 4π(cid:15) r2
0 0
or
r = x(x+L).
(cid:112)
7
E25-18 You must solve Ex. 16 and Ex. 17 before solving this problem.
If all charges are positive then moving q off axis will result in a net force away from the axis.
0
That’s unstable.
If q =−Q then both q and Q are on the same side of q . Moving q closer to q will result in the
0 0
attractive force growing faster than the repulsive force, so q will move away from equilibrium.
0
E25-19 We can start with the work that was done for us on Page 577, except since we are
concerned with sinθ =z/r we would have
1 q λdz z
dF =dF sinθ = 0 .
x 4π(cid:15)0(y2+z2) y2+z2
(cid:112)
We will need to take into consideration that λ changes sign for the two halves of the rod. Then
q λ 0 −zdz L/2 +zdz
Fx = 4π0(cid:15) (cid:32)(cid:90) (y2+z2)3/2 +(cid:90) (y2+z2)3/2(cid:33),
0 −L/2 0
q λ L/2 zdz
= 0 (cid:90) ,
2π(cid:15) (y2+z2)3/2
0 0
L/2
q λ −1
= 0 (cid:12) ,
2π(cid:15)0 y2+z2(cid:12)(cid:12)
(cid:112) (cid:12)0
(cid:12)
q λ 1 1
= 0 (cid:32) − (cid:33).
2π(cid:15)0 y y2+(L/2)2
(cid:112)
E25-20 Use Eq. 25-15 to find the magnitude of the force from any one rod, but write it as
1 qQ
F = ,
4π(cid:15)0r r2+L2/4
(cid:112)
where r2 =z2+L2/4. Thecomponentofthisalongthez axisis F =Fz/r. Sincethereare4 rods,
z
we have
1 qQz 1 qQz
F = ,= ,
π(cid:15)0r2 r2+L2/4 π(cid:15)0(z2+L2/4) z2+L2/2
(cid:112) (cid:112)
Equating the electric force with the force of gravity and solving for Q,
π(cid:15) mg
Q= 0 (z2+L2/4) z2+L2/2;
qz (cid:112)
putting in the numbers,
π(8.85×10−12C2/N·m2)(3.46×10−7kg)(9.8m/s2)
((0.214m)2+(0.25m)2/4) (0.214m)2+(0.25m)2/2
(2.45×10−12C)(0.214m) (cid:112)
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.
8
E25-22 (a) Use Eq. 25-4:
(8.99×109N·m2/C2)(2)(90)(1.60×10−19C)2
F = =290N.
(12×10−15m)2
(b) a=(290N)/(4)(1.66×10−27kg)=4.4×1028m/s2.
E25-23 Use Eq. 25-4:
(8.99×109N·m2/C2)(1.60×10−19C)2
F = =2.89×10−9N.
(282×10−12m)2
E25-24 (a) Use Eq. 25-4:
(3.7×10−9N)(5.0×10−10m)2
q =(cid:115) =3.20×10−19C.
(8.99×109N·m2/C2)
(b) N =(3.20×10−19C)/(1.60×10−19C)=2.
E25-25 Use Eq. 25-4,
1 q q (11.6×10−19C)(11.6×10−19C)
F = 1 2 = 3 3 =3.8N.
4π(cid:15) r2 4π(8.85×10−12C2/N·m2)(2.6×10−15m)2
0 12
E25-26 (a) N =(1.15×10−7C)/(1.60×10−19C)=7.19×1011.
(b)The pennyhas enoughelectrons tomake atotalchargeof −1.37×105C. The fractionis then
(1.15×10−7C)/(1.37×105C)=8.40×10−13.
E25-27 Equate the magnitudes of the forces:
1 q2
=mg,
4π(cid:15) r2
0
so
(8.99×109N·m2/C2)(1.60×10−19C)2
r =(cid:115) =5.07m
(9.11×10−31kg)(9.81m/s2)
E25-28 Q=(75.0kg)(−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 0.250kg of water is 1.34×107C. Mary’s imbalance is then
q =(52.0)(4)(1.34×107C)(0.0001)=2.79×105C,
1
while John’s imbalance is
q =(90.7)(4)(1.34×107C)(0.0001)=4.86×105C,
2
The electrostatic force of attraction is then
1 q q (2.79×105)(4.86×105)
F = 1 2 =(8.99×109N·m2/C2) =1.6×1018N.
4π(cid:15) r2 (28.0m)2
0
9