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Chapter 1:
Electric Charges, Electric Fields, Gauss’ Law
Homework 1:
3, 6, 9, 10, 16, 23, 33 (page 575-577)
3. What must be the distance between point charge q1 = 26.0 Cand point charge q2 = -47 C for the electrostatic forcebetween them to have a magnitude of 5.70 N?
221
r
qqkF
F
qqkr
21
(m) 39.17.5
)1047)(1026(1099.8 669
r
CC 6101
6. Two equally charged particles are held 3.2x10-3 m apart andthen released from rest. The initial acceleration of the firstparticle is observed to be 6.0 m/s2 and that of the second to be9.0 m/s2. If the mass of the first particle is 6.3x10-7 kg, whatare (a) the mass of the second particle and (b) the magnitude ofthe charge of each particle.
1121 amF 2212 amF
:2112 FF 12
12 m
a
am
11221
2112 amr
qqkFF
)(106.6
1099.8
102.36103.6 11
9
237211 C
k
ramq
(pC) 66q
9. Two identical conducting spheres, fixed in place, attract eachother with an electrostatic force of 0.108 N when their center-to-center separation is 50.0 cm. The spheres are then connected by athin conducting wire. When the wire is removed, the spheres repeleach other with an electrostatic force of 0.036 N of the initialcharges on the spheres, with a positive net charge, what was (a) thenegative charge on one of them and (b) the positive charge on theother?
Using the shell theorem:
Before:
After: the net charge is positive
)(100.3 12212
211 Cqq
r
qqkF
)(100.22 6
212
221
2
2
2 Cqqr
kr
qkF
The solutions:
1st:
2nd:
Note: if the net charge is negative
the solutions should be:
0100.3100.2 121
621 qq
)(100.1)(100.3 62
61 CqCq
)(100.3)(100.1 62
61 CqCq
)(100.2 621 Cqq
)(100.3)()(100.1)( 612
621 CqqCqq or or
10. In the figure as shown, four particles form a square. The chargesare q1 = q4 = Q and q2 = q3 = q. (a) What is Q/q if the netelectrostatic force on particles 1 and 4 is zero? (b) Is there anyvalue of q that makes the net electrostatic force on each of the fourparticles zero? explain.
(a) q1 & q4 have the same sign,all three forces act on q1
as shown
(b) if the net force acting on particle 3 is also zero: this is inconsistent with (a), so the answer is NO
312141 FFF
2141 2FF
83.222
2
2/3
22
2
q
Q
a
qQk
a
Qk
83.2Q
q
16. See the figure as shown, particle 1 (of charge q1) and particle 2(of charge q2) are fixed in place on an x axis, 8.0 cm apart. Particle3 (of charge q3 = +6.0x10-19 C) is to be placed on the line betweenparticles 1 and 2 so that they produce a net electrostatic force F3,net
on it. The diagram gives the x component of that force versus thecoordinate x at which particle 3 is placed. What are (a) the sign ofcharge q1 and (b) the ratio q2/q1?
● at 2 cm: Fnet=0 so 1 and 2 must havethe same sign.● when q3 approaches q1,F13
increases in magnitude. Fnet
increases in the positive x direction,so F13 is a repulsive force,q1 > 0
232
32
132
31
r
qqk
r
qqk
92
622
13
23
1
2
r
r
q
q
23. See the figure, particles 1 and 2 of charge q1 = q2 =+3.2x10-19 C are on a y axis at distance d = 17.0 cm from theorigin. Particle 3 of charge q3 = +6.4x10-19 C is moved graduallyalong the x axis from x = 0 to x = +5.0 m. At what values of xwill the magnitude of the electrostatic force on the third particlefrom the other two particles be (a) minimum and (b) maximum?what are the (c) minimum and (d) maximum magnitudes?The net force acting on particle 3:
cos223
r
qqkFnet
rr
xdxr cos;22
2/32232
dx
xkqqFnet
maximum: 2/020 22' dxxdF net
minimum :0 x
http://www.function-grapher.com/index.php
33. Calculate the number of coulombs of positive charge in 250cm3 of (neutral) water. (Hint: A hydrogen atom contains oneproton; an oxygen atom contains eight protons)
The mass of the sample:
The number of moles:
The positive charge:
)(2502501 gVm
9.1318
250
molarM
mn
)(1034.1106.11010023.69.13 71923 C
qnNQ A
Homework 2:
1, 5, 14, 15, 19, 23, 27, 31, 35, 44, 54, 56, 59 (page 598-603)
1. Sketch qualitatively the electric field lines both between andoutside two concentric conducting spherical shells when a uniformpositive charge q1 is on the inner shell and a uniform negativecharge –q2 is on the outer. Consider the cases q1 > q2, q1 = q2,and q1 < q2.
q1 = q2
E=0
E=0
q1 > q2 q1 < q2
q1 q2
5. What is the magnitude of a point charge whose electricfield 50 cm away has the magnitude 2.0 N/C?
2r
qkE
)(106.51099.8
5.02 11
9
22
Ck
Erq
14. See the figure, particle 1 of charge q1 = -4.0q and particle2 of charge q2 = +2.0q are fixed to an x axis. (a) As a multipleof distance L, at what coordinate on the axis is the net electricfield of the particles zero? (b) Sketch the net electric fieldlines.
- + E1E2
22
22
11
Lx
qkE
x
qkE
Lx
x
Lx41.3
4
2
2
2
15. The three particles are fixed in place and have charges q1 =q2 = +e and q3 = +2e. Distance a = 6.0 m. What are the (a)magnitude and (b) direction of the net electric field at point Pdue to the particles?
3
21 0
EE
EE
net
3E
2
2;
2
23a
OPOP
ekE
122
199
3100.6
106.141099.8
E
033 45),();/(160 OxECNE
19. The figure shows an electric dipole. What are the (a)magnitude and (b) direction (relative to the positive direction ofthe x axis) of the dipole’s electric field at point P, located atdistance r >> d?
• If x>>d:
• points in the negative direction of the y axis
32
cos2r
kqd
r
qkEnet
moment dipole electric :qdp
netE
2/3
22
2
xd
kpEnet
3x
kpEnet
netE
23. The figure shows two parallel nonconducting rings with theircentral axes along a common line. Ring 1 has uniform charge q1
and radius R; ring 2 has uniform charge q2 and the same radiusR. The rings are separated by distance d = 3.0R. The netelectric field at point P on the common line, at distance R fromring 1, is zero. What is the ratio q1/q2?
Note: q1 and q2 must have the same sign to produce a netelectric field equal to zero
2/3220 )(4 Rz
qzE
2/3222
2/3221
)4(
2
)( RR
Rkq
RR
Rkq
51.055
24
2
1 q
q
27. Two curved plastic rods, one of charge +q and the other ofcharge –q, form a circle of radius R = 8.5 cm in an xy plane. The xaxis passes through both of the connecting points, and the charge isdistributed uniformly on both rods. If q = 15.0 pC, what are the (a)magnitude and (b) direction (relative to the positive direction of thex axis) of the electric field produced at P, the center of the circle?
Consider a differential charge dq:
For two rods:
dq
Ed
1rod dθR
k
R
dskdE
coscos
)(
2
)( Rddsdq
2rods dθR
kdE
cos2
2
90
90
44cos2
R
kq
R
kdθ
R
kE
2rods
)/(8.23105.814.3
10151099.84
42
129
CNE
2rods
31. A nonconducting rod of length L=8.15 cm has charge –q=-4.23 fCuniformly distributed along its length. (a) What is the linear chargedensity of the rod? What are the (b) magnitude and (c) direction(relative to the positive direction of the x axis) of the electric fieldproduced at point P, at a = 12.0 cm from the rod? What is theelectric field magnitude at a = 50 m by (d) the rod and (e) a particleof charge –q = -4.23 fC that replaces the rod?
(a)
(b)
(c) the negative direction of the x axis
)/(10519,01015.8
1023.4 13
2
15
mCL
q
2xaL
dxkdEx
)/(106.1 3 CN
aLa
qkEx
2a
qkEx
:aL
)/(105.1 8 CNEx
(d) for a distant point, the rod acts like a pointcharge, so the electric field of the point charge isthe same as that of the rod:
)/(105.1 8 CNEx
35. At what distance along the central perpendicular axis of auniformly charged plastic disk of radius 0.6 m is the magnitudeof the electric field equal to one-half the magnitude of thefield at the center of the surface of the disk?
At the center (very close to the center):
220
12 Rz
zEz
02
cE
:2
1cz EE
2
11
22
Rz
z
)(35.03/ mRz
44. An alpha particle (the nucleus of a helium atom) has amass of 6.64 x 10-27 kg and a charge of +2e. What are the(a) magnitude and (b) direction of the electric field that willbalance the gravitational force on the particle?
gF
E
mgqE
e
mgE
2
54. An electron is shot at an initial speed of v0=4.0x106 m/s, at
angle 0 = 400 from an x axis. It moves through a uniform electricfield . A screen for detecting electrons is positionedparallel to the y axis, at distance x = 3.0 m. In unit vectornotation, what is the velocity of the electron when it hits thescreen?
So, we ignore gravity
Time to hit the screen:
jE ˆ)0.5( N/C
F
)(108
)(109
19
30
NF
NFg
)/(1078.8
1011.9
0.5106.1
211
31
19
sm
m
eEay
)(1098.040cos104
3
cos
6
0600
sv
xt
)/(1071.1sin 600 smatvtavv yyy
jsmismv ˆ)/(1071.1ˆ)/(1006.3 66
56. An electric dipole consists of charges +2e and -2e separatedby 0.85 nm. It is in an electric field of strength 3.4 x 106 N/C.Calculate the magnitude of the torque on the dipole when the dipolemoment is (a) parallel to, (b) perpendicular to, and (c) antiparallelto the electric field.
(a)
(b)
(c)
Fr
sinpE
)0(0 net
6919 104.31085.0106.122 edEnet
).(103.9 22 mNnet
)180(0 0 net
59. How much work is required to turn an electric dipole1800 in a uniform electric field of magnitude E = 46.0 N/Cif p = 3.02 x 10-25 C.m and the initial angle is 640?
cospEU
UWWEapplied
000 64cos)64180cos(00
pEpEUUWapplied
438.00.461002.32 25 appliedW
)(1022.1 23 JWapplied
064cos2pEWapplied
Homework 3: 1, 7, 14, 17, 21, 22, 36, 39, 43, 44, 51, 52 (Page 622-626)
1. The square surface as shown measures3.2 mm on each side. It is immersed in auniform electric field with E = 1800 N/Cand with field lines at an angle of = 350
with a normal to the surface. Calculate theelectric flux through the surface.
CNm /221051.1
1450
A
E
7. A point charge of 1.8 C is at the center of a cubicalGaussian surface 55 cm on edge. What is the net electric fluxthrough the surface?
Using Gauss’s law:
enclosedq0
CNmNmC
Cq/100.2
/1085.8
108.1 25
2212
6
0
enclosed
14. A charged particle is suspended at the center of twoconcentric spherical shells that are very thin and made ofnonconducting material. Figure a shows a cross section. Figure bgives the net flux through a Gaussian sphere centered on theparticle, as a function of the radius r of the sphere. (a) Whatis the charge of the central particle? What are the net chargesof (b) shell A and (c) shell B?
(a) For r < rA (region 1):
(b) For rA < r < rB (region 2):
(c) For rB < r (region 3):
enclosedq0
10 particleenclosed1 qq
C)(
(C)particle
81
1077.11021085.8 6512
.
q
1
2
3
20 Aqqq particleenclosed2
C or (C) /C)(Nm 2 3.5103.5104 652
Aq
B2/C)(Nm q 5
3 106
17. A uniformly charged conducting sphere of 1.2 m diameterhas a surface charge density of 8.1 C/m2. (a) Find the netcharge on the sphere. (b) What is the total electric flux leavingthe surface of the sphere?
(a) charge = area x surface density
(b) We choose a Gaussian surface covers whole the sphere, usingGauss’ law:
)(107.3101.86.014.344 5622 Crq
CNmq
/102.41085.8
107.3 26
12
5
0
enclosed
21. An isolated conductor of arbitrary shape has a net chargeof +10x10-6 C. Inside the conductor is a cavity within which is apoint charge q = +3.0x10-6 C. What is the charge (a) on thecavity wall and (b) on the outer surface of the conductor?
(a) Consider a Gaussian surface withinthe conductor that covers the cavity wall,in the conductor, E = 0:
(b) the total charge of the conductor:
0 pointwall qq
C Cqq 3103 6 or pointwall
μCCqqq 1310131010 66 or outerouterwall
22. An electron is released from rest at a perpendiculardistance of 9 cm from a line of charge on a very longnonconducting rod. That charge is uniformly distributed , with4.5 C per meter. What is the magnitude of the electron’sinitial acceleration?
Electric field at point P:
Force acting on the electron:
R = 9 cm = 0.09 m = 4.5 C/m = 4.5x10-6 C/m
RE
02
mR
eama
R
eeEF
00 22
36. The figure shows cross sections through two large, parallel,nonconducting sheets with identical distributions of positive chargewith surface charge density = 2.31x10-22 C/m2. In unit-vectornotation, what is at points (a) above the sheets, (b) betweenthem, and (c) below them?
For one non-conducting sheet:
Using the superposition to calculate E due to two sheets:(a)
The net electric field direction is upward(b) E = 0(c)The direction is downward
E
02
E
)/(1061.21085.8
1031.2
22 11
12
22
00
CNE
jCNE ˆ)/(1061.2 11
jCNE ˆ)/(1061.2 11
39. A small, nonconducting ball of mass m = 1 mg and chargeq = 2x10-8 C hangs from an insulating thread that makes anangle = 300 with a vertical, uniformly charged nonconductingsheet. Considering the gravitational force on the ball andassuming the sheet extends far vertically and into and out ofthe page, calculate the surface charge density of the sheet.
If the ball is in equilibrium:
F
gF
0 TFF g
T
02tan
mg
q
mg
qE
F
F
g
)/(105tan2 290 mC
q
mg
44. The figure gives the magnitude of the electric field insideand outside a sphere with a positive charge distributed uniformlythroughout its volume. What is the charge on the sphere?
• At r = 2 cm, E is maximum,so R = 2 cm
R)(r
r
R
qE
304
204 R
qE
)(102.2
)105()02.0(1085.814.34
6
7212
C
q
51. A nonconducting spherical shell of inner radius a = 2 cm andouter radius b = 2.4 cm has a positive volume charge density =A/r, where A is a constant and r is the distance from thecenter of the shell. In addition, a small ball of charge q = 45 fCis located at that center. What value should A have if theelectric field in the shell (a r b) is to be uniform?Key idea: First, we need to calculate Einside the shell, if the field is uniform,so E is independent of distance from thecenter
qshell is the enclosed charge in the shell ofthickness rG-a:
Gaussian surface
204
1
r
qEr
total
rG
shellqqq total
drrdVdqshell24
)(24 222 arAdrrr
Aq G
r
a
shell
G
++ + +
+ +++ ++
+ +
r
dr
dV = 4pr 2drdq= r ´dV
+
Using Gauss’ law:
We rewrite:
If E is uniform in the shell:
totalq0
totalqrE G 20 4
20
22
20 4
)(2
4 G
G
G r
arAq
r
qE
total
2
2
00
1
22
1
2Gr
AaqA
E
2
2
20
2 a
qAAa
q
)/(1079.1)02.0(14.32
1045 211
2
15
mCm
CA
52. The figure below shows a spherical shell with uniform volumecharge density = 1.56 nC/m3, inner radius a = 10 cm, andouter radius b = 2a. What is the magnitude of the electric fieldat radial distances (a) r = 0, (b) r = a/2, (c) r = a,(d) r = 1.5 a, (e) r = b, and (f) r = 3b?
For (a), (b), (c) using Gauss’s law, we findE = 0For (d), (e)The enclosed charge:
The electric field:
For (f):
:bra
33
3
4
3
4arVqenc
r
204
1
r
qE enc
2
33
02
0
)(3
4
4
1
4
1
r
ab
r
qE total
2
33
03 r
abE
