Download Electric Forces and Electric Fields

Chapter 15
Electric Forces and
Electric Fields
Problem Solutions
15.1
FR
Since these are like charges (both positive), the force is F6 and F3 .
15.2
Particle A exerts a force tow ard the right on particle B. By N ew ton’ s third law , particle
B w ill then exert a force tow ard the left back on particle A . The ratio of the final
m agnitud e of the force to the original m agnitud e of the force is
F6
8.99 109
3.00 10
F3
6
8.99 109
FR
6.00 10
N m2
C2
9
C 5.00 10
0.300 m
9
C
2
so
N
3.00 10
N m2
C2
F6
2
9
C 5.00 10
0.100 m
F3
2
1.38 10
5
2
9
C
1.35 10
5
N
N
The final vector force that B exerts on A is 1.57 N directed to the left .
1
2
CH APTER 15
15.3
(a) When the balls are an equilibrium d istance apart, the tension in the string equals the
m agnitud e of the repulsive electric force betw een the balls. Thus,
ke q 2q
F
r2
8.99 109
F2
or
2.16 10
7
2.50 N r 2
q2
2.50 N
2 ke
6.00 10
N m2
C2
9
C 2.00 10
0.707 m
9
C
2
N
(b) The charges ind uce opposite charges in the bulkhead s, but the ind uced charge in
the bulkhead near ball B is greater, d ue to B’ s greater charge. Therefore, the system
N m2
8.99 10
C2
9
F3
15.4
6.00 10
9
C 3.00 10
0.707 m
(a) The tw o ions are both singly charged
F
ke q1 q2
r
2
q
9
C
3.24 10
2
N
1e , one positive and one negative. Thus,
8.99 109 N m2 C2 1.60 10
ke e 2
r2
7
0.50 10
9
m
2
19
C
2
9.2 10
10
N
Since these are unlike charges, the force is attractive .
(b)
15.5
(a)
No. The electric force d epend s only on the m agnitud es of the tw o charges and the
d istance betw een them .
8.99 109 N m2 C2 2.00 10
F1
0.500 m
6
C 4.00 10
6
C
2
(b) The m ass of an alpha particle is F1 0.288 N , w here u 1.66 10
m ass unit. The acceleration of either alpha particle is then
a
F
m
36.8 N
4.002 6 1.66 10
27
kg
5.54 1027 m s 2
27
kg is the unified
Electric Forces and Electric Fields
15.6
The attractive force betw een the charged end s tend s to com press the m olecule. Its
m agnitud e is
F
ke 1e
2
N m2
8.99 10
C2
9
r2
19
1.60 10
2.17 10
6
C
m
2
4.89 10
2
17
N.
The com pression of the “ spring” is
Fy
F2 sin 60.0
so the spring constant is F
15.7
0.436 N ,
0.503 N sin 60.0
Fx2
Fy2
0.0365 N
2
0.436 N
2
0.438 N
1.00 g of hyd rogen contains Avogad ro’ s num ber of atom s, each containing one proton
and one electron. Thus, each charge has m agnitud e
Fy
0.436 N
tan 1
tan 1
85.2 . The d istance separating these charges is
Fx
0.0365 N
85.2 below
F
x-axis , w here RE is Earth’ s rad ius. Thus,
ke N A e
2 RE
2
2
N m2
8.99 109
C2
6.02 1023 1.60 10
6
4 6.38 10 m
19
2
2
C
5.12 105 N
3
4
CH APTER 15
15.8
Refer to the sketch at the right. The m agnitud es of the
ind ivid ual forces are:
50.0 cm x
2
q2
q1
x 2 and 50.0 cm
x
x
5.80 nC
3.00 nC
1.39 x
The com ponents of the resultant force on charge q are
Fx
F1
and
F3 cos 45
Fy
F1
2 1.50cos 45 ke
F3 sin 45
q2
a2
3.06ke
2 1.50sin 45 ke
The m agnitud e of the resultant force is
50.0 cm 20.9 cm 29.1 cm to the left of the
q2
a2
q2
a2
3.06ke
q2
a2
5.80 nC charge.
and it is d irected at Fe .
15.9
(a) The spherically sym m etric charge d istributions behave as if all charge w as located
at the centers of the spheres. Therefore, the m agnitud e of the
Fy 0
T cos5.0 mg force is
T
mg
cos5.0
(b) When the spheres are connected by a cond ucting w ire, the net charge
qnet q1 q2
6.0 10 9 C w ill d ivid e equally betw een the tw o id entical spheres.
Thus, the force is now
F
or
ke qnet 2
r2
ke q2
2 L sin5.0
2
2
N m2
8.99 10
C2
9
mg tan5.0
6.0 10
9
4 0.30 m
C
2
2
Electric Forces and Electric Fields
15.10
The forces are as show n in the sketch
at the right.
F1
F2
F3
ke q1q2
r122
ke q1 q3
2
13
r
ke q2 q3
r232
8.99 109
N m2
C2
6.00 10
6
3.00 10-2 m
N m2
C2
6.00 10
N m2
8.99 10
C2
1.50 10
8.99 109
9
The net force on the 6 C charge is F6
C 1.50 10
6
C 2.00 10
89.9 N
6
C
2
C 2.00 10
2.00 10-2 m
F1 F2
C
2
5.00 10-2 m
6
6
6
C
2
46.7 N (to the left)
The net force on the 1.5 C charge is F1.5
F1
F3
157 N (to the right)
The net force on the
F2
F3
111 N (to the left)
2 C charge is F 2
43.2 N
67.4 N
5
6
CH APTER 15
15.11
In the sketch at the right, FR is the resultant of the
forces F6 and F3 that are exerted on the charge at
the origin by the 6.00 nC and the –3.00 nC charges
respectively.
F6
3.00 10
F3
N m2
C2
8.99 109
6
15.12
FR
N m2
C2
8.99 109
F6
5
1.38 10
9
C 5.00 10
0.300 m
9
C
9
C
2
N
The resultant is FR
or
6.00 10
2
3.00 10
9
C 5.00 10
0.100 m
F3
2
1.38 10
N at 77.5 below
5
1.35 10
2
tan
N at
1
F3
F6
5
N
77.5
x axis
Consid er the arrangem ent of charges show n in the sketch at
the right. The d istance r is
r
2
0.500 m
0.500 m
2
0.707 m
The forces exerted on the 6.00 nC charge are
F2
N m2
8.99 10
C2
9
2.16 10
and
F3
7
9
C 2.00 10
0.707 m
9
C
9
C
2
N
N m2
8.99 10
C2
9
6.00 10
6.00 10
9
C 3.00 10
0.707 m
2
3.24 10
7
N
Electric Forces and Electric Fields
Thus,
Fx
F2
F3 cos45.0
and
Fy
F2
F3 sin 45.0
3.81 10
7
7.63 10
N
8
N
The resultant force on the 6.00 nC charge is then
or
15.13
2
FR
Fx
FR
3.89 10
2
Fy
7
3.89 10
7
tan
N at
1
Fy
Fx
11.3
N at 11.3 below +x axis
Please see the sketch at the righ t.
F1
8.99 109 N m2 C2 2.00 10
0.500 m
or
F2
F1
6
C
6
C 7.00 10
6
C
0.288 N
0.500 m
F2
C 4.00 10
2
8.99 109 N m2 C2 2.00 10
or
6
2
0.503 N
The com ponents of the resultant force acting on the 2.00 C charge are:
Fx
and
Fy
F1 F2 cos60.0
0.288 N
F2 sin 60.0
0.503 N cos60.0
0.503 N sin 60.0
3.65 10 2 N
0.436 N
The m agnitud e and d irection of this resultant force are
F
at
Fx2
tan
1
Fy2
Fy
Fx
0.0365 N
tan
1
2
0.436 N
0.0365 N
0.436 N
2
0.438 N
85.2 or 85.2 below
x-axis
7
8
CH APTER 15
15.14
If the forces exerted on the positive
third charge by the tw o negative
charges are to be in opposite
d irections as they m ust, the third
charge m ust be located on the line
betw een the tw o negative charges
as show n at the right.
If the tw o forces are to ad d to zero, their m agnitud es m ust be equal. These m agnitud es
are
q q
q2 q3
F1 ke 1 2 3 and
F2 ke
2
x
50.0 cm x
w here q1
5.80 nC .
3.00 nC and q2
50.0 cm x
Equating and canceling com m on factors gives
q2
2
q1
x2
or
50.0 cm
x
x
5.80 nC
3.00 nC
1.39 x
giving 2.39 x 50.0 cm
and
Thus, the third charged should be located 20.9 cm to the right of the
50.0 cm 20.9 cm 29.1 cm to the left of the
15.15
x 20.9 cm
3.00 nC charge or
5.80 nC charge.
Consid er the free-bod y d iagram of one of the spheres given
at the right. H ere, T is the tension in the string and Fe is the
repulsive electrical force exerted by the other sphere.
Fy
0
T cos5.0
Fx
0
Fe
mg
cos5.0
mg , or T
T sin5.0
mg tan5.0
At equilibrium , the d istance separating the tw o spheres is r
Thus, Fe
q
ke q2
mg tan5.0 becom es
2
2 L sin5.0
2 L sin 5.0
2 L sin5.0 .
mg tan5.0 and yield s
mg tan 5.0
ke
2 0.300 m sin 5.0
0.20 10
3
kg 9.80 m s 2 tan 5.0
8.99 109 N m 2 C2
7.2 nC
Electric Forces and Electric Fields
15.16
In the sketch at the right,
rBC
tan
and
(a)
FAC
x
(b)
FAC
y
3.00 m
4.00 m
1
2
5.00 m
36.9
0
FAC
qA qC
ke
2
rAC
FBC
2
8.99 10 N m C
y
qB qC
9
r
4
C 1.00 10
x
(e)
FBC
y
(f)
FR
x
FAC
x
FBC
x
(g)
FR
y
FAC
y
FBC
y
FBC cos
21.6 N cos 36.9
FBC sin
and
2
x
C
2
6.00 10
2
4
4
FR
tan
2
1
FR
y
FR
x
2
tan
24.3 N at 44.5° above the
21.6 N
13.0 N
17.0 N
17.0 N
1
C
17.3 N
30.0 13.0 N
17.3 N
y
2
4
17.3 N
21.6 N sin 36.9
0 17.3 N
30.0 N
C 1.00 10
5.00 m
FBC
FR
2
8.99 10 N m C
2
BC
(d )
or F R
3.00 10
2
3.00 m
ke
(h) FR
15.17
3.00 m
9
FAC
(c)
2
4.00 m
17.0 N
17.3 N
2
24.3 N
44.5°
x-direction
In ord er for the object to “ float” in the electric field , the electric force exerted on the
object by the field m ust be d irected upw ard and have a m agnitud e equal to the w eight
of the object. Thus, Fe qE mg , and the m agnitud e of the electric field m ust be
E
mg
q
3.80 g
9.80 m s2
18 10
6
C
1 kg
103 g
2.07 103 N C
9
10
CH APTER 15
The electric force on a negatively charged object is in the d irection opposite to that of the
electric field . Since the electric force m ust be d irected upw ard , the electric field m ust be
directed downward.
15.18
(a) Taking to the right as
positive, the resultant
electric field at point P is
given by
ER
E1
E3
ke q1
2
1
E2
ke q3
r
This gives ER
(b) F qER
or F
r22
r
8.99 109
or ER
k e q2
2
3
N m2
C2
6.00 10
6
0.020 0 m
C
2
2.00 10
6
0.030 0 m
2.00 107 N C
2.00 107 N C to the right
2.00 10
6
C 2.00 107 N C
40.0 N to the left
40.0 N
C
2
1.50 10
6
0.010 0 m
C
2
Electric Forces and Electric Fields
15.19
11
We shall treat the concentrations as point charges. Then, the resultant field consists of
tw o contributions, one d ue to each concentration.
The contribution d ue to the positive ch arge at 3 000 m altitud e is
E
ke
q
r
N m2
C2
8.99 109
2
40.0 C
1000 m
3.60 105 N C
2
downward
The contribution d ue to the negative charge at 1 000 m altitud e is
E
ke
q
r
N m2
C2
8.99 109
2
40.0 C
1000 m
3.60 105 N C
2
downward
The resultant field is then
E E
15.20
(a) The m agnitud e of the force on the electron is F
a
(b) v v0
15.21
7.20 105 N C downward
E
F
me
at
1.60 10
eE
me
19
C 300 N C
9.11 10
31
5.27 1013 m s2
kg
5.27 1013 m s2 1.00 10
0
q E eE , and the acceleration is
8
s
5.27 105 m s
N ote that at the point m id w ay betw een the tw o charges ( x 2.00 m, y 0 ), the field
contribution E1 (d ue to the negative charge at the origin) and the contribution E 2 (d ue
to the positive charge at x 4.00 m ) are both in the negative x-d irection. The m agnitud e
of the resultant field at this poin t is therefore
Enet
E1
Thus, Enet
E2
ke
q1
2
1
r
q2
r22
8.99 109 N m2 C2
27.0 N C in the negative x-direction .
5.00 10
2.00 m
9
C
2
7.00 10
2.00 m
9
C
2
27.0 N C
12
CH APTER 15
15.22
The force an electric field exerts on a positive change is in the d irection of the field . Since
this force m ust serve as a retard ing force and bring the proton to rest, th e force and
hence the field m ust be in the direction opposite to the proton 's velocity .
The w ork-energy theorem , Wnet
qE
15.23
1.60 10
or
19
a
F
m
qE
mp
(b) t
v
a
1.20 106 m s
6.12 1010 m s 2
(a)
(c)
x
v 2f
v02
2a
1
m p v 2f
2
(d ) KE f
15.24
x 0 KEi
KEi , gives the m agnitud e of the field as
KE f
KEi
q x
E
C 640 N C
-27
1.673 10
kg
1.96 10
1.20 106 m s
2
5
0
2 6.12 1010 m s 2
1
1.673 10
2
27
s
3.25 10
-19
1.60 10
15
J
1.63 104 N C
C 1.25 m
6.12 1010 m s2
19.6 s
11.8 m
kg 1.20 106 m s
2
1.20 10
15
J
(a) Please refer to the solution of Problem 15.13 earlier in this chapter. There it is show n
that the resultant electric force experienced by the 2.00 C located at the origin is
F 0.438 N at 85.2° below the x-axis . Since the electric field at a location is d efined
as the force per unit charge experienced by a test charge placed in that location, the
electric field at the origin in the charge configuration of Figure P15.13 is
E
F
q0
0.438 N
at
2.00 10-6 C
85.2°
2.19 105 N C at 85.2° below
x-axis
(b) The electric force experienced by the charge at the origin is d irectly proportional to
the m agnitud e of that charge. Thus, d oubling the m agnitud e of this charge w ould
double the magnitude of the electric force. H ow ever, the electric field is the force per
unit charge and the field would be unchanged if the charge was doubled. This is easily
seen in the calculation of part (a) above. Doubling the m agnitud e of the charge at
the origin w ould d ouble both the num erator and the d enom inator of the ratio F q0 ,
but the value of the ratio (i.e., the electric field ) w ould be unchanged .
Electric Forces and Electric Fields
15.25
13
The alpha particle (w ith a charge 2e and m ass of 6.63 10 27 kg ) w ill experience a
constant electric force in the y-direction (the d irection of the electric field ) once it enters
the tube. This force gives the alpha particle a constant acceleration w ith com ponents
ax
0 and
qEy
Fe
m
ay
3.20 10
m
19
C
4.50 10
6.64 10
27
4
NC
2.17 104 m s2
kg
The particle w ill strike the tube w all w hen its vertical d isplacem ent is
y 0.500 cm ,
0 , the elapsed tim e w hen this occurs is
2
and from
y v0 y t ayt 2 w ith v0 y
2
t
2
y
0.500 10
4
2
m
2.17 10 m s 2
ay
6.79 10
4
s
During this tim e, the d istance the alpha particle travels parallel to the axis of the tube is
1 2
ax t
2
x v0 x t
15.26
1 250 m s 6.79 10
4
s
0
0.849 m
If the resultant field is to be zero, the contributions
of the tw o charges m ust be equal in m agnitud e
and m ust have opposite d irections. This is only
possible at a point on the line betw een the tw o
negative charges.
Assum e the point of interest is located on the
y-axis at 4.0 m y 6.0 m . Then, for equal
m agnitud es,
0
or
E
Solving for y gives
Qinside 0 E 4
4
8.85 10
EA E 4 r 2
0
Er 2
12
spherically symmetric
C2 N m2 575 N C 0.230 m
2
3.38 10
9
C
3.38 nC
, or
14
CH APTER 15
15.27
If the resultant field is zero, the
contributions from the tw o charges m ust
be in opposite d irections and also have
equal m agnitud es. Choose the line
connecting the charges as the x-axis, w ith
the origin at the –2.5 C charge. Then, the
tw o contributions w ill have opposite
d irections only in the regions net positive charge and
. For the m agnitud es to be equal, the point m ust be nearer the sm aller charge. Thus,
the point of zero resultant field is on the x-axis at x 0 .
0
Requiring equal m agnitud es gives
2.5 C
d2
E
Qnet
Q 2Q
0
0
Q
6.0 C
2
1.0 m d
Thus,
E
Qnet
0
Q Q
0
0
Solving for d yield s
d 1.8 m ,
15.28
or
1.8 m to the left of the
2.5 C charge
The altitud e of the triangle is
h
0.500 m sin60.0
0.433 m
and the m agnitud es of the field s d ue to
each of the charges are
E1
ke q1
h2
8.99 109 N m 2 C2 3.00 10
0.433 m
9
C
2
144 N C
E
and
EA
E3
1.80 104 N C
5.50 m 2.00 m
1.98 105 N m2 C
ke q3
8.99 109 N m2 C2 5.00 10 9 C
r32
0.250 m
2
719 N C
0
or
Electric Forces and Electric Fields
Thus, r
a and
Ey
E1
144 N C
giving
ER
Ex
2
2
Ey
1.88 103 N C
and
1
tan
Ex
tan
1
0.0769
4.40
ER 1.88 103 N C at 4.40° below the +x axis
H ence
15.29
Ey
From the sym m etry of the charge d istribution, stud ents
should recognize that the resultant electric field at the
center is
ER
0
If one d oes not recognize this intuitively, consid er:
E1 E2 E3 , so
ER
r
a
and
Ey
Thus,
15.30
E1 y
ER
E2 y
Ex2
E y2
E3
ke q
r
2
ke q
sin 30
r
2
sin 30
ke q
r2
0
0
The m agnitud e of q2 is three tim es the m agnitud e of q1 because 3 tim es as m any lines
em erge from q2 as enter q1 .
(a) Then,
(b)
q2
q1 q2
q2
3 q1
13
0 because lines em erge from it,
and E
8.99 109 N m2 C2
1.50 m
2.00 10
2
9
C
because lines term inate on it.
15
16
CH APTER 15
15.31
N ote in the sketches at the right that
electric field lines originate on
positive charges and term inate on
negative charges. The d ensity of
lines is tw ice as great for the 2q
charge in (b) as it is for the 1q
charge in (a).
15.32
Rough sketches for these charge configurations are show n below .
15.33
(a) The sketch for (a) is show n at
the right. N ote that four tim es
as m any lines should leave q1
as em erge from q2 although, for
clarity, this is not show n in this
sketch.
(b) The field pattern looks the sam e
here as that show n for (a) w ith
the exception that the arrow s
are reversed on the field lines.
15.34
(a) In the sketch for (a) at
the right, note that there
are no lines insid e the
sphere. On the outsid e
of the sphere, the field
lines are uniform ly
spaced and rad ially
outw ard .
Electric Forces and Electric Fields
17
(b) In the sketch for (b) above,
note that the lines are
perpend icular to the surface
at the points w here they em erge. They should also be sym m etrical about the
sym m etry axes of the cube. The field is zero insid e the cube.
15.35
(a)
r 2.50 m b net charge on each surface of the sphere.
Qinside
4 0 r2
(b) The negative charge low ered into the sphere repels E
and leaves
ke Qinside
surface,
r2
5 C on the inside surface of the sphere.
(c) The negative charge low ered insid e the sphere neutralizes the inner surface, leaving
zero charge on the inside . This leaves
5 C on the outside surface of the sphere.
(d ) When the object is rem oved , the sphere is left w ith
and
15.36
E
5.00 C on the outside surface
0.
(a) The d om e is a closed cond ucting surface. Therefore, the electric field is
Qinside Qinner
Qcenter 0 everyw here insid e it.
surface
At the surface and outsid e of this spherically sym m etric charge d istribution, the
field is as if all the charge w ere concentrated at the center of the sphere.
(b) At the surface,
E
ke q
R2
8.99 109 N m2 C2 2.0 10
1.0 m
(c) Outsid e the spherical d om e, E
E
4.0 m
15.37
C
2
ke q
. Thus, at r
r2
8.99 109 N m2 C2 2.0 10
2
4
4
C
1.8 106 N C
4.0 m ,
1.1 105 N C
For a uniform ly charged sphere, the field is strongest at the surface.
Thus,
sheet
or
sheet
2 ,
18
CH APTER 15
15.38
If the w eight of the d rop is balanced by the electric force, then z 0 or the m ass of the
d rop m ust be
Ez
2
Fe
0
2
ma
0
and the rad ius of the d rop is 0 z 2.00 cm
2
Ez
(a)
2
2
But, E z
15.39
0
0
0
2
1.67 10
0
27
kg 1.52 1012 m s2
2.54 10
15
N in the d irection of the
acceleration, or rad ially outw ard .
(b) The d irection of the field is the d irection of the force on a positive charge (such as
the proton). Thus, the field is d irected z 2.00 cm The m agnitud e of the field is
E
15.40
Fe
q
2.54 10
1.60 10
19
N
C
1.59 104 N C
2
The flux through an area is E z
, w here is the angle betw een the
2 0
0
d irection of the field E and the line perpend icular to the area A .
(a)
z 0
(b) In this case, Ez
15.41
15
Ez ,lower
Ez ,upper
2
The area of the rectangular plane is Ez
0
0
Ez ,lower
2
Ez ,upper
and 0 z 2.00 cm
0
2
0
0
3
2
(a) When the plane is parallel to the yz plane, z 2.00 cm , and the flux is
Ez
Ez ,lower
Ez ,upper
2
0
0
2
0
.
0
Electric Forces and Electric Fields
19
(b) When the plane is parallel to the x-axis, 35.0 mC and
Fe
(c)
15.42
Q2
ke 2
r
9
2
8.99 10 N m C
35.0 10
2
3
C
3.10 102 m
2
2
115 N
115 N
(a) Gauss’ s law states that the electric flux through any closed surface equals the net
charge enclosed d ivid ed by 0 . We choose to consid er a closed surface in the form
of a sphere, centered on the center of the charged sphere and having a rad ius
infinitesim ally larger than that of the charged sphere. The electric field then has a
uniform m agnitud e and is perpend icular to our surface at all points on that surface.
The flux through the chosen closed surface is therefore E EA E 4 r 2 , and
Gauss’ s law gives
Qinside 0 E 4
4
0
8.85 10
Er 2
12
C2 N m2 575 N C 0.230 m
2
3.38 10
9
C
3.38 nC
(b) Since the electric field d isplays spherical sym m etry, you can conclud e that the
charge d istribution generating that field is spherically symmetric . Also, since the
electric field lines are d irected outw ard aw ay from the sphere, the sphere m ust
contain a net positive charge .
15.43 From Gauss’ s law , the electric flux through any closed
surface is equal to the net charge enclosed d ivid ed by 0 .
Thus, the flux through each surface (w ith a positive flux
com ing outw ard from the enclosed interior and a negative
flux going inw ard tow ard that interior) is
15.44
For S1 :
E
Qnet
0
Q 2Q
For S 2 :
E
Qnet
0
Q Q
For S 3 :
E
Qnet
0
2Q Q Q
For S 4 :
E
Qnet
0
0
0
Q
0
0
0
0
0
2Q
0
0
The electric field has constant m agnitud e and is everyw here perp end icular to the
bottom of the car. Thus, the electric flux through the bottom of the car is
E
EA
1.80 104 N C
5.50 m 2.00 m
1.98 105 N m2 C
20
CH APTER 15
15.45
We choose a spherical Gaussian surface, concentric w ith the charged spherical shell and
of rad ius r. Then, EA cos
E 4 r 2 cos0 4 r 2 E .
(a) For r a (that is, outsid e the shell), the total charge enclosed by the Gaussian
surface is Q
q q 0 . Thus, Gauss’ s law gives 4 r 2 E 0, or E 0 .
(b) Insid e the shell, r
a , and the enclosed charge is Q
Therefore, from Gauss’ s law , 4 r 2 E
q
, or E
0
The field for r
15.46
a is E
q.
q
4
0
r
2
ke q
r2
ke q
directed radially outward .
r2
(a) The surface of the cube is a closed surface w hich surround s a total charge of
Q 1.70 102 C . Thus, by Gauss’ s law , the electric flux through the w hole surface
of the cube is
Qinside
E
0
1.70 10 4 C
8.85 10-12 C2 N m 2
1.92 107 N m 2 C
(b) Since the charge is located at the center of the cube, the six faces of the cube are
sym m etrically positioned around the location of the charge. Thus, one -sixth of the
flux passes through each of the faces, or
E
face
6
1.92 107 N m 2 C
6
3.20 106 N m 2 C
(c) The answ er to part (b) w ould change because the charge could now be at d ifferent
d istances from each face of the cube, but the answ er to part (a) w ould be unchanged
because the flux through the entire closed surface d epend s only on the total charge
insid e the surface.
15.47
N ote that w ith the point charge 2.00 nC positioned at
the center of the spherical shell, w e have com plete
spherical sym m etry in this situation. Thus, w e can expect
the d istribution of charge on the shell, as w ell as the
electric field s both insid e and outsid e of the shell, to also
be spherically sym m etric.
Electric Forces and Electric Fields
21
(a) We choose a spherical Gaussian surface, centered on the center of the cond ucting
shell, w ith rad ius r 1.50 m a as show n at the right. Gauss’ s law gives
E
EA
Qinside
E 4 r2
or E
0
so
E
Qinside
4 0 r2
8.99 109 N m2 C2
1.50 m
and
E
7.99 N C
ke Qcenter
r2
2.00 10
9
C
2
The negative sign m eans that the field is radial inward .
(b) All points at r 2.20 m are in the range a r b , and hence are located w ithin the
cond ucting m aterial m aking up the shell. Und er cond itions of electrostatic
equilibrium , the field is E 0 at all points insid e a cond ucting m aterial.
(c) If the rad ius of our Gaussian surface is r 2.50 m b , Gauss’ s law (w ith total
Qinside
ke Qinside
spherical sym m etry) lead s to E
just as in part (a). H ow ever,
2
4 0r
r2
now Qinside Qshell Qcenter
3.00 nC 2.00 nC 1.00 nC . Thus, w e have
E
8.99 109 N m2 C2
2.50 m
1.00 10
9
C
2
1.44 N C
w ith the positive sign telling us that the field is radial outward at this location.
(d ) Und er cond itions of electrostatic equilibrium , all excess charge on a cond uctor
resid es entirely on its surface. Thus, the sum of the charge on the inner surface of
the shell and that on the outer surface of the shell is Qshell
3.00 nC . To see how
m uch of this is on the inner surface, consid er our Gaussian surface to have a rad ius
r that is infinitesim ally larger than a. Then, all points on the Gaussian surface lie
w ithin the cond ucting m aterial, m eaning that E 0 at all points and the total flux
through the surface is E 0 . Gauss’ s law then states that
Qinside Qinner
Qcenter 0 , or
surface
Qinner
Qcenter
2.00 nC
2.00 nC
surface
The charge on the outer surface m ust be
Qouter
Qshell Qinner
3.00 nC 2.00 nC
surface
surface
1.00 nC
22
CH APTER 15
15.48
Please review Exam ple 15.8 in your textbook. There it is show n that the electric field d ue
to a noncond ucting plane sheet of charge parallel to the xy-plane has a constant
m agnitud e given by Ez
sheet 2 0 , w here
sheet is the uniform charge per unit area on
the sheet. This field is everyw here perpend icular to the xy-plane, is d irected aw ay from
the sheet if it has a positive charge d ensity, and is d irected tow ard the sheet if it has a
negative charge d ensity.
In this problem , w e have tw o plane sheets of charge, both parallel to the xy-plane and
separated by a d istance of 2.00 cm . The upper sheet has charge d ensity sheet
2 ,
w hile the low er sheet has sheet
. Taking upw ard as the positive z-d irection, the
field s d ue to each of the sheets in the three regions of interest are:
Low er sheet (at z =0)
Upper sheet (at z =2.00 cm)
Electric Field
Electric Field
Region
z 0
0 z 2.00 cm
z 2.00 cm
Ez
2
Ez
2
Ez
2
0
0
0
2
2
2
2
Ez
2
0
0
0
0
0
0
0
2
Ez
2
0
2
Ez
2
0
When both plane sheets of charge are present, the resultant electric field in each region is
the vector sum of the field s d ue to the ind ivid ual sheets for that region.
(a) For z 0 :
(b) For 0 z 2.00 cm :
(c) For z 2.00 cm :
Ez
Ez
Ez
Ez ,lower
Ez ,lower
Ez ,lower
Ez ,upper
Ez ,upper
Ez ,upper
2
0
0
2
0
0
0
2
0
0
3
2
2
0
0
2
Electric Forces and Electric Fields
15.49
23
The rad ius of each sphere is sm all in com parison to the d istance to the nearest
neighboring charge (the other sphere). Thus, w e shall assum e tha t the charge is
uniform ly d istributed over the surface of each sphere and , in its interaction w ith the
other charge, treat it as though it w ere a point charge. In this m od el, w e then have tw o
id entical point charges, of m agnitud e 35.0 mC , separated by a total d istance of 310 m
(the length of the cord plus the rad ius of each sphere). Each of these charges repels the
other w ith a force of m agnitud e
Fe
Q2
ke 2
r
9
2
8.99 10 N m C
35.0 10
2
3
C
3.10 102 m
2
2
115 N
Thus, to counterbalance this repulsion and hold each sphere in equilibrium , the cord
m ust have a tension of 115 N so it w ill exert a 115 N on that sphere, d irected tow ard
the other sphere.
15.50
(a) As show n in Exam ple 15.8 in the textbook, the electric field d ue to a noncond ucting
plane sheet of charge has a constant m agnitud e of E
is the
2 0 , w here
uniform charge per unit area on the sheet. The d irection of the field at all locations
is perpend icular to the plane sheet and d irected aw ay from t he sheet if
is
2
positive, and tow ard the sheet if
is negative. Thus, if
5.20 C m , the
m agnitud e of the electric field at all d istances greater than zero from the plane
(includ ing the d istance of 8.70 cm ) is
E
5.20 10
2
0
2 8.85 10
12
6
C m2
2
C N m
2
2.94 105 N C
(b) The field d oes not vary w ith d istance as long as the d istance is sm all com pared w ith
the d im ensions of the plate.
24
CH APTER 15
15.51
The three contributions to the resultant
electric field at the point of interest are
show n in the sketch at the right.
The m agnitud e of the resultant field is
ER
ER
15.52
E1 E2
ke q1
E3
ke q3
ke q2
2
1
2
2
r
ke
2
3
r
r
N m2
C2
ER
8.99 109
ER
24 N C , or ER
q1
q2
q3
2
1
2
2
r32
r
4.0 10
2.5 m
9
C
2
r
5.0 10
9
2.0 m
C
3.0 10
2
1.2 m
9
C
2
24 N C in the +x direction
Consid er the free-bod y d iagram show n at the right.
Since Fe
Fy
0
T cos
Fx
0
Fe
T sin
mg
cos
mg tan
qE , w e have
qE mg tan , or q
q
mg or T
mg tan
E
2.00 10 3 kg 9.80 m s 2 tan15.0
1.00 103 N C
5.25 10
6
C
5.25 C
Electric Forces and Electric Fields
15.53
(a) At a point on the x-axis, the contributions by the tw o
charges to the resultant field have equal m agnitud es
ke q
given by E1 E2
.
r2
The com ponents of the resultant field are
Ey
E1 y
and
Since
ER
ke q
sin
r2
E2 y
Ex
cos
r2
E1x
br
r2
ke 2q b
a2
b2
ke q
cos
r2
E2 x
b
r3
32
ke q
sin
r2
b
a
2
b2
32
0
ke q
cos
r2
ke 2q
r2
, the resultant field is
in the +x direction
cos
25
26
CH APTER 15
(b) N ote that the result of part (a) m ay be w ritten as ER
ke Q b
a
2
b
2 32
w here Q 2q is
the total charge in the charge d istribution generating t he field .
In the case of a uniform ly charged circular ring, consid er the ring to consist of a
very large num ber of pairs of charges uniform ly spaced around the ring. Each pair
consists of tw o id entical charges located d iam etrically opposite each other on the
ring. The total charge of pair num ber i is Qi . At a point on the axis of the ring, this
pair of charges generates an electric field contribution that is parallel to the axis and
kebQi
has m agnitud e Ei
.
32
2
a b2
The resultant electric field of the ring is the sum m ation of the contributions by all
pairs of charges, or
ER
ke b
Ei
a
w here Q
ER
15.54
2
b
kebQ
Qi
2 32
a
2
b2
32
Qi is the total charge on the ring.
keQb
a2
b2
32
in the +x direction
It is d esired that the electric field exert a retard ing force on the electrons, slow ing them
d ow n and bringing them to rest. For the retard ing force to have m axim um effect, it
should be anti-parallel to the d irection of the electrons m otion. Since the force an electric
field exerts on negatively charged particles (such as electrons) is in the d irection
opposite to the field , the electric field should be in the direction of the electron’s motion.
The w ork a retard ing force of m agnitud e Fe
m ove d istance d is W
then gives
eEd
KE f
Fe d cos180
KEi
Fe d
q E eE d oes on the electrons as they
eEd . The w ork-energy theorem W
0 K
and the m agnitud e of the electric field required to stop the electrons in d istance d is
E
K
ed
or
E
K ed
KE
Electric Forces and Electric Fields
15.55
27
(a) Without the electric field present, the ball com es to equilibrium w hen the upw ard
force exerted on the ball by the spring equals the d ow nw ard gravitational force
acting on it, or w hen kx mg . The d istance the ball stretches the spring before
reaching the equilibrium position in this case is
x
4.00 kg 9.80 m s2
mg
k
845 N m
4.64 10
2
m
4.64 cm
(b) With the electric field present, the positively charged ball experiences an upw ard
electric force in ad d ition to the upw ard spring force and d ow nw ard gravitational
force. When the ball com es to equilibrium , w ith the spring stretched a d istance x , the
total upw ard force equals the m agnitud e of the d ow nw ard force, or qE kx mg .
The d istance betw een the un stretched position and the new equilibrium position of
the ball is
x
15.56
4.00 kg 9.80 m s2
mg qE
k
0.050 0 C 355 N C
845 N m
2.54 10
2
m
2.54 cm
(a) The d ow nw ard electrical force acting on the ball is
Fe
qE
2.00 10
6
C 1.00 105 N C
0.200 N
The total d ow nw ard force acting on the ball is then
F
Fe
mg
0.200 N+ 1.00 10-3 kg 9.80 m s 2
0.210 N
Thus, the ball w ill behave as if it w as in a m od ified gravitational field w here the
effective free-fall acceleration is
F
m
"g"
0.210 N
1.00 10-3 kg
210 m s 2
The period of the pend ulum w ill be
T
2
L
"g"
2
0.500 m
210 m s 2
0.307 s
28
CH APTER 15
(b)
Yes . The force of gravity is a significant portion of the total d ow nw ard force
acting on the ball. Without gravity, the effective acceleration w ould be
Fe
m
"g"
giving T
0.200 N
1.00 10-3 kg
0.500 m
200 m s 2
2
200 m s 2
0.314 s
a 2.28% d ifference from the correct value w ith gravity includ ed .
15.57
The sketch at the right gives a free-bod y d iagram of the
positively charged sphere. H ere, F1
ke q
2
r 2 is the attractive
force exerted by the negatively charged sphere and F2
is exerted by the electric field .
Fy
Fx
0
0
T cos10
F2
mg
cos10
mg or T
F1 +T sin10
qE
ke q
or qE
r2
2
mg tan10
At equilibrium , the d istance betw een the tw o spheres is r
E
ke q
4 L sin10
2
mg tan10
q
8.99 109 N m 2 C2 5.0 10
4 0.100 m sin10
8
C
2.0 10
2
or the need ed electric field strength is
15.58
2 L sin10 . Thus,
3
kg 9.80 m s 2 tan10
5.0 10
E
8
C
4.4 105 N C
(a) At any point on the x-axis in the range 0 x 1.00 m , the contributions m ad e to the
resultant electric field by the tw o charges are both in the positive x d irection. Thus,
it is not possible for these contributions to cancel each other and yield a zero field .
Electric Forces and Electric Fields
29
(b) Any point on the x-axis in the range x 0 is located closer to the larger m agnitud e
charge q 5.00 C than the sm aller m agnitud e charge q 4.00 C . Thus, the
contribution to the resultant electric field by the larger charge w ill alw ays have a
greater m agnitud e than the contribution m ad e by the sm aller charge. It is not
possible for these contributions to cancel to give a zero resultant field .
(c) If a point is on the x-axis in the region x 1.00 m , the contributions m ad e by the tw o
charges are in opposite d irections. Also, a point in this region is closer to the sm aller
m agnitud e charge than it is to the larger charge. Thus, th ere is a location in this
region w here the contributions of these charges to the total field w ill have equal
m agnitud es and cancel each other.
(d ) When the contributions by the tw o charges cancel each other, their m agnitud es
m ust be equal. That is,
ke
5.00 C
x
ke
2
4.00 C
x 1.00 m
2
x
Thus, the resultant field is zero at
15.59
4
x
5
or x 1.00 m
1.00 m
1
45
9.47 m
We assum e that the tw o spheres have equal charges, so the
repulsive force that one exerts on the other has m agnitud e
Fe ke q2 r 2 .
From Figure P15.59 in the textbook, observe that the d istance
separating the tw o spheres is
r 3.0 cm 2 5.0 cm sin10
4.7 cm 0.047 m
From the free-bod y d iagram of one sphere given above, observe
that
and
Fy
0
T cos10
Fx
0
Fe
mg or T
T sin10
mg
cos10
mg cos10°
sin10
mg tan10
30
CH APTER 15
Thus, ke q2 r 2
or
mg tan10
q
giving q 8.0 10
15.60
2
0.015 kg 9.8 m s 2 0.047 m tan10
mgr 2 tan10
ke
8
8.99 109 N m2 C2
C or q ~ 10
7
C
The charges on the spheres w ill be equal in m agnitud e and opposite in sign. From
F ke q2 r 2 , this charge m ust be
F r2
ke
q
1.00 104 N 1.00 m
2
8.99 109 N m2 C2
1.05 10
3
C
The num ber of electrons transferred is
n
q
e
1.05 10 3 C
1.60 10 19 C
6.59 1015
The total num ber of electrons in 100-g of silver is
N
47
electrons
atom
6.02 1023
atoms
mole
1 mole
100 g
107.87 g
2.62 1025
Thus, the fraction transferred is
n
N
15.61
6.59 1015
2.62 1025
2.51 10
10
(that is, 2.51 out of every 10 billion).
Because of the spherical sym m etry of the
charge d istribution, any electric field
present w ill be rad ial in d irection. If a field
d oes exist at d istance R from the center, it is
the sam e as if the net charge located w ithin
r R w ere concentrated as a point charge
at the center of the inner sphere. Charge
located at r R d oes not contribute to the
field at r R .
(a) At r 1.00 cm , E
0 since static
electric field s cannot exist w ithin
cond ucting m aterials.
Electric Forces and Electric Fields
(b) The net charge located at r 3.00 cm is Q
8.00 C .
Thus, at r 3.00 cm ,
E
ke Q
r2
8.99 109 N m 2 C 2 8.00 10
3.00 10
(c) At r 4.50 cm , E
2
m
6
C
2
7.99 107 N C outward
0 since this is located w ithin cond ucting m aterials.
(d ) The net charge located at r 7.00 cm is Q
4.00 C .
Thus, at r 7.00 cm,
E
ke Q
r2
8.99 109 N m 2 C 2 4.00 10
7.00 10
2
m
2
6
C
7.34 106 N C outward
31
32
CH APTER 15
15.62
Consid er the free-bod y d iagram of the rightm ost charge given below .
and
But,
0
T cos
Fx
0
Fe =T sin
ke q 2
r12
Fe
Thus,
If
Fy
15.63
ke q 2
r22
5ke q 2
4 L2 sin 2
or
T
mg cos
ke q 2
mg tan
sin
mg tan
ke q 2
2
L sin
mg cos
or q
2
2L sin
5ke q 2
4L2 sin 2
4L2 mg sin 2 tan
5ke
45 , m 0.10 kg, and L 0.300 m then
4 0.300 m
q
or
mg
2
0.10 kg 9.80 m s 2 sin 2 45 tan 45
5 8.99 109 N m 2 C2
q 2.0 10
6
C
2.0 C
(a) When an electron (negative charge) m oves d istance
electric field , the w ork d one on it is
W
Fe
x cos
eE
x cos180
From the w ork-energy theorem Wnet
eE
x
KEi , or E
KEi
e x
eE
KE f
x
KEi w ith KE f
1.60 10
1.60 10
19
x in the d irection of an
17
J
C 0.100 m
0 , w e have
1.00 103 N C
Electric Forces and Electric Fields
33
(b) The m agnitud e of the retard ing force acting on the electron is Fe eE , and
N ew ton’ s second law gives the acceleration as a
Fe m
eE m . Thus, the tim e
required to bring the electron to rest is
t
v v0
a
0
2 KEi m
2m KEi
eE m
eE
or
2 9.11 10
t
1.60 10
19
31
kg 1.60 10
17
J
3.37 10
3
C 1.00 10 N C
8
s
33.7 ns
(c) After bringing the electron to rest, the electric force continues to act on it causing the
electron to accelerate in the direction opposite to the field at a rate of
a
15.64
eE
m
1.60 10
19
C 1.00 103 N C
9.11 10-31 kg
1.76 1014 m s2
(a) The acceleration of the protons is d ow nw ard (in the d irection of the field ) and
ay
Fe
m
1.60 10
eE
m
19
C 720 N C
1.67 10
27
kg
6.90 1010 m s2
The tim e of flight for the proton is tw ice the tim e required to reach the peak of the
arc, or
t
2t peak
2
v0 y
2v0 sin
ay
ay
The horizontal d istance traveled in this tim e is
R v0 x t
v0 cos
2v0 sin
v02 sin 2
ay
ay
34
CH APTER 15
Thus, if R 1.27 10 3 m , w e m ust have
6.90 1010 m s 2 1.27 10
ay R
sin 2
2
0
v
giving 2
9 550 m s
73.9 or 2
180
73.9
3
m
0.961
2
106.1° . H ence,
37.0 or 53.0°
(b) The tim e of flight for each possible angle of projection is:
For
37.0 :
t
2v0 sin
2 9 550 m s sin37.0
ay
6.90 1010 m s2
For
53.0 :
t
2v0 sin
2 9 550 m s sin53.0
ay
10
6.90 10
ms
2
1.66 10
7
s
2.21 10
7
s