Download Chapter 23 Energy is conserved, so the change in potential energy

Chapter 23
1. Energy is conserved, so the change in potential energy is the opposite of the change in kinetic
energy. The change in potential energy is related to the change in potential.
U  qV  K 
V 
K
q

K initial  K final
q

mv 2
2q
 9.11  10 kg 5.0  10 m s 

2  1.60  10 C 
31
5
19
2
 0.71V
The final potential should be lower than the initial potential in order to stop the electron.
2. The work done by the electric field can be found from Eq. 23-2b.
Vba  
Wba
q


 Wba   qVba   1.60  1019 C  55V  185V   3.84  1017 J
3. The kinetic energy gained by the electron is the work done by the electric force. Use Eq. 23-2b
to calculate the potential difference.
Vba  
Wba
q

5.25  1016 J
 1.60  10
19
C

 3280 V
The electron moves from low potential to high potential, so plate B is at the higher potential.
4. By the work energy theorem, the total work done, by the external force and the electric field
together, is the change in kinetic energy. The work done by the electric field is given by Eq. 232b.
Wexternal  Welectric  KE final  KEinitial  Wexternal  q Vb  Va   KEfinal 
Vb  Va  
Wexternal  KE final
q

7.00  104 J  2.10  104 J
 53.8 V
9.10  106C
Since the potential difference is negative, we see that
Va  Vb
.
6. The distance between the plates is found from Eq. 23-4b, using the magnitude of the electric field.
E 
Vba
d
 d
Vba
E

45 V
1300 V m
 3.5  102 m
14. (a) The potential at the surface of a charged sphere is derived in Example 23-4.
V0 

Q
4 0 r0
Q
Area

 Q  4 0 rV

0 0
Q
4 r
2
0

4 0 rV
0 0

4 r
2
0
V0 0
r0
 680 V  8.85  1012

 0.16 m 
C2 /Nm 2
  3.761  10
8
C m2
 3.8  108 C m 2
(b) The potential away from the surface of a charged sphere is also derived in Example 23-4.
V
Q
4 0 r

4 0 rV
0 0

4 0 r
rV
0 0
 r
r
rV
0 0
V

 0.16 m 680 V 
 4.352 m 
 25V 
4.4 m
22. Because of the spherical symmetry of the problem, the electric field in each region is the same as
that of a point charge equal to the net enclosed charge.
r  r2 : E 
(a) For
For
1 Qencl
4 0 r
r1  r  r2 : E  0 ,
0  r  r1 : E 
For
(b) For
r  r2
2

3
2
1
Q
4 0 r
2

3 Q
8 0 r 2
because the electric field is 0 inside of conducting material.
1 Qencl
4 0 r
2

1
2
1
Q
4 0 r
2

1 Q
8 0 r 2
, the potential is that of a point charge at the center of the sphere.
V
1 Q
4 0 r

1
3
2
Q
4 0 r

3 Q
8 0 r
, r  r2
r  r  r2
(c) For 1
, the potential is constant and equal to its value on the outer shell, because
there is no electric field inside the conducting material.
V  V  r  r2  
(d) For
0  r  r1
E d l  Edr.
3 Q
8 0 r2
, r1  r  r2
, we use Eq. 23-4a. The field is radial, so we integrate along a radial line so that
r
r
r
Vr  Vr    E d l    E dr   
1
r1
Vr  Vr 
1
r1
r1
1 Q
8 0 r
E E0
For
Q 1
1
 

8 0  r r1 
1
Q  1 1
Q  1 1
 
 


   , 0  r  r1
8 0  r r1  8 0  2 r1 r  8 0  r2 r 
(e) To plot, we first calculate
plot
and
dr 
Q 1
V0  V  r  r2  
V V0
2
as functions of
0  r  r1
3Q
8 0 r2
E0  E  r  r2  
and
For
1
3
 r r2 2
:
3 Q
3 Q
V 8 0 r r2
E 8 0 r 2 r22
1
2

   r r2  ;

 2   r r2 
3Q
3Q
V0
r
E0
r
2
8 0 r2
8 0 r2
5.0
4.0
3.0
V /V 0
Then we
3 Q
V 8 0 r2
E
0

1;

0
3Q
3Q
V0
E0
8 0 r2
8 0 r22
:
2.0
1.0
0.0
0.00
.
:
r1  r  r2
r  r2
8 0 r22
r r2 .
Q  1 1
1 Q
 

V 8 0  r2 r  1
E 8 0 r 2 1 r22
1

 3 1   r r2   ;

3 2 
3Q
3Q
V0
E0
r
8 0 r2
8 0 r22
For
3Q
0.25
0.50
0.75
1.00
r /r 2
1.25
1.50
1.75
2.00
5.0
4.0
E /E 0
3.0
2.0
1.0
0.0
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
r /r 2
26. (a) Because of the inverse square nature of the electric
field, any location where the field is zero must be closer to the
x
d
q2  0
q1  0
q
weaker charge  2  . Also, in between the two charges, the
fields due to the two charges are parallel to each other (both to the left) and cannot cancel. Thus the
only places where the field can be zero are closer to the weaker charge, but not between them. In the
diagram, this is the point to the left of
E
x
q2
1
4 0 x
2

q2
q1 
1
q1
4 0  d  x 
d
q2
2
q2 .
Take rightward as the positive direction.
q2  d  x   q1 x 2 
0 
2
2.0  106 C
3.4  106 C  2.0  106 C
 5.0cm  
16 cm left of q2
(b) The potential due to the positive charge is positive
d
x1
x2
everywhere, and the potential due to the negative charge is
negative everywhere. Since the negative charge is smaller in
q2  0
q1  0
magnitude than the positive charge, any point where the
potential is zero must be closer to the negative charge. So consider locations between the charges
(position
x1
) and to the left of the negative charge (position
Vlocation 1 
Vlocation 2 
x2
1 
q1

4 0   d  x1 

q2 
  0  x1
x1 
) as shown in the diagram.
 2.0  10 C  5.0 cm   1.852 cm


q  q 
 5.4  10 C 
6
q2 d
6
2
q1
q 
 20 

4 0   d  x2  x2 
x2
1
1 
 2.0  10 C  5.0 cm   7.143cm


q  q 
1.4  10 C 
6
q2 d
6
1
2
So the two locations where the potential is zero are 1.9 cm from the negative charge towards the
positive charge, and 7.1 cm from the negative charge away from the positive charge.
30. By energy conservation, all of the initial potential energy of the charges will change to kinetic energy
when the charges are very far away from each other. By momentum conservation, since the initial
momentum is zero and the charges have identical masses, the charges will have equal speeds in
opposite directions from each other as they move. Thus each charge will have the same kinetic energy.
1 Q2
Einitial  Efinal  U initial  K final 
v
1 Q2
4 0 mr
4 0 r
2

1
2
mv 2
8.99  10 N m C 5.5  10 C 
1.0  10 kg   0.065 m 
9

2
6
2
6


2
 2.0  103 m s
34. The potential at the corner is the sum of the potentials due to each of the charges, using Eq. 23-5.
V
1
 3Q 
4 0
l

1
Q
4 0
2l

1
 2Q 
4 0
l

1 Q
1 
1
2Q
1 

4 0 l 
2  4 0 2 l


2 1
40. For both parts of the problem, use Eq. 23-6b to find the potential of a continuous charge
distribution. Choose a differential element of length dx at position x along the rod. The charge on

 
the element is dq   dx  ax dx .
(a) The element is a distance
V
r
1
4 0
y
x2  y 2

dq
r

from a point on the y axis.
1
4 0
l

l
axdx
x2  y 2
r
0
x
l
l
dx
The integral is equal to 0 because the region of integration is “even” with
respect to the origin, while the integrand is “odd.” Alternatively, the antiderivative can be found, and
the integral can be shown to be 0. This is to be expected since the potential from points symmetric
about the origin would cancel on the y axis.
x
E   2 0 ,
43. The electric field from a large plate is uniform with magnitude
with the field pointing
away from the plate on both sides. Equation 23-4(a) can be integrated between two arbitrary points to
calculate the potential difference between those points.
x1
V   
x0

 ( x0  x1 )
dx 
2 0
2 0
Setting the change in voltage equal to 100 V and solving for
lines.
x0  x1 
2 0 V


x0  x1

2 8.85  1012 C2 /Nm2 100 V 

6
0.75  10 C/m2
gives the distance between field
 2.36  103 m  2 mm
e
54. Let the side length of the equilateral triangle be L. Imagine bringing the
electrons in from infinity one at a time. It takes no work to bring the first
electron to its final location, because there are no other charges present.
e
l
l
l
W 0
Thus 1
. The work done in bringing in the second electron to its final
location is equal to the charge on the electron times the potential (due to the

e
1 e
W2   e   

1 e2
 4 0 l  4 0 L . The
first electron) at the final location of the second electron. Thus
work done in bringing the third electron to its final location is equal to the charge on the electron times

W3   e   
 4 0 l
the potential (due to the first two electrons). Thus
work done is the sum
W1  W2  W3
W  W1  W2  W3  0 
1 e2
4 0 l

1 e

1 e

4 0 l 
1 2e 2
4 0 l
. The total
.
1 2e 2
4 0 l

1 3e 2
4 0 l



3 8.99  109 N m 2 C2 1.60  1019 C
1.0  10
10
m


2
1eV

  43eV
19 
 1.60  10 J 
 6.9  1018 J  6.9  1018 J 
55. The gain of kinetic energy comes from a loss of potential energy due to conservation of energy, and
the magnitude of the potential difference is the energy per unit charge. The helium nucleus has a
charge of 2e.
V 
U
q

K
q

125  103 eV
2e
 62.5kV
The negative sign indicates that the helium nucleus had to go from a higher potential to a lower
potential.
85.
(a) The voltage at x  0.20 m is obtained by inserting the given data directly into the voltage
equation.
B
V  0.20 m  
x
2
 R2

2

150 V m 4
 0.20 m  2   0.20 m  2 


2
 23 kV
(b) The electric field is the negative derivative of the potential.
E x  
d 

B
4 Bx ˆi

 ˆi 
2
2
2 3
dx   x 2  R 2  
x

R




Since the voltage only depends on x the electric field points in the positive x direction.
(c) Inserting the given values in the equation of part (b) gives the electric field at x  0.20 m
E(0.20 m) 

4 150 V m 4
  0.20 m  ˆi
 0.20 m    0.20 m  


2
2
3
 2.3  105 V m ˆi
q
86. Use energy conservation, equating the energy of charge 1 at its initial position to its final position
at infinity. Take the speed at infinity to be 0, and take the potential of the point charges to be 0 at
infinity.
Einitial  Efinal  K initial  U initial  K final  U final 
1
2
2
mv02    q1 Vinitial  12 mvfinal
   q1 Vfinal
point
1
2
mv02    q1 
1
2 q2
4 0
a b
2
2
 0  0  v0 
1
m 0
q1q2
a 2  b2
point