CHAPTER 19 ELECTRIC POTENTIAL ENERGY
AND THE ELECTRIC POTENTIAL
PROBLEMS
_____________________________________________________________________________________________
1.
SSM
REASONING AND SOLUTION Combining Equations 19.1 and 19.3, we have
WAB = EPE A – EPE B = q 0 (VA – V B ) = ( +1.6 ´ 10 –19 C)(0.070 V) = 1.1´ 10 –20 J
_____________________________________________________________________________________________
2.
REASONING The change in the proton’s electric potential energy, EPEA - EPEB, is equal to the
work WAB done by the electric force as the proton moves from point A to point B (see Equation 19.1).
According to Equation 6.1, the work done by a constant force is the product of the magnitude F of the
force, the magnitude s of the proton’s displacement, and the cosine of the angle between the force and
displacement; WAB = F s cos q. Since the magnitude of the electric force is equal to product of the
proton charge q and the magnitude E of the electric field, the work can be written as WAB = (qE) s cos
q.
SOLUTION
a. The displacement of the proton is in the same direction as the electric force, so the angle between
them is q = 0.0°:
EPE A - EPE B = W AB = (qE )s cosq
(
)(
)
= 1.6 ´ 10 -19 C 2.0 ´ 10 4 N/C (0.15 m )cos 0.0° = 4.8 ´ 10 -16 J
b. When the proton moves opposite the electric field, the angle between the proton’s displacement
and the electric force is q = 180°, so that cos 180° = -1. The solution proceeds in the same manner as
in part (a), except that the angle is 180°. Therefore, the change in the electric potential energy is
-4.8 ´ 10 -16 J
______________________________________________________________________________________________
3.
REASONING The number N of electrons that jumps from your hand (point A) to the door knob
(point B) is equal to the total charge q that jumps divided by the charge -e of one electron: N =
q/(-e), where e = 1.6 ´ 10-19 C. We can determine q by using Equation 19.4, which relates the work
WAB done by the electric force to the difference in the electric potentials, VB - VA, and the charge.
VB
VA
The
difference
in
the
potentials
is
given
as
=
2.0 ´ 104 V.
SOLUTION The number of electrons that jumps from your hand to the door knob is
578 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
N=
q
=
-e
- WAB
VB - V A
-e
-1.5 ´ 10 -7 J
2.0 ´ 10 4 V
=
=
-1.6 ´ 10 -19 C
4.7 ´ 10 7
______________________________________________________________________________________________
4.
REASONING AND SOLUTION
a. According to Equation 19.4, the work done by the electric force as the electron goes from point A
(the cathode) to point B (the anode) is
WAB = - q ( V B - VA ) = - ( -1.6 ´ 10 -19 C)(+125 000V ) = +2.00 ´ 10 -14 J
b. The only force that acts on the electron is the conservative electric force. Therefore, the total
energy of the electron is conserved as it moves from point A to point B:
1
1
2
2
2 mv A + EPE A = 2 mv B + EPE B
1442443 1442443
Total energy at
point A
Total energy at
point B
Since the electron starts from rest, vA = 0. The electric potential V is related to the electric potential
energy EPE by V = EPE/q (see Equation 19.3). With these changes, the equation above gives the
kinetic energy of the electron at point B (the anode) to be
1 mv 2
B
2
= - EPE B + EPE A
(
)
= - q (VB - V A )= - -1.60 ´ 10 -19 C (125 000 V) = 2.00 ´ 10 -14 J
______________________________________________________________________________________________
5.
REASONING The only force acting on the moving electron is the conservative electric
force. Therefore, the total energy of the electron (the sum of the kinetic energy KE and the electric
potential energy EPE) remains constant throughout the trajectory of the electron. Let the subscripts A
and B refer to the initial and final positions, respectively, of the electron. Then,
1
1
2
2
mvA + EPE A = mvB + EPE B
SSM
2
2
Solving for vB gives
2
vB = vA –
2
(EPE B – EPE A )
m
Since the electron starts from rest, vA = 0. The difference in potential energies is related to the
difference in potentials by Equation 19.4, EPE B - EPE A = q(VB - VA ) .
Chapter 19
579
SOLUTION The speed vB of the electron just before it reaches the screen is
vB =
2q
– (VB – VA ) =
m
2(–1.6 ´ 10 -19 C)
(25 000 V) = 9.4 ´ 10 7 m/s
–
–31
9.11 ´ 10
kg
_____________________________________________________________________________________________
6. REASONING AND SOLUTION The only force that acts on the a-particle is the conservative electric
force. Therefore, the total energy of the a-particle is conserved as it moves from point A to point B:
+ EPE A = 12 mv 2B + EPE B
1442443 1442443
1 mv 2
A
2
Total energy at
point B
Total energy at
point A
Since the a-particle starts from rest, vA = 0. The electric potential V is related to the electric potential
energy EPE by V = EPE/q (see Equation 19.3). With these changes, the equation above gives the
kinetic energy of the a-particle at point B to be
1 mv 2
B
2
= EPE A - EPE B = q (VA - V B )
Since an a-particle contains two protons, its charge is q = 2e = 3.2 ´ 10-19 C. Thus, the kinetic
energy (in electron-volts) is
1
mv B2
2
(
)
= q (V A - VB ) = 3.2 ´ 10 -19 C [+ 250 V - (- 150 V )]
æ 1.0 eV ö
= 1.28 ´ 10 -16 J ç
= 8.0 ´ 10 2 eV
-19 ÷
Jø
è 1.6 ´ 10
______________________________________________________________________________________________
7.
REASONING AND SOLUTION The power rating P is defined as the work WAB done by the battery
divided by the time t,
P=
WAB
t
The work done by the electric force as the charge moves from point A (the positive terminal), through
the electric motor, and to point B (the negative terminal) is
WAB = q (VA - V B )= (1300 C )(320 V ) = 4.2 ´ 10 5 J
The power rating is
P=
WAB
t
=
4.2 ´ 10 5 J
= 5.2 ´ 10 4 W
8.0 s
(19.4)
580 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
Since 746 W = 1 hp, the minimum horsepower rating of the car is
1 hp
=
(5.20 ´ 10 4 W )746
W
7.0 ´10 1 hp
______________________________________________________________________________________________
8.
REASONING AND SOLUTION As the charge in the lightning bolt moves from point A to point B,
the change in its electric potential energy is
(
)
EPE A - EPE B = q (V A - VB )= (29 C ) 1.4 ´ 10 8 V = 4.1 ´ 10 9 J
If this electric energy were converted into heat Q, the mass m of liquid water at 100 °C that is
converted into steam is given by m = Q/Lv (see Section 12.8), where Lv is the latent heat of
vaporization (Lv = 22.6 ´ 105 J/kg). The mass of water that is boiled away is
m =
Q
4.1 ´ 10 9 J
=
= 1.8 ´ 10 3 kg
5
Lv
22.6 ´ 10 J/kg
______________________________________________________________________________________________
9.
REASONING The only force acting on the moving charge is the conservative
electric force. Therefore, the total energy of the charge remains constant. Applying the principle of
conservation of energy between locations A and B, we obtain
SSM
WWW
1
2
mvA
2
+ EPE A =
1
2
2
mvB + EPE B
Since the charged particle starts from rest, vA = 0. The difference in potential energies is related to
the difference in potentials by Equation 19.4, EPE B - EPE A = q(VB - VA ). Thus, we have
1
(1)
q(VA - VB ) = mv 2B
2
Similarly, applying the conservation of energy between locations C and B gives
1
2
2
q(VC - VB ) = m(2vB )
Dividing Equation (1) by Equation (2) yields
VA – VB 1
=
VC – VB 4
This expression can be solved for VB .
SOLUTION Solving for VB , we find that
VB =
4VA – VC 4(452 V) – 791 V
=
= 339 V
3
3
(2)
Chapter 19
581
_____________________________________________________________________________________________
10. REASONING AND SOLUTION The electron, starting from rest at point A, accelerates to point B.
The only force acting on the electron is the conservative electric force. Therefore, the total energy of
the electron remains constant. Applying the principle of conservation of energy between locations A
and B, we obtain
1
mv2A
2
+ EPE A =
1
2
mv2B + EPE B
Solving for vB, the speed of the electron at point B, yields
= v 2A +
v electron
B
2
(EPE A - EPE B )
m
Since the electron starts from rest, vA = 0. The difference in potential energies is related to the
difference in potentials by Equation 19.4, EPE A - EPE B = q(V A - VB ) . Thus, the speed of the
electron at point B is
v electron
B
=
2q (VA - V B )
m
=
(
)
2 -1.6 ´ 10 -19 C (V A - VB )
9.11´ 10 -31 kg
Since the electron accelerated from point A to point B, the proton, being positively charged,
accelerates from B to A. Using the conservation of energy, the speed of the proton at point A is
2
2
v proton
v
=
(EPE A - EPE B )
A
B
m
Since, vB = 0 (the proton starts from rest), and EPEA - EPEB = q(VA - VB),
proton
vA
=
-2 q(V A - VB )
m
=
(
)
-2 +1.6 ´ 10 -19 C (V A - VB )
1.67 ´ 10 -27 kg
The ratio of the electron speed at point B to the proton speed at point A is
(
)
2 -1.6 ´ 10 -19 C (V A - VB )
v electron
B
v Aproton
=
9.11 ´ 10
(
-31
)
kg
- 2 1.6 ´ 10 -19 C (VA - V B )
=
42.8
1.67 ´ 10 -27 kg
______________________________________________________________________________________________
11. REASONING The gravitational and electric forces are conservative forces, so the total energy of
the particle remains constant as it moves from point A to point B. Recall from Equation 6.5 that the
582 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
gravitational potential energy (GPE) of a particle of mass m is GPE = mgh, where h is the height of
the particle above the earth’s surface. The conservation of energy is written as
1
2
2 mv A
+ mgh A + EPE A =
1
2
2 mv B
+ mgh B + EPE B
(1)
We will use this equation several times to determine the initial speed vA of the negatively charged
particle.
SOLUTION
When the negatively charged particle is thrown upward, it attains a maximum height of h. For this
particle we have:
vA = ?
vB = 0 (at maximum height)
EPEB = (-q)VB
hB = h (the maximum height)
EPEA = (-q) VA
hA = 0 (ground level)
Solving the conservation of energy equation, Equation (1), for vA and substituting in the data above
gives
(2)
2
vA =
mgh + (- q )(VB - V A )
m
[
]
Equation (2) cannot be solved as it stands because the height h and the potential difference (VB - VA)
are not known.
We now make use of the fact that a positively charged particle, when thrown straight upward with an
initial speed of 30.0 m/s, also reaches the maximum height h. For this particle we have:
vA = 30.0 m/s
vB = 0 (at maximum height)
EPEA = (+q) VA
EPEB= (+q)VB
hB = h
hA = 0 (ground level)
Solving the conservation of energy equation, Equation (1),
(VB - VA) and substituting in the data above gives
VB - V A =
1
+q
[ m (30.0 m/s)
1
2
2
- mgh
]
for the potential difference
(3)
Substituting Equation (3) into Equation (2) gives, after some algebraic simplifications,
v A = 4gh - (30.0 m/s )
2
Equation (4) cannot be solved because the height h is still unknown.
(4)
Chapter 19
583
We now make use of the fact that the uncharged particle, when thrown straight upward with an initial
speed of 25.0 m/s, also reaches the maximum height h. For this particle we have:
vA = 25.0 m/s vB = 0 (at maximum height)
EPEA = qVA = 0 (since q = 0)
hA = 0 (ground level)
EPEB = qVB = 0 (since q = 0)
hA = h
Solving Equation (1) with this data for the maximum height h yields
h=
(25.0 m/s )2
2g
=
(25.0 m/s )2
(
2 9.80 m/s 2
)
= 31.9 m
Substituting h = 31.9 m into Equation (4) gives vA = 18.7 m/s .
______________________________________________________________________________________________
12. REASONING AND SOLUTION First, we need to find the magnitude of the charge. Since
V = kq/r, we have
q=
Vr
(164 V )(0.20 m)
=
= 3.6 ´ 10 -9 C
k
9.0 ´ 10 9 N ×m 2 /C 2
Thus, at a distance of 0.80 m the potential would be
V=
kq
(9.0 ´ 10 9 N ×m 2 /C 2 )(3.6 ´ 10 -9 C)
=
= 41V
r
0.80 m
______________________________________________________________________________________________
13.
REASONING AND SOLUTION The electric potential V at a distance r from a point charge
q is given by Equation 19.6, V = kq / r . Solving this expression for q, we find that
(0.25 m)(+130 V)
rV
=
= +3.6 ´ 10 – 9 C
q=
k
9.0 ´ 10 9 N × m 2 / C2
SSM
______________________________________________________________________________________________
14. REASONING The potential V at a distance r from a proton is V = k(+e)/r (see Equation 19.6), where
+e is the charge of the proton. When an electron (q = -e) is placed at a distance r from the proton, the
electric potential energy is EPE = -eV, as per Equation 19.3.
SOLUTION The difference in the electric potential energies when the electron and proton are
separated by rfinal = 5.29 ´ 10-11 m and when they are very far apart (rinitial = ¥) is
584 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
EPE final - EPE initial =
(- e)ke (- e )ke
r
r
final
initial
(
= - (8.99 ´ 10 9 N × m 2 / C 2 ) 1.60 ´ 10 -19 C
æ
1
1ö
´ çç
- ÷÷ =
è 5.29 ´ 10 -11 m ¥ ø
)
2
-4.35 ´ 10 -18 J
______________________________________________________________________________________________
15. REASONING AND SOLUTION The initial and final separations of the charges are the same. Since
the electric potential energy depends only on the separation of the charges, there is no change in the
electric potential energy, and, hence,
no work is done .
______________________________________________________________________________________________
16. REASONING AND SOLUTION The initial and final electric potential energies of the configuration
are, respectively,
EPE A =
kq1 q 2
EPE B =
and
rA
kq1 q2
rB
Since EPEB = 2 EPEA, we have
k q1 q2
rB
rB =
or
rA
2
=
= 2
kq1 q2
0.74 m
=
2
rA
0.37 m .
______________________________________________________________________________________________
17.
REASONING Initially, suppose that one
charge is at C and the other charge is held fixed at B. The
charge at C is then moved to position A. According to
Equation 19.4, the work WCA done by the electric force as
the charge moves from C to A is WCA = q(VC - VA ) ,
where, from Equation 19.6, VC = kq / d and VA = kq / r .
From the figure at the right we see that
SSM
2
2
d = r + r = 2r . Therefore, we find that
æ kq kq ö kq 2 æ 1
ö
=
=
WCA q
–
–1
è
è 2r r ø
r
2 ø
C
A
WWW
d
r
B
r
Chapter 19
585
SOLUTION Substituting values, we obtain
(8.99 ´ 109 N × m 2 / C2 )(3.0 ´ 10 – 6 C)2 æ 1
– 1ö = –4.7 ´ 10 –2 J
WCA =
è
ø
0.500 m
2
______________________________________________________________________________________________
18. REASONING Each object contributes an electric potential at point A given by Equation 19.6 as V =
kq/r, where q is the charge and r is the distance between the object and point A. The total electric
potential Vtotal at point A due to the three charges is the sum of the individual potentials. When a
charge q0 is placed at point A, the electric potential energy is EPE = q0 Vtotal, as per Equation 19.3.
SOLUTION
(a) The total electric potential at point A is
Vtotal =
kq1
r1
+
kq 2
r2
+
kq 3
r3
æ +3.6 ´ 10 -6 C -20.0 ´ 10 -6 C +6.0 ´ 10 -6 C ö
= (9.0 ´ 10 9 N × m 2 / C 2 )ç
+
+
÷
è 2.0 ´ 10 -2 m
4.0 ´ 10 -2 m
3.0 ´ 10 -2 m ø
= - 1.1 ´ 10 6 V
(b) When a +3.5 ´ 10-6 C charge is placed at A, the electric potential energy is
(
)(
)
EPE = q 0 Vtotal = +3.5 ´ 10 -6 C -1.1´ 10 6 V = -3.8 J
______________________________________________________________________________________________
19. REASONING AND SOLUTION Let s be the length of the side of the square and Q be the value of
the unknown charge. The potential at either of the vacant corners is
V =0=
k (9q ) k (- 8q )
kQ
+
+
æ s ö
s
s
ç
÷
è 2ø
so
Q =
–q
2
______________________________________________________________________________________________
20. REASONING AND SOLUTION Let the first spot where the potential is zero be a distance x to the
left of the negative charge. Then,
k (2q)
kq
=
d-x
x
or
x=
d
3
586 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
Let the second spot where the potential is zero be a distance x to the right of the negative charge.
Then,
k (2q)
kq
=
or
x= d
d+x
x
______________________________________________________________________________________________
21.
REASONING The only force acting on the moving charge is the conservative electric force.
Therefore, the sum of the kinetic energy KE and the electric potential energy EPE is the same at
points A and B:
SSM
1
2
mvA
2
+ EPE A =
1
2
2
mvB + EPE B
Since the particle comes to rest at B, vB = 0 . Combining Equations 19.3 and 19.6, we have
æ kq ö
EPE A = qVA = q ç 1 ÷
è d ø
and
æ kq ö
EPE B = qVB = q ç 1 ÷
è r ø
where d is the initial distance between the fixed charge and the moving charged particle, and r is the
distance between the charged particles after the moving charge has stopped.
Therefore, the
expression for the conservation of energy becomes
1
mv A2
2
+
kqq1
d
=
kqq1
r
This expression can be solved for r. Once r is known, the distance that the charged particle moves
can be determined.
SOLUTION Solving the expression above for r gives
r=
=
kqq1
kqq1
1
mvA2 +
2
d
(8.99 ´ 10 9 N ×m 2 / C 2 )(–8.00 ´ 10 – 6 C)(–3.00 ´10 – 6 C)
(8.99 ´10 9 N ×m 2 / C 2 )(–8.00 ´ 10 – 6 C)(–3.00 ´ 10 – 6 C)
1
(7.20 ´ 10 –3 kg)(65.0 m/s) 2 +
2
0.0450 m
= 0.0108 m
Therefore, the charge moves a distance of 0.0450 m – 0.0108 m = 0.0342 m .
______________________________________________________________________________________________
Chapter 19
587
22. REASONING Initially, the four charges are infinitely far apart. We will proceed as in Example 8 by
adding charges to the square, one at a time, and determining the electric potential energy at each step.
According to Equation 19.3, the electric potential energy EPE is the product of the charge q and the
electric potential V at the spot where the charge is placed, EPE = qV. The total electric potential
energy of the group is the sum of the energies of each step in assembling the group.
SOLUTION Let the corners of the square be numbered clockwise as 1, 2, 3 and 4, starting with the
top-left corner. When the first charge (q = 5.0 µC) is placed at a corner 1, the charge has no electric
potential energy, EPE1 = 0. This is because the electric potential V1 produced by the other three
charges at corner 1 is zero, since they are infinitely far away.
Once q is in place, the electric potential V2 that it creates at corner 2 is
V2 =
kq
r21
where r21 = 0.50 m is the distance between corners 2 and 1. When the second charge is placed at
corner 2, its electric potential energy EPE2 is
æ kq ö
EPE 2 = qV2 = q ç
÷ =
è r21 ø
(9.0 ´ 10 9 N × m 2 /C 2 )(5.0 ´ 10 -6 C )2
0.50 m
= 0.45 J
The electric potential V3 at the third corner is the sum of the potentials due to the two charges that are
already in place on corners 1 and 2:
V3 =
kq kq
+
r31 r32
where r31 = 0.71 m, and r32 = 0.50 m. When the third charge is placed at corner 3, the electric
potential energy EPE3 is
æ kq
æ 1
kq ö
1 ö
+
+
EPE 3 = qV3 = q ç
÷ = k q2 ç
÷
è r31 r32 ø
è r31 r32 ø
(
= 9.0 ´ 10 9 N × m 2 /C 2
æ
ö
)(5.0 ´ 10 -6 C )2 çè 0.711 m + 0.501 m ÷ø
= 0.77 J
The electric potential V4 at the fourth corner is the sum of the potentials due to the three charges that
are already in place on corners 1, 2, and 3:
588 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
V4 =
kq
kq
kq
+
+
r41 r42 r 43
where r41 = 0.50 m, r42 = 0.71 m, and r43 = 0.50 m. When the fourth charge is placed at corner 4, the
electric potential energy EPE4 is
æ kq kq
æ 1
kq ö
1
1 ö
EPE 4 = qV 4 = q ç
+
+
+
+
÷ = kq 2 ç
÷
è r41 r42 r43 ø
è r 41 r42 r43 ø
(
= 9.0 ´ 10 9 N ×m 2 /C 2
æ
ö
)(5.0 ´ 10 -6 C )2 çè 0.501 m + 0.711 m + 0.501 m ÷ø
= 1.2 J
The electric potential energy of the entire array is
EPE = EPE 1 + EPE 2 + EPE 3 + EPE 4 = 0 + 0.45 J + 0.77 J + 1.2 J =
2.4 J
______________________________________________________________________________________________
23. REASONING The only force acting on each proton is the conservative electric force. Therefore, the
total energy (kinetic energy plus electric potential energy) is conserved at all points along the motion.
For two points, A and B, the conservation of energy is
+ 12 mv A2 + EPE A = 12 mv 2B + 12 mv 2B + EPE B
1442443 123
1
2
2 mv A
Kinetic energy of
the two protons
Electric
potential
energy
The electric potential energy of two protons (charge = +e) that are separated by a distance r is EPE =
ke2/r. By substituting this expression for EPE into the conservation of energy, we will be able to
determine the distance of closest approach.
SOLUTION When the protons are very far apart (rA = ¥), so that EPEA = 0. At the distance rB of
closest approach, the speed of each proton is momentarily zero (vB = 0). With these substitutions, the
conservation of energy equation reduces to
1 mv 2
A
2
+ 12 mv A2 =
ke 2
rB
Solving for rB, the distance of closest approach is
(
)(
9.0 ´ 10 9 N ×m 2 / C 2 1.6 ´ 10 -19 C
ke 2
rB =
=
2
mv 2A
1.67 ´ 10 -27 kg 1.5 ´ 10 6 m/s
(
)(
)
)2
= 6.1 ´ 10 -14 m
Chapter 19
589
______________________________________________________________________________________________
24. The figure at the right shows two identical charges, q,
fixed to diagonally opposite corners of a square. The
potential at corner A is caused by the presence of the two
charges. It is given by
(V A ) 0 =
A
C
d
r
kq kq 2kq
+
=
r
r
r
B
r
Since both charges are the same distance from corner B,
this is equal to the potential at corner B as well.
If a third charge, q3, is placed at the center of the square, the potential at corner A (as well as
corner B) becomes
kq 3
kq kq
(VA ) f =
+
+
r
r (d / 2)
We can find d by applying the Pythagorean theorem to the geometry in the figure above:
d = r2 + r2 =
2r
so that
d
=
2
r
2
The expression for (VA)f becomes
(VA ) f =
kq kq kq 3 2
+
+
.
r
r
r
From the problem statement, we know that the addition of q3 causes the potential at A and B to
change sign without changing magnitude.
In other words, (VA)f = -(VA)0, or
kq kq kq 3 2
2 kq
+
+
= r
r
r
r
Solving for q3 gives
q 3 = - (2 2 )q = - (2 2 )(1.7 ´ 10 –6 C) = - 4.8 ´ 10 –6 C
______________________________________________________________________________________________
25.
REASONING Initially, the three charges are infinitely far apart. We will proceed as in
Example 8 by adding charges to the triangle, one at a time, and determining the electric potential
energy at each step. According to Equation 19.3, the electric potential energy EPE is the product of
the charge q and the electric potential V at the spot where the charge is placed, EPE = qV. The total
electric potential energy of the group is the sum of the energies of each step in assembling the group.
SSM
590 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
SOLUTION Let the corners of the triangle be numbered clockwise as 1, 2 and 3, starting with the
top corner. When the first charge (q1 = 8.00-µC) is placed at a corner 1, the charge has no electric
potential energy, EPE1 = 0. This is because the electric potential V1 produced by the other two
charges at corner 1 is zero, since they are infinitely far away.
Once the 8.00-mC charge is in place, the electric potential V2 that it creates at corner 2 is
V2 =
kq1
r21
where r21 = 5.00 m is the distance between corners 1 and 2, and q1 = 8.00 mC. When the 20.0-mC
charge is placed at corner 2, its electric potential energy EPE2 is
æ kq ö
EPE 2 = q2 V2 = q 2 ç 1 ÷
è r 21 ø
(
= 20.0 ´
10 -6
(
)(
)
é 8.99 ´ 10 9 N ×m 2 /C 2 8.00 ´ 10 -6 C ù
ú = 0.288 J
C ê
5.00 m
êë
úû
)
The electric potential V3 at the remaining empty corner is the sum of the potentials due to the two
charges that are already in place on corners 1 and 2:
V3 =
kq1
r31
+
kq 2
r32
where q1 = 8.00 mC, r31 = 3.00 m, q2 = 20.0 mC, and r32 = 4.00 m. When the third charge
(q3 = -15.0 mC) is placed at corner 3, its electric potential energy EPE3 is
æ kq
æq
kq ö
q ö
EPE 3 = q3 V3 = q 3 ç 1 + 2 ÷ = q3 k ç 1 + 2 ÷
è r31 r32 ø
è r31 r32 ø
æ 8.00 ´ 10 -6 C 20.0 ´ 10 -6 C ö
= -15.0 ´ 10 -6 C 8.99 ´ 10 9 N ×m 2 /C 2 ç
+
÷ = - 1.034 J
3.00 m
4.00 m
è
ø
(
)(
)
The electric potential energy of the entire array is given by
EPE = EPE 1 + EPE 2 + EPE 3 = 0 + 0.288 J + (-1.034 J) = -0.746 J
______________________________________________________________________________________________
26. REASONING AND SOLUTION The information about the electric field requires that
Chapter 19
kq2/(1.00 m)2 = kq1/(4.00 m)2
so
591
q2 = (1/16.0) q1
Let x be the distance of the zero-potential point from the negative charge. Then
kq1/(d + x) = kq2/x
if the point is to the right of q2 and
kq1/(d - x) = kq2/x
if the point is to the left of q2. Solving for x gives
x = 0.200 m to the right of the negative charge
x = 0.176 m to the left of the negative charge
______________________________________________________________________________________________
27. REASONING AND SOLUTION The electrical potential energy of the group of charges is
EPE = kq1q2/d + kq1q3/(2d) + kq2q3/d = 0
so
q1q2 + (1/2)q1q3 + q2q3 = 0
a. If q1 = q2 = q, then
q + (1/2)q3 + q3 = 0
or
q 3 = - 23 q
b. If q1 = q and q2 = -q then
-q + (1/2)q3 - q3 = 0
or
q 3 = - 2q
______________________________________________________________________________________________
28. REASONING The only force acting on each particle is the conservative electric force. Therefore, the
total energy (kinetic energy plus electric potential energy) is conserved as the particles move apart. In
addition, the net external force acting on the system of two particles is zero (the electric force that
each particle exerts on the other is an internal force). Thus, the total linear momentum of the system
is also conserved. We will use the conservation of energy and the conservation of linear momentum
to find the initial separation of the particles.
SOLUTION For two points, A and B, along the motion, the conservation of energy is
592 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
1 m v2
1
2
A + 2 m 2 v 2 ,A
214
14
1, 4
24443
Initial kinetic energy
of the two particles
kq 1 q 2
+
rA
12
3
Initial electric
potential energy
=
1 m v2
2
+ 1 m 2 v 2,B
2 4
14
1,B
1
42244
43
Final kinetic energy
of the two particles
+
kq1 q 2
rB
12
3
Final electric
potential energy
Solving this equation for 1/rA and setting v1,A = v2,A = 0 since the particles are initially at rest, we
obtain
1
1
1
=
+
rA
rB
kq1 q 2
2
2
+ 12 m2 v 2,B
(12 m1v1,B
)
(1)
This equation cannot be solved for the initial separation rA, because the final speed v2,B of the second
particle is not known. To find this speed, we will use the conservation of linear momentum:
m1 v 1, A + m2 v 2 ,A = m1 v1,B + m 2 v 2,B
144
42444
3 144
42444
3
Initial linear momentum
Final linear momentum
Setting v1,A = v2,A = 0 and solving for v2,B gives
v 2 ,B = -
m1
m2
v1,B = -
3.00 ´ 10 -3 kg
6.00 ´ 10 -3 kg
(125 m/s )
= - 62.5 m/s
Substituting this value for v2,B into Equation (1) yields
1
1
=
+
r
0.100 m
A
1
(8.99 ´ 10 9 N × m 2 / C 2 )(8.00 ´ 10 -6 C )2
´ éë 1 (3.00 ´ 10 -3 kg )(125 m/s )2 + 1 (6.00 ´ 10 -3 kg )(-62.5 m/s )2 ùû
2
2
r A = 1.41 ´ 10 -2 m
______________________________________________________________________________________________
29.
REASONING AND SOLUTION Since all points on the equipotential surface are the same
distance r from the point charge, the potential is given by Equation 19.6,
SSM
V=
kq (8.99 ´ 10 9 N × m 2 /C 2 )(3.0 ´ 10 -7 C)
=
= 18 000 V
r
0.15 m
______________________________________________________________________________________________
30.
REASONING AND SOLUTION We know V = kq/r, so
Chapter 19
kq
r =
=
V
(9.0 ´ 10 9 N × m 2 / C 2 )(2.0 ´ 10 -10 C ) =
12 V
593
0.15 m
______________________________________________________________________________________________
DV
, where DV
Ds
is the potential difference between the two metal conductors of the spark plug, and Ds is the distance
between the two conductors. We can use this relation to find DV.
31. REASONING The magnitude E of the electric field is given by Equation 19.7 as E =
SOLUTION The potential difference between the conductors is
(
)(
)
DV = EDs = 2.8 ´ 10 6 V/m 0.75 ´ 10 -3 m = 2.1´ 10 3 V
______________________________________________________________________________________________
DV
, where DV is the
Ds
potential difference between the two surfaces of the membrane, and Ds is the distance between them.
If A is a point on the positive surface and B is a point on the negative surface, then DV = VA - VB =
0.070 V. The electric field between the surfaces is
32. REASONING AND SOLUTION From Equation 19.7 we know that E = -
E= =
V - VA
V - VB
DV
= - B
= A
Ds
Ds
Ds
0.070 V
= 8.8 ´ 10 6 V/m
9
8.0 ´ 10 m
_____________________________________________________________________________________________
33.
REASONING
AND
SOLUTION
As described in the
problem statement, the charges jump
between your hand and a doorknob. If we
assume that the electric field is uniform,
Equation 19.7 applies, and we have
SSM
WWW
E=–
E
+x
Hand
(Low potential)
Knob
(High potential)
V
– V hand
DV
= – knob
Ds
Ds
Therefore, solving for the potential difference between your hand and the doorknob, we have
Vknob – Vhand = – E Ds = –(–3.0 ´ 10 6 N / C)(3.0 ´ 10 –3 m) = +9.0 ´ 10 3 V
______________________________________________________________________________________________
34. REASONING AND SOLUTION The ratio of the potentials of the two equipotential surfaces
surrounding the positive point charge is
594 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
VA
=
VB
kq / r A
kq / r B
=
rB
(1)
rA
The equipotential surfaces around a single point charge are concentric spheres; therefore, the area of
any equipotential surface around a point charge is equal to 4pr2, where r is the distance from the
point charge to the equipotential surface.
Since surface A has twice the area of surface B,
=
4p r A2
2( 4p rB2 )
so that
r B2
r A2
=
1
2
and
rB
rA
=
1
2
Then, from Equation (1)
VA
VB
=
rB
rA
=
1
= 0.707
2
______________________________________________________________________________________________
35. REASONING AND SOLUTION
The drawing shows the
electric field and the three points, A, B, and C, around the point
P. We choose the upward direction as being positive. Thus, E =
-3.0 ´ 103 V/m, since the electric field points straight down.
A
0.0080 m
a. The electric potential at point A can be determined from
Equation 19.7 as •••
••
E
{
-3.0 ´ 10 3 V/m
= -
V - VP
VA - 135 V
DV
=- A
= Ds
Ds
8.0 ´ 10 -3 m - 0
C
P
-0.0033 m
E
B
so that V A = 159 V .
b. The electric potential at point B is found in a similar manner:
E
{
-3.0 ´ 10 3 V/m
= -
V - VP
VB - 135 V
DV
= B
= Ds
Ds
-3.3 ´ 10 -3 m - 0
so that VB = 125 V
c. DV = 0, since the path is perpendicular to the electric field, and no work is done in
moving a charge along such a path: VC = 135 V
______________________________________________________________________________________________
Chapter 19
595
36. REASONING The electric field E that exists between two points in space is, according to Equation
19.7, proportional to the electric potential difference DV between the points divided by the distance
Dx between them: E = - DV/ Dx.
SOLUTION
a. The electric field in the region from A to B is
E= -
5.0 V - 5.0 V
DV
= = 0 V/m
0.20 m - 0 m
Dx
b. The electric field in the region from B to C is
E= -
3.0 V - 5.0 V
DV
= = 1.0 ´ 10 1 V /m
0.40 m - 0.20 m
Dx
c. The electric field in the region from C to D is
E= -
1.0 V - 3.0 V
DV
= = 5.0 V/m
0.80 m - 0.40 m
Dx
______________________________________________________________________________________________
37.
REASONING
SSM
The total energy of the particle on the equipotential surface A is
1
EA = mv2A + qVA . Similarly, the total energy of the particle when it reaches equipotential surface B
2
1
2
is EB = mv B + qVB . According to Equation 6.8, the work W done by the outside force is equal to
2
the final total energy EB minus the initial total energy EA , W = E B - EA .
SOLUTION The work done by the outside force in moving the particle from A to B is,
(
)(
) (
)
1
1
1
2
2
2
2
W = E B - EA = mvB + qVB – mv A + qVA = m v B - v A + q(VB - VA )
2
2
2
1
2
[
= (5.00 ´ 10 –2 kg) (3.00 m/s) 2 - (2.00 m/s) 2
]
+ (4.00 ´ 10 –5 C)[ 7850 V - 5650 V ] = 0.213 J
______________________________________________________________________________________________
38. REASONING AND SOLUTION Let E1 represent the magnitude of the electric field on the first
equipotential surface, and E2 represent the magnitude of the electric field when it has shrunk to onehalf of its initial value. Then
E1 =
kq
r1 2
and
E2 =
kq
r2 2
596 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
Since E2 =
1
E ,
2 1
we have
kq
1 kq
=
2
r2
2 r1 2
so that
r2
r1
=
2
Thus, the electric field shrinks to one-half its original value at a distance of r2 = 2 r1.
The potential difference between the first equipotential surface and the locus of points where the
electric field has dropped to one-half of its original magnitude is
DV =
kq
kq
kq æ 1
ö
=
ç
- 1÷
r2
r1
r1 è 2
ø
(8.99 ´ 10 9 N × m 2 / C 2 )(2.00 ´ 10 –6 C) æ 1
ö
ç
- 1÷ = - 3290 V
=
1.60 m
è 2
ø
Since the potential difference between two successive surfaces is 1.00 ´ 103 V, we have
Magnitude of
DV from r 1 to r2
DV between successive
equipotential surfaces
=
3290 V
= 3.29
1.00 ´ 10 3 V
Therefore, the number of equipotential surfaces crossed in going radially outward from r1 to r2 is
3 .
____________________________________________________________________________________________
39. REASONING AND SOLUTION The capacitance, voltage, and charge are related by
V=
q
7.2 ´ 10 -5 C
=
= 12 V
C
6.0 ´ 10 -6 F
(19.8)
_____________________________________________________________________________________________
40. REASONING AND SOLUTION For an empty capacitor
A =
Cd (1.00 F )(1.00 m )
=
= 1.13 ´ 1011 m 2
12
e0
F/m
8.85 ´ 10
(19.10)
In square miles, this area is
(
A = 1.13 ´ 1011 m 2
1 mi 2
= 4.36 ´ 10 4 mi 2
6
2
2.59 ´ 10 m
)
______________________________________________________________________________________________
Chapter 19
41.
597
REASONING AND SOLUTION Equation 19.10 gives the capacitance for a parallel plate
capacitor filled with a dielectric of constant k : C = k e 0 A / d . Solving for k , we have
Cd (7.0 ´ 10 – 6 F) (1.0 ´ 10 –5 m)
=
= 5.3
k=
e 0 A (8.85 ´ 10 –12 F/m)(1.5 m 2 )
SSM
______________________________________________________________________________________________
42. REASONING AND SOLUTION The capacitance is given by
C=
ke 0 A
d
=
(
)(
5 8.85 ´ 10 -12 F/m 5 ´ 10 -6 m 2
1´ 10 -8 m
)=
2 ´ 10 -8 F
______________________________________________________________________________________________
43. REASONING AND SOLUTION The total charge transferred is given by
q = CV = (2.5 ´ 10 -8 F )(450 V)
The number of electrons transferred is, then,
Number of electrons =
q
(2.5 ´ 10 -8 F )(450 V )
=
= 7.0 ´ 10 13
-19
e
´
1.60 10
C
______________________________________________________________________________________________
44. REASONING AND SOLUTION The capacitance is
C=
q
2.7 ´ 10 -6 C
=
= 1.8 ´ 10 -6 F
V
1.5 V
The plate separation is given by
d =
ke 0 A
C
=
(
)(
5.4 8.85 ´ 10 -12 F/m 3.8 m 2
1.8 ´
10 -6
F
)=
1.0 ´ 10 -4 m
______________________________________________________________________________________________
45.
REASONING The charge that resides on the outer surface of the cell membrane is q = CV ,
according to Equation 19.8. Before we can use this expression, however, we must first determine the
capacitance of the membrane. If we assume that the cell membrane behaves like a parallel plate
capacitor filled with a dielectric, Equation 19.10 ( C = k e 0 A / d ) applies as well.
SSM
SOLUTION The capacitance of the cell membrane is
C=
k e0 A
d
=
(5.0)(8.85 ´10 –12 F/m)(5.0 ´ 10 –9 m 2 )
1.0 ´ 10
–8
m
= 2.2 ´10 –11 F
598 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
a. The charge on the outer surface of the membrane is, therefore,
q = CV = (2.2 ´ 10 –11 F)(60.0 ´ 10 –3 V) = 1.3 ´ 10 –12 C
+
b. If the charge in part (a) is due to K ions with charge +e (e = 1.6 ´ 10-19 C), the number of ions
present on the outer surface of the membrane is
Number of 1.3 ´10 – 12 C
=
= 8.1´ 10 6
– 19
K + ions
1.6 ´ 10
C
______________________________________________________________________________________________
46. REASONING The capacitance of a capacitor is defined by Equation 19.8 as C = q/V, where q is the
magnitude of the charge on one of the plates and V is the potential difference between the plates. The
potential difference is related to the magnitude E of the electric field between the plates by E = V/Ds
(Equation 19.7), where Ds is the distance between the plates. Eliminating V from these two equations
gives C = q/(EDs).
SOLUTION The capacitance of the parallel plate capacitor is
q
1.9 ´ 10 -5 C
=
=
C =
E Ds
(640 V/m ) 11 ´ 10 -3 m
(
)
2.7 ´ 10 -6 F
______________________________________________________________________________________________
47. REASONING AND SOLUTION
a. We know that
Energy =
1
2
2 CV
= 12 (750 ´ 10 -6 F )(330 V ) 2 =
41 J
b. The power developed by the flash is
P=
Energy
41 J
=
=
Time
5.0 ´ 10 -3 s
8200 W
______________________________________________________________________________________________
48. REASONING AND SOLUTION The electric energy stored in the region between the metal spheres
is
Energy =
1
2
k e 0 E 2 ( Volume )
(19.12)
However, the magnitude E of the electric field is related to the potential difference DV between the
spheres and the distance Ds between them via Equation 19.7, E = DV/ Ds. Thus,
Chapter 19
Energy =
1
2
=
1
2
599
æç DV ö÷ 2
ke0
( Volume)
è Ds ø
æ
C2
(2.1)ç 8.85 ´ 10 -12
N× m2
è
ö æ 3.0 V ö 2
4p (0.1500 m )2 (0.0020 m )
÷
֍
ø è 0.0020 m ø 1444442444443
Volume
[
]
= 1.2 ´ 10 -8 J
______________________________________________________________________________________________
49.
SSM
REASONING
According to Equation 19.11, the energy stored in a capacitor with a
1
2
capacitance C and potential V across its plates is Energy = 2 CV . Once we determine how much
energy is required to operate a 75-W light bulb for one minute, we can then use the expression for the
energy to solve for V.
SOLUTION The energy stored in the capacitor, which is equal to the energy required to operate a
75-W bulb for one minute (= 60 s), is
Energy = Pt = (75 W)(60 s) = 4500 J
Therefore, solving Equation 19.11 for V, we have
V=
2(Energy)
=
C
2(4500 J)
= 52 V
3.3 F
______________________________________________________________________________________________
50. REASONING AND SOLUTION From Equations 19.9 and 18.4, we have
A=
q
=
k e0E
1.7 ´ 10 –7 C
= 3.9 ´ 10 –4 m 2
2 ö
æ
C
1.4 ´ 10 7 N/C
(3.5 )ç 8.85 ´ 10 –12
2÷
N×m ø
è
(
)
Since A = pr2 , the radius of the plates is
r =
A
=
p
3.9 ´ 10 -4 m 2
= 1.1 ´ 10 -2 m
p
______________________________________________________________________________________________
51. REASONING AND SOLUTION The potential difference is constant so V0 = q0 /C0 = q/C = q/(kC0).
Thus,
q = kq 0 = kC 0 V0 = ( 4.0)(2.7 ´ 10 -6 F )(12 V ) = 1.3 ´ 10 -4 C
600 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
The original charge is q0 = C0V0 = 3.2 ´ 10-5 C. The surface charge Q is, therefore,
Q = q - q 0 = 1.0 ´ 10 -4 C
______________________________________________________________________________________________
52.
REASONING AND SOLUTION The electron and proton experience forces of the same magnitude
in the same electric field, since they have the same magnitude of charge. Newton's second law gives
m pap = meae
where ap and ae are the accelerations of the proton and the electron. Kinematics gives
ap = 2xp /t2
so
and
ae = 2xe /t2
9.11´ 10 -31 kg
(0.0400 m ) = 2.18 ´ 10 -5 m
xp =
xe =
27
mp
1.67 ´ 10
kg
me
______________________________________________________________________________________________
53.
REASONING According to Equation 19.10, the capacitance of a parallel plate capacitor
filled with a dielectric is C = k e 0 A / d , where k is the dielectric constant, A is the area of one plate,
and d is the distance between the plates.
SSM
From the definition of capacitance (Equation 19.8), q = CV . Thus, the charge on a parallel plate
capacitor that contains a dielectric is given by Equation 19.10 as q = (k e 0 A / d )V . Since each
dielectric occupies one-half of the volume between the plates, the area of each plate in contact with
each material is A/2. Thus,
q1 =
k 1 e 0 (A / 2)
d
V =
k 1e 0 A
2d
V
k 2e0 ( A/2 )
q2 =
and
d
V =
k 2e0 A
2d
V
According to the problem statement, the total charge stored by the capacitor is
q1 + q 2 = CV
(1)
where q1 and q2 are the charges on the plates in contact with dielectrics 1 and 2, respectively.
Using the expressions for q1 and q2 above, Equation (1) becomes
CV =
k 1e 0 A
2d
V+
k 2e0 A
2d
This expression can be solved for C.
V =
k 1e 0 A + k 2 e 0 A
2d
V =
(k 1 + k 2 )e 0 A
2d
V
Chapter 19
SOLUTION Solving for C, we obtain C =
601
e 0 A(k 1 + k 2 )
2d
______________________________________________________________________________________________
54. REASONING AND SOLUTION The electric potential is given by
(9.0 ´ 10 9 N × m 2 / C 2 )(1.0 C )
kq
V =
=
r
1.0 ´ 10 3 m
=
9.0 ´ 10 6 V
______________________________________________________________________________________________
55. REASONING AND SOLUTION The potential at point A is
kq
=
VA =
rA
(8.99 ´ 10 9 N × m 2 /C 2 )(-3.00 ´ 10 -8 C ) =
2.00 m
- 135 V
Similarly, the potential at point B is VB = kq/rB = -89.9 V
The difference in the potentials is
VB - VA = +45 V
Point B is at the higher potential.
______________________________________________________________________________________________
56. REASONING AND SOLUTION
a. The change in the electric potential energy is
EPE A - EPE B = WAB =
5.80 ´ 10 -3 J
b. The potential difference between the points is
V A - VB =
c.
EPE A - EPE B
q
=
5.80 ´ 10 -3 J
= 32.2 V
1.80 ´ 10 -4 C
Point A has the higher potential.
______________________________________________________________________________________________
57.
SSM
REASONING
According to Equation 19.11, the energy stored in a capacitor with
1
2
capacitance C and potential V across its plates is Energy = 2 CV .
SOLUTION Therefore, solving Equation 19.11 for V, we have
602 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
2(Energy)
=
C
V=
2(73 J)
= 1.1 ´10 3 V
–6
120 ´ 10 F
______________________________________________________________________________________________
58. REASONING AND SOLUTION
a. The only force that acts on the particle is the conservative electric force. Therefore, the total
energy of the particle is conserved as it moves from point A to point B:
+ EPE A = 12 mv 2B + EPE B
1442443 1442443
1 mv 2
A
2
Total energy at
point B
Total energy at
point A
Since the particle starts from rest, vA = 0. The electric potential V is related to the electric potential
energy EPE by V = EPE/q (see Equation 19.3). With these changes, we can solve the equation above
for the potential difference VB - VA:
VB - V A =
(
1
0 - 12 mv 2B
q
)=
[ (
)
1
2
0 - 12 2.5 ´ 10 -6 kg (42 m/s )
-6
-1.5 ´ 10 C
]=
1500 V
b.
Point B is at the higher potential, because a negative charge accelerates from a lower potential
to a higher potential.
______________________________________________________________________________________________
59. REASONING AND SOLUTION The capacitance is C = q0/V0 = q/V. The new charge q is, therefore,
q =
q0 V
V0
=
(5.3 ´ 10 -5 C )(9.0 V ) =
6.0 V
8.0 ´ 10 -5 C
______________________________________________________________________________________________
60. REASONING AND SOLUTION Let d = 2.0 m be the separation between the charges, and let y be
the distance from the negative charge to the point where the potential is zero. Then
kq
=
y
so y = ±
2 kq
y2 + d 2
d
= ± 1.2 m . The points are located
3
1.2 m on either side of the negative charge .
______________________________________________________________________________________________
61.
SSM
REASONING AND SOLUTION Equation 19.7 gives the result directly:
Chapter 19
E = –
V - VA
28 V – 95 V
DV
= – B
= –
=
0.016 m
Ds
Ds
603
4.2 ´ 10 3 V/m
The electric field points from high potential to low potential. Thus, it points from A to B .
______________________________________________________________________________________________
62. REASONING AND SOLUTION The charge on the empty capacitor is q0 = C0V0. With the dielectric
in place, the charge remains the same. However, the new capacitance is C = kC0 and the new voltage
is V. Thus,
q0 = CV = kC0 V = C0V0
Solving for the new voltage yields
V = V0 /k = (12.0 V)/ 2.8 = 4.3 V
The potential difference is 12.0 - 4.3 = 7.7 V . The change in potential is a decrease .
______________________________________________________________________________________________
63. REASONING AND SOLUTION Let point A be on the x-axis where the potential is 515 V. Let
point B be on the x-axis where the potential is 495 V. From Equation 19.7, the electric field is
E = -
V - VA
495 V - 515 V
DV
= - B
= = - 1.7 ´ 10 3 V/m
-3 m
Ds
Ds
6.0
10
-2
´
(
)
The magnitude of the electric field is 1.7 ´ 10 3 V/m . Since the electric field is negative, it points to
the left , from the high toward the low potential.
______________________________________________________________________________________________
64. REASONING The only force acting on each particle is the conservative electric force. Therefore, the
total energy (kinetic energy plus electric potential energy) is conserved as the particles move apart. In
addition, the net external force acting on the system of two particles is zero (the electric force that
each particle exerts on the other is an internal force). Thus, the total linear momentum of the system
is also conserved. We will use the conservation of energy and the conservation of linear momentum
to find the final speed of each particle.
SOLUTION For two points, A and B, along the motion, the conservation of energy is
1 mv 2
+ 12 mv 22 ,A
2141,A
4
424
44
3
Initial kinetic energy
of the two particles
+
kq 1 q 2
rA
12
3
Initial electric
potential energy
=
1 mv 2
+ 12 mv 22,B
2 41,B
1
4
424
44
3
Final kinetic energy
of the two particles
+
kq 1 q 2
rB
12
3
Final electric
potential energy
604 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
Setting v1,A = v2,A = 0 since the particles are initially at rest, and letting r B = 12 r A , the conservation of
energy equation becomes
1 mv 2
2
1,B
2
+ 12 mv 2,B
= -
kq1 q2
rA
(1)
This equation cannot be solved for v1,B because the final speed v2,B of the second particle is not
known. To find this speed, we will use the conservation of linear momentum:
mv1,A + mv 2,A =
14
42443
Initial linear momentum
mv1,B + mv 2 ,B
1442443
Final linear momentum
Setting v1,A = v2,A = 0 and solving for v2,B gives v2,B = - v1,B. Substituting this result into Equation (1)
and solving for v1,B yields
v 1,B =
=
- kq1 q 2
m rA
(
)(+5.0 ´ 10 -6 C )(-5.0 ´ 10 -6 C ) =
(6.0 ´ 10 -3 kg )(0.80 m )
- 9.0 ´ 10 9 N ×m 2 / C 2
6.8 m/s
This is also the speed of v2,B.
______________________________________________________________________________________________
65.
REASONING If we assume that the motion of the proton and the electron is
horizontal in the +x direction, the motion of the proton is determined by Equation 2.8,
1
x = v0 t + 2 apt 2 , where x is the distance traveled by the proton, v0 is its initial speed, and ap is its
acceleration. If the distance between the capacitor places is d, then this relation becomes
1
1
2
d = v0 t + 2 apt , or
2
SSM
WWW
2
d = 2v0t + apt
(1)
We can solve Equation (1) for the initial speed v0 of the proton, but, first, we must determine the
time t and the acceleration ap of the proton . Since the proton strikes the negative plate at the same
instant the electron strikes the positive plate, we can use the motion of the electron to determine the
time t.
For the electron,
1
2
d = 2 aet 2 , where we have taken into account the fact that the electron is released
1
from rest.
Solving this expression for t we have t = d / ae . Substituting this expression into
Equation (1), we have
Chapter 19
d çæ ap ö
÷ d
+
ae è a e ø
d = 2v0
605
(2)
The accelerations can be found by noting that the magnitudes of the forces on the electron and proton
are equal, since these particles have the same magnitude of charge. The force on the electron is
F = eE = eV / d , and the acceleration of the electron is, therefore,
F
eV
=
me
med
ae =
(3)
Newton's second law requires that me ae = mpap , so that
ap
=
ae
me
(4)
mp
Combining Equations (2), (3) and (4) leads to the following expression for v0, the initial speed of the
proton:
m ö eV
1æ
v 0 = çç 1 – e ÷÷
mp ø me
2è
SOLUTION Substituting values into the expression above, we find
v0 =
9.11´ 10 –31 kg ö
1æ
1–
ç
÷
2 è 1.67 ´ 10 –27 kg ø
(1.6 ´ 10 –19 C)(175 V)
9.11´ 10
–31
kg
= 2.77 ´ 10 6 m/s
______________________________________________________________________________________________
66. REASONING AND SOLUTION
a. Let d be the distance between the charges. The potential at the point x1 = 4.00 cm to the left of the
negative charge is
V= 0=
kq1
d - x1
-
kq2
x1
which gives
q1
q2
=
d
-1
x1
Similarly, at the point x2 = 7.00 cm to the right of the negative charge we have
(1)
606 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
V= 0 =
kq1
x2 + d
-
k q2
x2
which gives
q1
q2
=
d
+ 1
x2
(2)
Equating Equations (1) and (2) and solving for d gives d = 0.187 m .
b. Using the above value for d in Equation (1) yields
q1
q2
=
3.67 .
______________________________________________________________________________________________