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CHAPTER
19 ELECTRIC POTENTIAL ENERGY
AND THE ELECTRIC POTENTIAL
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION The work done in moving a charge q0 from A to B is given by
Equation 19.4: WAB  q0 (VA  VB) . For the cases in the drawing, we have
Case 1:
WAB  q0 (150 V 100 V)  q0 (50 V)
Case 2:
WAB  q0 [25 V  (–25 V)]  q0 (50 V)
Case 3:
WAB  q0 [–10 V (–60 V)]  q0 (50 V)
The work done on the charge by the electric force is the same in each case.
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2.
REASONING AND SOLUTION The potential at a point in space that is a distance r from a point
charge q is given by Equation 19.6: V  kq/ r . When more than one point charge is present, the
total potential at any location is the algebraic sum of the individual potentials created by each charge
at that location.
A positive point charge and a negative
point charge have equal magnitudes. One
L
q = +Q
of the charges is fixed to one corner of a
square. If the other charge is placed
opposite to the first charge along the
diagonal of the square, then each charge
L
L
will be the same distance L from the
empty corners. The potential at each of
the empty corners will be
q = –Q
L
k(Q) k(Q)
V

0
L
L
Therefore, if the potential at each empty corner is to be the same, then the charges must be placed at
diagonally opposite corners as shown in the figure.
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3.
REASONING AND SOLUTION The potential at a point in space that is a distance r from a point
charge q is given by Equation 19.6: V  kq/ r . When more than one point charge is present, the total
potential at any location is the algebraic sum of the individual potentials created by each charge at
that location.
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ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
Three point charges have identical
magnitudes, but two of the charges are
positive and one is negative. These
charges are fixed to the corners of a
square, one to a corner, as shown in the
figure.
The potential at the empty corner is
given by
q
q
q
V 1 2  3
L L 2 L
q
q
1
2
L
L 2
q
L
3
Using q to denote the magnitude of each charge, we have the following possibilities:
q1 and q2 are positive: V 
kq
kq
kq
kq



L L 2 L
L 2
q1 and q3 are positive: V 
kq
kq
kq kq 
1 
2 




L L 2 L
L 
2 
q2 and q3 are positive: V  
kq
kq
kq
kq



L L 2 L L 2
In each case, the potential at the empty corner is positive.
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4.
REASONING AND SOLUTION Four
q
q
1
2
point charges of equal magnitude are
placed at the corners of a square as shown
r
r
in the figure at the right.
The electric field at the center of the
r
r
square is the vector sum of the electric
field at the center due to each of the
q
q
4
3
charges individually. The potential at the
center of the square is equal to the
algebraic sum of the potentials at the
center due to each of the charges
individually.
All four charges are equidistant from the center (a distance r in the figure). If two diagonal
charges have the same magnitude and sign, then the electric field at the center due to these two
charges have equal magnitude and opposite directions. Their resultant is, therefore, zero. Thus, the
electric field at the center will be zero if each diagonal pair of charges has the same magnitude and
sign.
If a diagonal pair of charges has the same magnitude and sign, they will give rise to a non-zero
potential at the center. Thus, the potential due to one diagonal pair of charges must cancel the
potential due to the other diagonal pair of charges. This will be the case if the two pairs of diagonal
charges have opposite signs.
Chapter 19 Conceptual Questions
105
Thus, both the electric field and the electric potential will be zero at the center of the square if all
four charges have the same magnitude, q1 and q3 have the same sign, and q2 and q4 have the same
sign, which is opposite to the signs of q1 and q3.
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5.
REASONING AND SOLUTION
Positive charge is spread uniformly
around a circular ring, as shown in the
figure, with q representing the total
amount of charge on the ring. Equation
19.6 gives the correct potential at points
along the line perpendicular to the plane
Center of ring
of the ring at its center. The equation
does not, however, give the correct
potential at points that do not lie on this
line.
We can consider the ring to be composed of point charges of magnitude q spread uniformly
around the ring so that q   q . According to Equation 19.6, each "point charge" gives rise to a
potential given by V = kq/r at a distance r from the "point charge." The potential due to the ring at
any point in space is the algebraic sum of the potentials of all the "point charges". For any point
along the line perpendicular to the plane of the ring at its center, the distance r is the same for all
"point charges" on the ring. Therefore, for any point on that line, V   V   kq / r  kq/ r ; thus,
we obtain Equation 19.6. For points that are not on the line, the distance r will, in general, be
different for each "point charge" that makes up the total charge q; therefore, the potential V at that
point will not be given by Equation 19.6 with a single value for r.
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6.
REASONING AND SOLUTION The electric field at a single location is zero. This does not
necessarily mean that the electric potential at the same place is zero. According to Equation 19.7,
E  –V / s . Therefore, the fact that E = 0 means that the potential gradient, V / s , is zero, and
the potential difference does not change with distance. In other words, in the vicinity of a point
where E is zero, the electric potential is constant, but not necessarily zero
To illustrate this fact, consider the situation of two identical point charges. The electric field at
each location in the vicinity of the charges is the resultant of the electric field at that location due to
each of the two charges. The magnitude of the field due to each charge is given by Equation 18.3:
E  kq/ r2 . At the point exactly halfway between the charges,
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ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
along the line that joins them, r is the
same for the field contribution from each
charge; therefore, the magnitudes of the
field contributions at this point are the
same. Since the charges are identical, the
directions of the field contributions at this
point are in opposite directions. The
figure at the right shows the situation for
both positive and negative charges. The
resultant of the two field contributions is
zero. Thus, the electric field is zero
exactly halfway between the charges on
the line that joins them. The electric
potential at this point, however, is the
algebraic sum of the potential at that point
due to each individual charge.
Positive point charges
E
1
q
2
E
2
1
r
r
q
Negative point charges
1
q
E
1
E
2
2
q
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107
Therefore, the potential at the point exactly halfway between the charges is
V
kq kq 2kq


r
r
r
which is clearly non-zero.
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7.
REASONING AND SOLUTION An electric potential energy exists when two protons are separated
by a certain distance. It is equal to the work that must be done by an external agent to assemble the
configuration. Suppose that we imagine assembling the system, one particle at a time. If there are no
other charges in the region, there are no existing electric fields; therefore, no work is required to put
the first proton in place. That proton, however, gives rise to an electric field that fills the region. Its
magnitude at a distance r from the proton is given by Equation 18.3, E  ke/ r2 , where +e is the
magnitude of the charge on the proton. Since the region contains an electric field due to the first
proton, there also exists an electric potential, and the external agent must do work to place the second
proton at a distance d from the first proton. The electric potential energy of the final configuration is
equal to the work that must be done to bring the second proton from infinity and place it at a distance
d from the first proton. The electric potential at a distance d from the first proton is Vproton  ke/ d
(Equation 19.6). According to Equation 19.3, the electric potential energy of the final configuration
is therefore
EPE  Vproton (e)  
ke2
d
a. If both protons are replaced by electrons, similar arguments apply. However, since the electron
carries a negative charge (–e), the electric potential at a distance d from the first electron is
Velectron  ke/ d . The electric potential energy of the final configuration is now given by
EPE  Velectron (e)  
ke
ke2
(e)  
d
d
Therefore, if both protons are replaced by electrons, the electric potential energy remains the same.
b. When only one of the protons is replaced by an electron, we find that
2
ke
 ke
EPE  Vproton (e)    (e)  
 d 
d
Thus, when only one of the protons is replaced by an electron, the electric potential energy decreases
2
2
from ke / d to ke / d .
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8.
REASONING AND SOLUTION A proton (charge = +e) is fixed in place. An electron (charge = –
e) is released from rest and allowed to collide with the proton. Since the electron is released from
rest, its initial kinetic energy is zero. Since the electron is in the vicinity of the proton, it has electric
potential energy. The initial total energy of the electron is, therefore, electric potential energy. After
the electron is released, the potential energy decreases, while the kinetic energy increases. Assuming
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ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
that the charges are point charges and that ri is the initial separation between the charges, while rf is
the separation just before the collision, the conservation of energy gives
Einitial = Efinal
 ke2  ke2 1

 melectron v 2
2
ri
rf
 1
1 
1
ke 2     (EPE)  2 m electronv 2
r f r i 
 1
1 
where (EPE)  ke 2    is the change in the electric potential energy.
r f r i 
Therefore, the speed of the electron when it collides with the proton is
v
2(EPE)
melectron
(1)
When the roles of the electron and proton are interchanged and the same experiment repeated, similar
arguments apply.
It can be shown that the speed of the proton when it collides with the electron is
v
2(EPE)
m proton
(2)
Since the mass of the electron is much less than the mass of the proton, we can conclude from
Equations 1 and 2 that the electron is traveling faster when the collision occurs.
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9.
REASONING AND SOLUTION The electric potential is constant throughout a given region of
space. Since the potential does not change from point to point in this region, the potential gradient,
V / s , is zero throughout this region. From Equation 19.7, we see that the electric field is
E  V / s and must also be zero throughout this region.
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10. REASONING AND SOLUTION In a region where the electric field is constant, we see from
Equation 19.7, E  V / s , that the potential gradient is constant. In other words, the rate at which
the electric potential changes with distance in the region is constant. The potential, therefore, is not
zero, but varies from point to point in a linear fashion with distance.
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11. REASONING AND SOLUTION A positive charge is placed in an electric field. At each location
in the electric field, the field is perpendicular to an equipotential surface. Therefore, in order for the
Chapter 19 Conceptual Questions
109
charge to experience a constant electric potential, it must be moved along a path that is perpendicular
to the electric field.
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12. REASONING AND SOLUTION The electric potential at any location in the vicinity of the charges
is given by
V
k( q) k(q)

r
r
where r+ is the distance from the positive charge to the point in question and r– is the distance from
the negative charge to the point in question. All points for which r+ = r–correspond to points where
the potential is zero, according to the above expression. Such points would lie on a plane that is
midway between the two charges. Therefore, we can conclude that line B in the figure is the edge-on
view of an equipotential surface.
As stated in the text, the electric field created by a group of charges is everywhere perpendicular to
the associated equipotential surfaces. Lines A and C can be ruled out as possible edge-on views of
equipotential surfaces, because they are not perpendicular to the electric field lines that would arise
from the dipole configuration (see Figure 19.15).
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13. REASONING AND SOLUTION
identical point charges.
A positive test charge is moved along the line between two
a. If the two identical point charges are positive, then the test charge experiences a repulsive force
from each of the two fixed charges. Halfway between the charges, the resultant of these two
repulsive forces will be zero since the distance from the test charge to either fixed charge is the same.
As the test charge is moved from the midpoint, it will feel a net repulsive force which tends to push
the charge back to the midpoint. The work done by this repulsive force in pushing the positive charge
back to the midpoint is positive. According to the discussion in Section 19.4, the charge therefore
goes from a higher potential (away from the midpoint) toward a lower potential (at the midpoint).
Thus, with regard to the electric potential, the midpoint is analogous to the bottom of a valley.
b. If the two identical point charges are negative, then the test charge experiences an attractive force
from each of the two fixed charges. Halfway between the charges, the resultant of these two
attractive forces will be zero since the distance from the test charge to either fixed charge is the same.
As the test charge is moved from the midpoint, it will feel a net attractive force which tends to pull
the charge away from the midpoint. The work done by this attractive force in pulling the positive
charge away from the midpoint is positive. According to the discussion in Section 19.4, the charge
therefore goes from a higher potential (at the midpoint) toward a lower potential (away from the
midpoint). Thus, with regard to the electric potential, the midpoint is analogous to the top of a hill.
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14. REASONING AND SOLUTION
identical point charges.
A negative test charge is moved along the line between two
a. If the two identical point charges are positive, then the test charge experiences an attractive force
from each of the two fixed charges. Halfway between the charges, the resultant of these two
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ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
attractive forces will be zero since the distance from the test charge to either fixed charge is the same.
As the test charge is moved from the midpoint, it will feel a net attractive force which tends to pull
the charge away from the midpoint. The work done by this attractive force in pulling the test charge
away from the midpoint is positive. According to the discussion in Section 19.4, the charge therefore
goes from a higher potential (at the midpoint) toward a lower potential (away from the midpoint).
Thus, with regard to the electric potential, the midpoint is analogous to the top of a hill.
b. If the two identical point charges are negative, then the test charge experiences a repulsive force
from each of the two fixed charges. Halfway between the charges, the resultant of these two
repulsive forces will be zero since the distance from the test charge to either fixed charge is the same.
As the test charge is moved from the midpoint, it will feel a net repulsive force which tends to push
the charge toward the midpoint. The work done by this repulsive force in pushing the test charge
back to the midpoint is positive. According to the discussion in Section 19.4, the charge therefore
goes from a higher potential (away from the midpoint) toward a lower potential (at the midpoint).
Thus, with regard to the electric potential, the midpoint is analogous to the bottom of a valley.
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15. REASONING AND SOLUTION From the definition of electric potential, we know that
1 joule/coulomb. Therefore
1 volt =
volt
joule/coulomb
joule
1
newton meter
1
newton






meter
meter
coulomb meter
coulomb
meter coulomb
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16. REASONING AND SOLUTION Since both particles are released from rest, their initial kinetic
energies are zero. They both have electric potential energy by virtue of their respective positions in
the electric field between the plates. Since the particles are oppositely charged, they move in opposite
directions toward opposite plates of the capacitor. As they move toward the plates, the particles gain
kinetic energy and lose potential energy. Using (EPE)0 and (EPE)f to denote the initial and final
electric potential energies of the particle, respectively, we find from energy conservation that
EPE 0  12 mparticle v2f  EPE f
The final speed of each particle is given by
vf 

2 EPE0  EPE f

mparticle
Since both particles travel through the same distance between the plates of the capacitor, the change
in the electric potential energy is the same for both particles. Since the mass of the electron is smaller
than the mass of the proton, the final speed of the electron will be greater than that of the proton.
Therefore, the electron travels faster than the proton as the particles move toward the respective
plates. The electron, therefore, strikes the capacitor plate first.
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17. REASONING AND SOLUTION A parallel plate capacitor is charged up by a battery. The battery
is then disconnected, but the charge remains on the plates. The plates are then pulled apart.
Chapter 19 Conceptual Questions
111
a. The capacitance of the system is given by Equation 19.10 with   1 , C   0 A / d , where A is the
area of either plate and d is the distance between the plates. Since d increases when the plates are
pulled apart and A remains the same, the capacitance decreases.
b. According to Equation 19.8, the potential difference between the plates of the capacitor is given by
V  q / C . Since q remains the same and C decreases, the potential difference V must increase.
c. The magnitude of the electric field between the plates is given by E  V / d . Since V  q / C , and
C   0 A / d , the electric field between the plates is given by
E
(q / C) q(d /  0 A)
q


d
d
0 A
Since both q and A remain the same, the electric field remains the same.
2
d. The energy stored in the capacitor is given by Equation 19.11, Energy  (1/ 2)CV .
V  q / C , we have
2
2
1
1q
2 1 q 
Energy  CV  C

2
2 C 
2 C
Since
Since q remains the same and C decreases, the electric potential energy stored by the capacitor
increases. This increase will be equal to the work that is done by the external agent in pulling the
plates apart.
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