Chapter 23: Electric Potential Hints & Answer- Updated 9/19/08
1.
a) Fill in the following chart: Ug stands for gravitational potential energy and Wg is the work done by the
force of gravity. Remember that potential energy is relative to an arbitrarily chosen U=0.
Reference
Relative to
table-top
Relative to
the floor
Relative to
the ceiling
Ug A
0
Ug B
0
Ug C
30 J
UAB
0
UAC
30 J
UBC
30 J
Wg AB
0
Wg AC
-30 J
Wg BC
-30 J
20 J
20 J
50J
0
30 J
30 J
0
-30 J
-30 J
-40J
-40J
-10J
0
30 J
30 J
0
-30 J
-30 J
b) This chart illustrates the definition of potential energy because it shows that U= - Wg, which is the
definition of potential energy and potential energy requires a conservative force.
c) The work done by and against gravity depends only of vertical displacement.
d) K= - U= 50 J = mv2/2.
A
a
E
2. The following is an electric version of the problem above. A +2 C charge
B
C
is moved from point A to point B then to point C inside a uniform electric
0.04 m
field E=200 N/C. The distance from Cto B is 4 cm. The angle is 370.
a) & b)Fill in the following chart. Here Ue represents the electric potential energy and We the work done by
the electric force. The symbol UAB represents the change in U from A to B = UB UA.
Reference
Relative to
A
Relative to
B
Relative to
C
Ue A
0; 0
Ue B
0; 0
“
“
-16J;
+24J
-16J;
+24J
Ue C
+16J;
-24J
“
UAB
0; 0
UBC
+16J;
-24J
“
We AB
0; 0
“
UAC
+16J;
-24J
“
0; 0
“
“
“
“
We AC
-16J;
+24J
“
We BC
-16J;
+24J
“
“
“
“
c) The potential differences are: VAB=0, VAC=8 v, and VBC=8 v.
d) You are not encumbered by the variations in + and –energies that result from the two different types of
charge that can move in an electric field. Like electric field, electric potential describes the energy properties
of space itself.
3. Enough positive and equal negative charge is accumulated on two flat parallel plates until the voltage
between them is 36 volts. Define the negative plate as zero volts (that is, “ground”).
a) The broken lines drawn parallel to the plates are called equipotentials because being perpendicular to
the electric field the potential cannot change along these lines. Plates are themselves equipotentials for
the same reason. Here E is constant so V is proportional to s and the lines must be equally spaced.
b) VA= 6v, VB= 12v, etc....(V= 6v)
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c) VAB= 6v, VAE= 24v, VEC= -12v, & VDB= -12v. Calculate the following:
d)
(i) 18J
(ii) -6J.
(iii) 12e J
(iv) -36J & 36J
e) E= 600N/C &  = o600 C/m2.
f)
g) The answers to b) would differ, they would be negative starting from the positive plate. Also the shape of
the graph in (f) would be the same but V=0 at the positive plate and the other potential values would be (-).
4. Consider a configuration of three charged flat plates set parallel to each other. As you go from one plate to
the other the potential changes according to the graph shown below the plates.
a) Taking the negative of the slopes of V, the electric fields are zero outside,
EAB= -200i & EBC= 1200i.
b) Draw a graph of the electric field vs distance. Remember V decreases in the
Direction of the field.
B
A
C
a)
+1200N/C
V +100v
E
0.50m
0.75m
0.50m
0.75m
-200N/C
-200v
c) To find acceleration, a= eE/m. To find speed use K=Vq = mv2/2.
d) The electron will not make it to the other plate because the electric field is too strong. It will be
decelerated to v=0 and turned around after moving 1/6 of the way to C.
5. Consider two parallel conducting plates of area 0.028 m2 and different amounts of charge. One plate has a
net charge of +6pC (10-12C) and the other -12pC and they are 6 cm apart.
a) Using Gauss’ law you can show that Eout= ( +  )/2 o= 12 N/C . and that Ein= 3Eout=36 N/C.
b) Remember V decreases in the direction of the field.
-
E
+
+36N/C
V
0.06m
0.06m
-12N/C
6. Identify the following statements as either true of false. Justify your choice.
a)
b)
c)
d)
e)
f)
Equipotential lines are always perpendicular to lines of force. True
Equipotential lines can intersect in space. False
Equipotential lines spread out as the electric field gets weaker. True
The electric potential inside a conductor is always zero. False
If you touch a high-voltage wire you will invariably be electrocuted. False
Electrons will freely move toward higher potentials. True
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g) If an electron and a proton are accelerated through the same potential difference, they will have the
same acceleration (False )and they will acquire the same amount of energy (True).
7. The figure shows a number of equipotentials curves in a cross-section of space.
EA
a) At B the equipotentials are denser.
b) On diagram.
c) The electric field intensity is given approximately by (V/x):
at point A is about 500 N/C and at point B is about 1000N/C.
d) The electric field vectors are parallel to the line of force but their directions
depend on whether the source at the center is positive of negative, and that is not
given in this problem.
EB
1 cm
8. Since equipotentials are perpendicular to the lines of force they tend to follow the surface contour of the
source charge. They are closer together near points and sharp corners because charges accumulate more
densely at those locations. Like lines of force they can blend but not cross. Note how the equipotentials follow
the “contour” of the charged objects.
-2q
+
Q
+q
+Q
+q
+q
-Q
-q
+q
+
Q
These are concentric spheres
9. The potential in a certain region in space is given by V=axy, for x & y >0, and a is a constant.
This is conceptually challenging problem.
a) Ex = dV/dx = -ay; and Ey =dV/dy = -ax
b) Equipotentials would be a series of hyperbolas with different constant V’s.
c) Two equally (-) charged plates at right angles to each other could do something like this.
-Q
-Q
10. The following problem will review the gravitational potential energy for spherically symmetric
gravitational fields such as those from planets and stars. A rocket of mass 105 kg moves from the surface of the
earth to a distance of 3 earth-radii above the surface of the earth. Recall that we define the zero of gravitational
potential energy at infinity in these cases.
a) Uinit=-GMEm/RE= -6.23 x 1012 J and Ufinal= -GMEm/4RE= -1.56 x 1012 J.
b) Ug= +4.67 x 1012 J and Wg = -4.67 x 1012 J.
c) Ug= -GMEm/RE (1/8 -1/4) = +0.78 x 1012 J. The reason it is so much less is that further from the earth
the force of gravity is much less therefore energy changes require more distance.
d) Graphs:
-U
-Fg
r
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e) We chose Ug=0 at infinity because it gives us the simplest potential energy function Ug= -GMm/r.
11. Now for an electric version of the problem above. When the electron in the hydrogen atom is “excited” it
rises from its ground energy state position (r=0.52 x 10-10 m) to a position 4 times further.
a) Uinit= -ke2/R= -4.4 x 10-18 J= -27.7 ev, and Ufinal= - ke2/4R = -1.1 x 10-18 J= -6.9 ev. For potentials
divide by the electron charge: Vinit= +27.7 v, and Vfinal= +6.9 v, the + sign here reflects the fact that
the potential source is a positive charge.
b) UE= +3.3 x 10-18 J= 20.8 ev and WE = -3.3 x 10-18 J.
c) Graphs:
-U
+V
r
r
d) The energy values would change sign but not the potential values.
e) Same answer as 2d.
12. Consider the following examples of spherically symmetric electric fields. Fill in the following chart
summarizing important properties. Some of these were done in the previous problem set. Remember Voo=0.
Example
a)
Equipotentials
illustration
Point charge
+q

V
kq/r2
kq/r
V at relevant
surfaces
Graph of E
vs. r
Graph of V
vs. r
+V
E
r
b)
c)
d)
2
Conducting
sphere of
charge Q with
radius R
Eout=kQ/r
Vout =kQ/r
Vsurface
=kQ/R
Ein=0
Vin =kQ/R
Charged metal
sphere of
radius a
surrounded by
un-charged
conducting
shell with radii
b to c.
Nonconducting
sphere with
radius R &
charge Q
uniformly
distributed
Eout=kq/r2
Vout =kq/r
Vc =kq/c
Ebetween=
kq/r2
Ein=0
Vbetween
=kq[(1/r)
+(1/c)(1/b)]
Vb =kq/c
Eout=kQ/r2
Vout =kQ/r
VR =kQ/R
Ein=
kQr/R3
Vin
=(3kQ/2R)
(kQr2/2R3)
Vcener
=3kQ/2R
E
+V
R
R
E
+V
a b c
Va
=(kq/c)+Vba
a
2-4
b c
+V
E
R
13. Two conducting sphere of different radii a & b have equal amounts of charge Q on them.
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a)
b)
c)
d)
Va >Vb.
Charges will move until the potentials are equal.
The final charge ratio qa/qb=a/b
The final surface density ratio a/b =b/a. This shows final surface densities are NOT equal which is a
common misconception.
e) If b=2a, then qa/qb=1/2; a/b =2/1. It is also possible to figure out how the original 2Q charge has
been divided. Since qa+qb= 2Q, you can figure out that qa=2Q/3 and that qb=4Q/3.
14.
a) Because potential is not a vector you don’t have to worry about directions. Add up potentials from both
charges and set equal to zero to find answers:
(4q/x) + q/(d-x) =0, the two possible places are x=4d/3 and x=4d/5.
4q
-q
These are only two points in an equipotential surface that
(x-d)
(d-x)
0
encircles the –q charge.
d
V=0
b) There is only one point where E=0, and that’s at x=2d. The reason the zero
points are not the same is simply that E=-dV/dx and just because a function
has a zero value it doesn’t mean its derivative is also zero. Remember that E
changes in value as you pass over the E=0 point.
15. Energy conservation requires that Uo + Ko= U+K. In this problem the Uo=-kq1q2/d, Ko=0, U=0, and
K=m1v12/2 + m2v22/2. Momentum conservation requires that po1 + po2= p1 + p2. The particles start at rest, so 0
+0= m1v1 - m2v2.
Using the givens you can combine these equations and get expressions for the final speeds of the particles. I
get 7.24 x105 m/s and 4.24 x107 m/s.
16.
a) Vy=0. Vx= kp/(x2-a2) for |x|>|a|. For |x|<|a|,Vx= 2kqx/(a2-x2).
b) The electron here has no energy with respect to infinity but it experiences a +i horizontal electric force
due to the electric field. The electron will accelerate and curve toward the x-axis, it will cross the axis
with max speed and begin to decelerate and curve toward the (-)y-axis. It should come to a (temporary)
rest at the same distance below zero as it started above zero. Then it will retrace the path back to its
original location.
y
c) & d)Begin with the approximation Vxy= kq[(r-acos) -(r+acos) ].
Since a<<r, this expression can be simplified to
Vxy= kq2acos/(r2-a2cos2) = kpx/r3.
-1
-1
-q
r

-a
+q
x
a
e) This is only worth doing if you want to practice your math skills: Ex= -dV/dx in addition Ey= -dV/dy,
where r=(x2+y2)-1/2. Cranking through the derivatives gives

kp
E  3 3 cos 2   1i  3 sin  cos j
r
This show that it is possible to get expressions for complex electric fields if you first determine the
potential which is easier to do since it is a scalar.


y
17. Consider a set of two +q charges a distance 2a apart on the x-axis.
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+q
+q
-a
a
x
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a) Vx= 2kqx/(x2-a2) for |x|>|a|.
b) Vy= 2kq/(y2+a2)-1/2.
c) In both cases a becomes negligible and you get Vx= 2kq/x, Vy= 2kq/y, which are the point potentials for
a 2q charge.
18. Another important symmetry to study is the cylindrical spreading electric field. Imagine a long rod with
linear charge density  and cross-section radius a.
b
a
a) You will have to integrate the electric field from a to b:
Vab= -S 2kdr/r=2kln(a/b).
b) It doesn’t make sense to define the zero of potential at infinity here because it doesn’t simplify the
potential function. If we arbitrarily define the zero potential at b, then V(r) = 2kln(b/r). This means
that for r>b, V<0 and for r<b, V>0.
c) Since E= -dV/dr = 2k/r, which is the correct expression for the electric field of a rod.
d) Graphs: of V vs r and of E vs r, again taking V=0 at b.
E
+V
a
b
r
e) The potential between the rod and cylinder would be the same as for the rod alone. Outside, the
potential is the same as on the surface of the cylinder since Eout=0.
19. Determine the potential function for a thin rod of charge Q and length L along the x-axis a distance “x”
from the center of the rod.
l
a) Here it is better to integrate the potential contributions of each
L/2
x
element of charge dQ: V= S kdl /(x-l) = kln[(x+.5L)/(x-.5L)]
-L/2
b) Set-up is similar to problem above but resulting integral in not a common one:
V= kCd/(x-) , and you would need a table of integrals to determine expression.
c) Show that the results in (a) above approaches that for a point charge as x>>a, but not in (b) above.
20. Consider a ring of radius R and charge Q.
a) The integral here is trivial because potential is a scalar, Vx=kQ/(R2 + x2)1/2.
b) Again, Ex= -dV/dx= kQx/(R2 + x2)3/2.
x
R
c) Obviously V is a maximum at x=0. E is a maximum at x= + 0.71R as derived
in the previous chapter.
+V
E
d) Draw graphs of V vs r and of E vs r.
.71R
21. Consider a circular plate of charge Q and radius R.
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a) & b) This problem is similar to the one we did to find the field. Divide the plate into infinitesimal rings
of charge dQ, radius r, and potential dV=kdQ/(r2 + x2)1/2. Using the charge density dQ=2rdr. It is fairly
easy to integrate dV and get V= 2kQ[(R2 + x2)1/2-x]/R2.
22. Some of the more “tricky” problems involve concentric spheres carrying different charges. Consider the
following cases. Assume both sphere and shell are conductors and that the shell’s thickness is negligible.
a)
Example
Equipotential
illustration
Vba=Vb–Va
V(r)
Inner sphere
charge is +q
and outer
sphere of -q
Equipotentials are
all spherical
surfaces..
Vba =
kq(1/a -1/b)
Vout =0;
V at a and at
b w/infinity
Vb=0;
Vin=
kq(1/r -1/b)
Va=
kq(1/a -1/b)
a
Graph of
E vs. r
Graph of V
vs. r
+V
E
a
b
a
b
b)
Inner sphere
charge is +q
and outer
sphere of +q
Vba =
kq(1/a -1/b)
a
Vout =k2q/r;
Vb =k2q/b;
Vin=
kq(1/r +1/b)
Va=
kq(1/a + 1/b)
Vout =18/r;
Vb =72 v;
b
+V
E
a
b
a
b
b
c)
d)
Inner sphere
charge is 2nC and
outer sphere
of +4nC
Vba =
kq(1/a -1/b)
= -108 v
10 cm
+V
E
Vin=
144 - (18/r)
Va= -36 v
Vout =
k(q1+ q2)/r;
Vb =
k(q1+ q2)/b;
Vin=
k(q1/r + q2/b)
Va=
k(q1/a+ q2/b)
a
b
a
b
25 cm
Inner sphere
charge is q1
and outer
sphere of q2
Vba =
kq1(1/a -1/b)
Depends on
charges
Depends on
charges
a
b
23. This change does not affect the potentials outside the shell, but the potential difference between inner and
outer spheres will be less than before and the potential of the inner sphere will be less also: Vb =72 v; Vba =
-90 v; Va = -18 v.
23.5 Assuming a vertical orientation of the plate, E= (x/o)i, where x is the distance from the center of the
plate perpendicular to the surface. Integrating the field inside the plate gives V=-(D2/2o). The potential
decreases parabolically from the center.
24. a) For any two point a & b in an electric field,
 E  ds  V
ab
 Vba  V ab  V ab  0 , qed.
b) Start out by tracing out a rectangular closed path in the electric field that goes parallel
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and perpendicular to the lines of force as shown. You can argue that the integral on this
closed path cannot be zero because the field is stronger on the upper part of the path. This
contradicts the requirement that the potential be path independent (conservative) so a field
that increases in magnitude without changing direction is not possible.
25. This is mostly a kinematics problem. The electron spends time t1= L/vx between the plates accelerating
toward the positive plate at rate a=eE/m= eVd/md. It travels vertically y1=at12/2, and comes out with a vertical
component of velocity vy=at1. After exiting the plates, the electron has no acceleration but it spends time t2=
D/vx moving toward the screen and traveling vertically and additional y2= vy t2.
a) The final vertical position of the beam when it strikes the screen is Y=y1 +y2. Putting everything in
terms of the given quantities you get, Y= (eVdL/mdvx2)(L/2 + D), which is proportional to Vd.
b) Because the vertical position of the beam is proportional to Vd of the plates, one can connect an
external potential to the plates and measure it with the beam displacement.
c) The horizontal sweep is accomplished by having a second set of parallel plates perpendicular to the
first that deflects the beam back and forth at a particular rate.
y
Va
d, Vd
x
vx
L
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