Chapter 23: Electric Potential
Objectives
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Electric Potential and Electric Potential Energy—what are
they and how are they related?
What is Voltage and how does it relate to electric potential?
How are the electric potential and the electric field related?
The Electron Volt, a unit of energy—how do I calculate it?
What is the electric potential around a point charge or point
charges?
How do I use this to calculate the potential surrounding an
Electric Dipole?
Cathode Ray Tubes: TVs, Monitors and Oscilloscopes, how
are they made, and what do they do?
If I know V how do I find E?
Lecture Note
1. Introduction
The electrostatic force is a conservative force. This means that the
work it does on a particle depends only on the initial and final position
of the particle, and not on the path followed. With each conservative
force, a potential energy can be associated. The introduction of the
potential energy is useful since it allows us to apply conservation of
mechanical energy which simplifies the solution of a large number of
problems.
The potential energy U associated with a conservative force F is
defined in the following manner
(1)
where U(P0) is the potential energy at the reference position P0
(usually U(P0) = 0) and the path integral is along any convenient path
connecting P0 and P1. Since the force F is conservative, the integral in
eq.(1) will not depend on the path chosen. If the work W is positive
(force and displacement pointing in the same direction) the potential
energy at P1 will be smaller than the potential energy at P0. If energy
is conserved, a decrease in the potential energy will result in an
increase of the kinetic energy. If the work W is negative (force and
displacement pointing in opposite directions) the potential energy at P1
will be larger than the potential energy at P0. If energy is conserved,
an increase in the potential energy will result in an decrease of the
kinetic energy. If In electrostatic problems the reference point P0 is
usually chosen to correspond to an infinite distance, and the potential
energy at this reference point is taken to be equal to zero. Equation
(1) can then be rewritten as:
(2)
To describe the potential energy associated with a charge distribution
the concept of the electrostatic potential V is introduced. The
electrostatic potential V at a given position is defined as the potential
energy of a test particle divided by the charge q of this object:
(3)
In the last step of eq.(3) we have assumed that the reference point P0
is taken at infinity, and that the electrostatic potential at that point is
equal to 0. Since the force per unit charge is the electric field (see
Chapter 23), eq. (3) can be rewritten as
(4)
The unit of electrostatic potential is the volt (V), and 1 V = 1 J/C = 1
Nm/C. Equation (4) shows that as the unit of the electric field we can
also use V/m.
A common used unit for the energy of a particle is the electron-volt
(eV) which is defined as the change in kinetic energy of an electron
that travels over a potential difference of 1 V. The electron-volt can be
related to the Joule via eq.(3). Equation (3) shows that the change in
energy of an electron when it crosses over a 1 V potential difference is
equal to 1.6 . 10-19 J and we thus conclude that 1 eV = 1.6 . 10-19 J.
2. Calculating the Electrostatic Potential
A charge q is moved from P0 to P1 in the vicinity of charge q' (see
Figure 1). The electrostatic potential at P1 can be determined using eq.
(4) and evaluating the integral along the path shown in Figure 1. Along
the circular part of the path the electric field and the displacement are
perpendicular, and the change in the electrostatic potential will be
zero. Equation (4) can therefore be rewritten as
(5)
If the charge q' is positive, the potential increases with a decreasing
distance r. The electric field points away from a positive charge, and
we conclude that the electric field points from regions with a high
electrostatic potential towards regions with a low electrostatic
potential.
Figure 1. Path followed by charge q between P0 and P1.
From the definition of the electrostatic potential in terms of the
potential energy (eq.(3)) it is clear that the potential energy of a
charge q under the influence of the electric field generated by charge
q' is given by
(6)
Example Problem 1: Electrostatic Potential of a Charged Rod
A total charge Q is distributed uniformly along a straight rod of length
L. Find the potential at point P at a distance h from the midpoint of the
rod (see Figure 2).
The potential at P due to a small segment of the rod, with length dx
and charge dQ, located at the position indicated in Figure 3 is given by
(7)
The charge dQ of the segment is related to the total charge Q and
length L
(8)
Combining equations (7) and (8) we obtain the following expression
for dV:
(9)
Figure 2. Problem 21.
Figure 3. Solution of Problem 21.
The total potential at P can be obtained by summing over all small
segments. This is equivalent to integrating eq.(9) between x = - L/2
and x = L/2.
(10)
Example Problem 2: Distance of Closest Approach
An alpha particle with a kinetic energy of 1.7 x 10-12 J is shot directly
towards a platinum nucleus from a very large distance. What will be
the distance of closest approach ? The electric charge of the alpha
particle is 2e and that of the platinum nucleus is 78e. Treat the alpha
particle and the nucleus as spherical charge distributions and disregard
the motion of the nucleus.
The initial mechanical energy is equal to the kinetic energy of the
alpha particle
(11)
Due to the electric repulsion between the alpha particle and the
platinum nucleus, the alpha particle will slow down. At the distance of
closest approach the velocity of the alpha particle is zero, and thus its
kinetic energy is equal to zero. The total mechanical energy at this
point is equal to the potential energy of the system
(12)
where q1 is the charge of the alpha particle, q2 is the charge of the
platinum nucleus, and d is the distance of closest approach. Applying
conservation of mechanical energy we obtain
(13)
The distance of closest approach can be obtained from eq.(13)
(14)
3. The Electrostatic Field as a Conservative Field
The electric field is a conservative field since the electric force is a
conservative force. This implies that the path integral
(15)
between point P0 and point P1 is independent of the path between
these two points. In this case the path integral for any closed path will
be zero:
(16)
Equation (16) can be used to prove an interesting theorem:
" within a closed, empty cavity inside a homogeneous conductor, the
electric field is exactly zero ".
Figure 4. Cross section of cavity inside spherical conductor.
Figure 4 shows the cross section of a possible cavity inside a spherical
conductor. Suppose there is a field inside the conductor and one of the
field lines is shown in Figure 4. Consider the path integral of eq.(16)
along the path indicated in Figure 4. In Chapter 24 it was shown that
the electric field within a conductor is zero. Thus the contribution of
the path inside the conductor to the path integral is zero. Since the
remaining part of the path is chosen along the field line, the direction
of the field is parallel to the direction of the path, and therefore the
path integral will be non-zero. This obviously violates eq.(16) and we
must conclude that the field inside the cavity is equal to zero (in this
case the path integral is of course equal to zero).
4. The Gradient of the Electrostatic Potential
The electrostatic potential V is related to the electrostatic field E. If the
electric field E is known, the electrostatic potential V can be obtained
using eq.(4), and vice-versa. In this section we will discuss how the
electric field E can be obtained if the electrostatic potential is known.
Figure 5. Calculation of the electric field E.
Consider the two points shown in Figure 5. These two nearly identical
positions are separated by an infinitesimal distance dL. The change in
the electrostatic potential between P1 and P2 is given by
(17)
where the angle [theta] is the angle between the direction of the
electric field and the direction of the displacement (see Figure 5).
Equation (17) can be rewritten as
(18)
where EL indicates the component of the electric field along the L-axis.
If the direction of the displacement is chosen to coincide with the xaxis, eq.(18) becomes
(19)
For the displacements along the y-axis and z-axis we obtain
(20)
(21)
The total electric field E can be obtained from the electrostatic
potential V by combining equations (19), (20), and (21):
(22)
Equation (22) is usually written in the following form
(23)
where --V is the gradient of the potential V.
In many electrostatic problems the electric field of a certain charge
distribution must be evaluated. The calculation of the electric field can
be carried out using two different methods:
1. the electric field can be calculated by applying Coulomb's law and
vector addition of the contributions from all charges of the charge
distribution.
2. the total electrostatic potential V can be obtained from the algebraic
sum of the potential due to all charges that make up the charge
distribution, and subsequently using eq.(23) to calculate the electric
field E.
In many cases method 2 is simpler since the calculation of the
electrostatic potential involves an algebraic sum, while method 1 relies
on the vector sum.
Example Problem 3: Electric Field derived from Electrostatic
Potential
In some region of space, the electrostatic potential is the following
function of x, y, and z:
(24)
where the potential is measured in volts and the distances in meters.
Find the electric field at the points x = 2 m, y = 2 m.
The x, y and z components of the electric field E can be obtained from
the gradient of the potential V (eq.(23)):
(25)
(26)
(27)
Evaluating equations (25), (26), and (27) at x = 2 m and y = 2 m
gives
(28)
(29)
(30)
Thus
(31)
Example Problem 4: Potential and Field of a Charged Annulus
An annulus (a disk with a hole) made of paper has an outer radius R
and an inner radius R/2 (see Figure 6). An amount Q of electric charge
is uniformly distributed over the paper.
a) Find the potential as a function of the distance on the axis of the
annulus.
b) Find the electric field on the axis of the annulus.
We define the x-axis to coincide with the axis of the annulus (see
Figure 7). The first step in the calculation of the total electrostatic
potential at point P due to the annulus is to calculate the electrostatic
potential at P due to a small segment of the annulus. Consider a ring
with radius r and width dr as shown in Figure 7. The electrostatic
potential dV at P generated by this ring is given by
(32)
where dQ is the charge on the ring. The charge density [rho] of the
annulus is equal to
(33)
Figure 6. Problem 36.
Using eq. (33) the charge dQ of the ring can be calculates
(34)
Substituting eq.(34) into eq.(32) we obtain
(35)
The total electrostatic potential can be obtained by integrating eq.(35)
over the whole annulus:
(36)
Figure 7. Calculation of electrostatic potential in Problem 36.
Due to the symmetry of the problem, the electric field will be directed
along the x-axis. The field strength can be obtained by applying
eq.(23) to eq.(36):
(37)
Since the electrostatic field and the electrostatic potential are related
we can replace the field lines by so called equipotential surfaces.
Equipotential surfaces are defined as surfaces on which each point has
the same electrostatic potential. The component of the electric field
parallel to this surface must be zero since the change in the potential
between all points on this surface is equal to zero. This implies that
the direction of the electric field is perpendicular to the equipotential
surfaces.
5. The Potential and Field of a Dipole
Figure 8 shows an electric dipole located along the z-axis. It consists of
two charges + Q and - Q, separated by a distance L. The electrostatic
potential at point P can be found by summing the potentials generated
by each of the two charges:
(38)
Figure 8. The electric dipole.
If the point P is far away from the dipole (r >> L) we can make the
approximation that r1 and r2 are parallel. In this case
(39)
and
(40)
The electrostatic potential at P can now be rewritten as
(41)
where p is the dipole moment of the charge distribution. The electric
field of the dipole can be obtained from eq.(41) by taking the gradient
(see eq.(23)).
Electric Potential and Capacitance
Recall that a conservative force is one in which the work
done on a particle that moves between two points depends
only on the two points and not on the path followed. With this
property, we were able to define a potential energy that could be
associated with the force involved. Similarly, we can define a
conservative vector field as a field that generates a conservative
force. This allows us to introduce the concept of a potential for the
field, in direct analogy to the potential energy for a force. Specifically,
we define the potential, Δ, of a vector field, ƒ, by the relation
 
   f  d
(17)
Notice that the potential is a scalar quantity. Instead of three (or
more) equations to solve, there is only one. It is this fact that makes
the potential easier to work with in most calculations. Also, just as
with potential energy, the potential of a field is not an exact quantity.
Instead, it only has physical meaning when changes in the potential
are considered. In fact, the field potential is related to the potential
energy in exactly the same way the field is related to the force (up to
a change in sign), i.e., by
 
U
W

q0
q0
(18)
where q0 is the charge associated with the field under consideration.
Equipotential Surfaces
When ΔW = 0, we have a special situation. By equation (18),
we have that Δ = 0. The locus of points which have Δ = 0 is
called an equipotential surface. All equipotential surfaces are at
right angles to the field lines everywhere. This fact makes it very easy
to find the equipotential surfaces once the field lines are known (or
vice versa).
Example:
Find the equipotential surfaces for two charges of equal
magnitude but opposite signs separated by a distance a.
When dealing with electromagnetism, it is common to denote the
electric potential by the symbol, V, which has units of volts. In terms
of more fundamental units, a volt is defined as
1 volt = 1 joule / coulomb.
Since electromagnetism is a long range force, it is common to take the
reference point, Va, to be located at infinity, so that Va = 0. When this
choice is made, then ΔW is just the work required to move a test
charge in from infinity to the point in question. Notice though, that
there is no requirement that Va be at infinity; occasionally, the
inherent symmetry of the problem will dictate another choice for the
location of Va. An example of this will be given later.
Let us now consider a single point charge. What is the potential
due to this charge? Recall that the electric field for the charge is given
by

q
E  k 2 r
r
Then we find, for a point charge
Vk
q
r
(19)
If we have a collection of discrete charges. then the potential for
the system is just the sum of the individual potentials:
n
V  k
i 1
qi
ri
Example:
What is the potential of an electric dipole?
(20)
+q
r

1
r
2a
r
2
-q
Assume that the two charges have a magnitude of q and are
separated by a distance 2a. For simplicity, place the center of the
dipole at the origin. Then for a point at a distance r from the origin
and at an angle θ from the dipole direction, we have
q q
V  V1  V2  k   
 r1 r2 
where
r1  r 2  a 2  2ra cos 
and
r2  r 2  a 2  2ra cos 
In the limit r >> 2a, this reduces to
Vk
p cos 
r2
(21)
where p = 2aq is the electric dipole moment. Notice that the dipole
is strongest along it's axis of orientation (θ = 0, π) for any radius.
The electric dipole moment is found in many molecules and
atoms. It is formed when the distribution of the charge is not
symmetric about the physical distribution of the object. In these
cases, there is an excess of positive charge at one side, and negative
charge at the other. When the object is given a dipole moment by an
external electric field, then the dipole is known as an induced electric
dipole moment. When the external field is removed, the induced
dipole moment will also disappear.
Example:
What is the potential of a spherical shell of radius R with a
charge Q?
For a spherical charge density, the electric field is given by

Qr
E  k 3 r
R
Since we are working with a shell instead of a solid sphere, this is split
into three sections: r > R, r = R, and r < R. Let us look at each one
separately.
Case 1: r > R.
In this case, the electric field is given by

Q
E  k 2 r
r
Since this is the electric field of a point charge, we have already found
the potential. It is
Vk
Q
r
Case 2: r < R.
In this case, the electric field is zero (all of the charge is on the
surface of the shell). So we see that
Vb - Va = 0  Vb = Va = const
where the constant is yet to be determined.
Case 3: r = R.
This case is what is called a boundary condition. Since V is a
continuous function, the values of V(r>R) and V(r<R) must be equal at
r = R. Looking at V(r>R), we see that, at r = R, we have
Vk
Q
R
Thus, we see that the unknown constant for case 2 is just
Vk
Q
R
Summing all this up, we have that, for a spherical shell, the potential
is given by

k
V 
k

Q
R
Q
r
rR
rR
(22)
In this example we can see an interesting phenomenon. We can
write Q as a density times a volume. As the volume is decreased, the
density will increase. This causes a buildup of large potentials and
fields at sharp points and corners. If the charge buildup is large
enough, then the electric field generated by the charge will cause the
air around it to ionize. This effect is known as a corona discharge
and the best known example of it is lightning during a thunderstorm.
Example:
What is the potential of an infinite line charge?
The change in the voltage is given by
ΔV = - E · Δl.
The electric field of an infinite line charge is given by


E  2 k r
r
(23)
where λ is the charge per unit length. Taking Δl = ∫r r and summing
over all the Δr, we get
V  2 k ln r  Va
where Va is our reference potential. Recall that earlier I said that it is
common to take the reference potential, Va, to be located at infinity,
so that Va = 0. In this case, if we take r = ∞, instead of getting Va =
0 (as we would want) we would find Va = 2k λ lna. Thus to have Va =
0, we must take some arbitrary distance r = a. Then the potential
becomes
V  2 k ln
r
a
(24)
ELECTRIC ENERGY OF A SYSTEM OF POINT CHARGES
1 Introduction
Figure 1. System of three charges.
The electric potential energy U of a system of two point charges was
discussed in our previous Chapter and is equal to
(1)
where q1 and q2 are the electric charges of the two objects, and r is
their separation distance. The electric potential energy of a system of
three point charges (see Figure 1) can be calculated in a similar
manner
(2)
where q1, q2, and q3 are the electric charges of the three objects, and
r12, r13, and r23 are their separation distances (see Figure 1). The
potential energy in eq.(2) is the energy required to assemble the
system of charges from an initial situation in which all charges are
infinitely far apart. Equation (2) can be written in terms of the
electrostatic potentials V:
(3)
where Vother(1) is the electric potential at the position of charge 1
produced by all other charges
(4)
and similarly for Vother(2) and Vother(3).
Example Problem: Model of a Carbon Nucleus
According to the alpha-particle model of the nucleus some nuclei
consist of a regular geometric arrangement of alpha particles. For
instance, the nucleus of 12C consists of three alpha particles on an
equilateral triangle (see Figure 2). Assuming that the distance between
pairs of alpha particles is 3 x 10-15 m, what is the electric energy of
this arrangement of alpha particles ? Treat the alpha particles as
pointlike.
Figure 2. Alpha-particle model of
12C.
The electric potential at the location of each alpha particle is equal to
(5)
where d = 3.0 x 10-15 m. The electric energy of this configuration can
be calculated by combining eq.(5) and eq.(3):
(6)
2 Energy of a System of Conductors
Figure 3. The capacitor.
The electrostatic energy of a system of conductors can be calculated
using eq.(3). For example, a capacitor consists of two large parallel
metallic plates with area A. Suppose that charges +Q and -Q are
placed on the two plates (see Figure 3). Suppose the electrostatic
potential of plate 1 is V1 and the potential of plate 2 is V2. The
electrostatic energy of the capacitor is then equal to
(7)
The electric field E between the plates is a function of the charge
density [sigma]
(8)
The potential difference V1 - V2 between the plates can be obtained by
a path integration of the electric field
(9)
Combining eq.(9) and eq.(7) we can calculate the electrostatic energy
of the system:
(10)
This equation shows that electrostatic energy can be stored in a
capacitor. Equation (10) can be rewritten as
(11)
where Volume is the volume between the capacitor plates. The
quantity [epsilon]0 . E2/2 is called the energy density (potential energy
per unit volume).
Figure 4. Field lines at the edge of a capacitor.
In the calculation of the energy density carried out for the capacitor we
assumed that the electric field was homogeneous in the region
between the plates. In a real capacitor the field at the edge is not
homogeneous, and the calculation will have to be modified. Figure 4
shows a couple of field lines at the edge of a capacitor. Consider the
two small sections of the capacitor plates with charges dQ and -dQ,
respectively, shown in Figure 4. The contribution of these two sections
to the total electrostatic energy of the capacitor is given by
(12)
where V1 and V2 are the electrostatic potential of the top and bottom
plate, respectively. The potential difference, V1 - V2, is related to the
electric field between the plates
(13)
The electric field E(l) can be related to the charges on the small
segments of the capacitor plates via Gauss' law. Consider a volume
with its sides parallel to the field lines (see Figure 5). The electric flux
through its surface is equal to
(14)
where E(l) is the strength of the electric field at a distance l from the
bottom capacitor plate (see Figure 5) and dS(l) is the area of the top
of the integration volume. The flux is negative since the field lines are
entering the integration volume. The flux through the sides of the
integration volume is zero since the sides are chosen to be parallel to
the field lines. The flux through the bottom of the integration volume is
also zero, since the electric field in any conductor is zero. Gauss' law
requires that the flux through the surface of any volume is equal to the
charge enclosed by that volume divided by [epsilon]0:
(15)
Figure 5. Integration volume discussed in the text.
Combining eq.(14) and eq.(15) we obtain
(16)
Equations (12), (13) and (16) can be combined to give
(17)
This calculation can be generalized to objects of arbitrary shapes, and
the electrostatic energy of any system can be expressed as the volume
integral of the energy density u which is defined as
(18)
Thus
(19)
where the volume integration extends over all regions where there is
an electric field.
Example Problem: Fission of Uranium
In symmetric fission, the nucleus of uranium (238U) splits into two
nuclei of palladium (119Pd). The uranium nucleus is spherical with a
radius of 7.4 x 10-15 m. Assume that the two palladium nuclei adopt a
spherical shape immediately after fission; at this instant, the
configuration is as shown in Figure 6. The size of the nuclei in Figure 6
can be calculated from the size of the uranium nucleus because
nuclear material maintains a constant density.
Figure 6. Two palladium nuclei right after fission of 238U.
a) Calculate the electric energy of the uranium nucleus before fission
b) Calculate the total electric energy of the palladium nuclei in the
configuration shown in Figure 6, immediately after fission. Take into
account the mutual electric potential energy of the two nuclei and also
the individual electric energy of the two palladium nuclei by
themselves.
c) Calculate the total electric energy a long time after fission when the
two palladium nuclei have moved apart by a very large distance.
d) Ultimately, how much electric energy is released into other forms of
energy in the complete fission process ?
e) If 1 kg of uranium undergoes fission, how much electric energy is
released ?
a) The electric energy of the uranium nucleus before fission can be
calculated using the known electric field distribution generated by a
uniformly charged sphere of radius R:
(20)
For the uranium nucleus q = 92e and R = 7.4 x 10-15 m. Substituting
these values into eq.(20) we obtain
(21)
b) Suppose the radius of a palladium nucleus is RPd. The total volume
of nuclear matter of the system shown in Figure 6 is equal to
(22)
Since the density of nuclear matter is constant, the volume in eq.(22)
must be equal to the volume of the original uranium nucleus
(23)
Combining eq.(23) and (22) we obtain the following equation for the
radius of the palladium nucleus:
(24)
The electrostatic energy of each palladium nucleus is equal to
(25)
where we have used the radius calculated in eq.(24) and a charge qPd
= 46e. Besides the internal energy of the palladium nuclei, the electric
energy of the configuration must also be included in the calculation of
the total electric potential energy of the nuclear system
(26)
where qPd is the charge of the palladium nucleus (qPd = 26e) and Rint is
the distance between the centers of the two nuclei (Rint = 2 RPd = 11.7
x 10-15 m). Substituting these values into eq.(26) we obtain
(27)
The total electric energy of the system at fission is therefore
(28)
c) Due to the electric repulsion between the positively charge
palladium nuclei, they will separate and move to infinity. At this point,
the electric energy of the system is just the sum of the electric
energies of the two palladium nuclei:
(29)
d) The total release of energy is equal to the difference in the electric
energy of the system before fission (eq.(21)) and long after fission
(eq.(29)):
(30)
e) Equation (30) gives the energy released when 1 uranium nucleus
fissions. The number of uranium nuclei in 1 kg of uranium is equal to
(31)
The total release of energy is equal to
(32)
To get a feeling for the amount of energy released when uranium
fissions, we can compare the energy in eq.(32) with the energy
released by falling water. Suppose 1 kg of water falls 100 m. The
energy released is equal to the change in the potential energy of the
water:
(33)
The mass of water needed to generate an amount of energy equal to
that released in the fission of 1 kg uranium is
Energy of the Electric Potential
Since the electric potential is defined in terms of the potential
energy, we can use our knowledge of the electric potential to
determine the energy stored in the electric field. In general this is
given by
U  qV
Example:
How much energy is required to move an electron through a
potential difference of 1 V?
Using the relation between energy and electric potential we have
U  eV  (1.6x10 -19 C)(1 V)  1. 6 x10 19 J
This combination shows up so frequently in particle physics that it has
been given it's own designation, the electron volt, or eV. We
frequently talk about particles with kinetic energies in the thousands,
millions, billions or even trillion eV, so these are denoted keV, MeV,
GeV and TeV respectively.