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PHYSICS 126
Lecture Notes Book 1
Prepared by Kai Wong
Table of Contents
1. Charging Processes
………………………………………………………… 2
2. Coulomb’s Law
…………………………………………………………. 7
3. Electric Field
……………………………………………………. 11
4. Electric Potential I
…………………………………………. 19
5. Electric Potential II
………………………………………………………… 27
6. Capacitance …………………………………………………………………… 33
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1. Charging Processes
Atomic Origin of Electricity
Atoms are composed of electrons and nuclei, which permanently carry negative and
positive electric charges respectively. The charge carried by a single electron is equal to
 e , where
e  1.6  1019 C
The symbol C stands for Coulomb, which is the SI unit for charge. A basic property of
electric charges is that like-charges repel and unlike-charges attract. The force weakens
with distances.
The atom is neutral as a whole, carrying as many units of positive as negative electric
charges. It becomes an ion when it loses or gains electrons. Aggregate matter is therefore
usually neutral, and becomes noticeably charged only if a significant amount of charge of
one kind has been transferred into or out of the matter, in the form of transfer of electrons
or ions.
Example: How much is the charge left on 1 cc of water after all the electrons have been
removed?
Solution: Since the density of water is 1 g/cm3, the mass is 1g.
From the molecular formula H2O for water, the molecular weight is
2 1  16  18 .
1
 number of moles 
 0.056
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number of molecules  0.056  6.022  1023  3.37  1022
The number of electrons per molecule = 2+8=10
 electric ch arg e  3.37  1022  10  1.6  1019 C  5.4  104 C
When a glass rod is rubbed with silk cloth, electrons are transferred from the rod to the
cloth. As a result, the glass rod becomes positively charged, while the silk cloth becomes
negatively charged. On the other hand, when a plastic rod is rubbed with fur, the charge
transferred is such that the plastic rod becomes negatively charged, while the fur is
positively charged. Because charging arises from separation and transfer of charges, no
net charge is created in the universe by the charging process. This fact is called the
conservation of electric charge, and can be expressed as
Q  Q
after
becore
2
where the summation is over all objects involved in the charge transfer process, and the
charge Q carries the proper sign.
Ordinary matter can be classified into conductors and insulators. All metals are
conductors. In a conductor, a large number of electrons are ionised from the atoms, and
are called free electrons, because they can move freely inside the conductor in response
to electric forces. Insulators are such materials as plastics, dried wood, and glasses. All
electrons in an insulator are bound to atomic nuclei , so that charges cannot flow freely.
If a conductor is isolated and contains an excess of electrons so that it becomes
negatively charged, the excess electrons tend to move to populate the surface of the
conductor because of their mutual repulsion. Likewise, if the conductor is depleted in
electrons so that it is positively charged as a whole, the excess positive charges also
reside on the surface.
Excess charge on an isolated conductor stays on its surface:
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Charging by Contact
When two conductors are brought into contact, they behave as a single conductor so that
electrons can flow freely among the two. In particular, a neutral conductor can be charged
up after contact with a charged conductor, and bears charge of the same kind. The earth
itself is a giant conductor. If a charged conductor is brought in contact with the earth, as
happens when it is touched by your hands and you are not wearing rubber shoes, the
charge on the conductor will be shared with the earth. Since the latter is so much bigger,
almost all the charge on the conductor is drained to the earth. This phenomenon is known
as grounding. To avoid grounding, the conductor should stand on an insulator to isolate
from the earth.
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Example: Two identical metal spheres initially contain charges  5q and  9q
respectively. They are brought into contact and then separated. What are the charges on
them after separation?
Solution: Since the two spheres are identical, the charges are equally distributed between
while in contact. Let this charge be Q . From charge conservation:
2Q  5q  9q
 Q  2q
Charging by Induction
When a positively charged object is brought near a neutral conductor but not in contact,
the free electrons in the conductor are attracted by the object and rush to reside on the
part of the surface near the object, leaving behind a net positive charge on the part of the
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surface away from the object. A polarization is said to be induced on the conductor.
When the object is removed, the free electrons will move to nullify the polarization so
that the conductor is restored to its original configuration with no charge on the surface.
On the other hand, if you touch the far end of the conductor while the positively charged
object is still in place, the positive charge on the far end is conducted away to ground. If
the object is now removed, there is a net negative charge left on the conductor, which,
because of mutual repulsion, will spread out to occupy the surface. The isolated
conductor is now negatively charged.
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Polarization also occurs when a charged object is brought near certain insulators. These
are insulators for which the charge distribution inside a molecule is skewed, so that one
end of the molecule is slightly more positive while the other end is slightly more
negative. The molecule is said to form an electric dipole. Ordinarily, the orientation of
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these dipoles are random, so that in any given volume of the material, the net charge is
zero. If a positively charged object is brought near such an insulator, the negative ends of
the dipoles are attracted to the object, resulting in a lining up of the dipoles. Inside any
given volume of the insulator, the net charge is still zero, because of cancellation between
positive and negative charges. However, on the surface facing the object, the negative
ends of the dipoles are not cancelled by positive charges from other dipoles. The net
charge on this part of the surface is therefore negative. Similarly, the far end of the
insulator becomes positively charged. Because the negative charge on the insulator is
closer to the positively charged object, the insulator is attracted to the object with a
stronger force than the repulsive force on its positive far end. If the insulator is light, it
can be levitated and stick to the object. This is why shreds of paper can be picked up by a
comb that you have rubbed against your hair.
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2. Coulomb’s Law
Two point charges at a distance r apart and carrying charges q1 and q2 respectively will
either repel or attract each other with a force whose magnitude is given by
F k
q1  q2
r2
The force is an attraction if q1 and q2 have opposite signs, otherwise it is a repulsion. If
the charges q1 and q2 are measured in C, the distance r is in meter, and the force F is in
Newton, the constant k , known as the Coulomb’s law constant, is given by
k  9.0  109 N  m2 / C 2
Note that the forces on the two charges are equal and opposite, and form an action and
reaction pair in the sense of Newton’s third law of motion. The force decreases with
distance, being inversely proportional to the distance squared, similar to the gravitational
force between two point masses.
For convenience, we also introduce the constant  0 known as the permitivity of free
space through the definition
k
1
4 0
 0  8.85  10 C 2 / N  m 2 
12
Example: Find the force between point charges  3nC and 2nC at a distance of 2m
apart.
Solution: The force is repulsive, with magnitude
3  109  2  109
F  9.0  109 
 13.5  10 9 N
2
2
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Example: In the Bohr model of hydrogen atom, the electron is at a distance of
r  5.29  1011 m from the nucleus. Find the electrostatic force of attraction on the
electron due to the nucleus. Compare it with the gravitational force, also due to the
nucleus.
Solution:
The electrostatic force is
Felectric

ke2 9.0  109  1.6  1019
 2 
2
r
5.26  1011



2
 8.2  108 N
Since
Fgravity 
Gme m p
r2
where the gravitational constant G  6.67  1011 N  m2 / kg2
the mass of electron me  9.1  1031 kg
mass of proton mp  1.67 1027 kg
the ratio
Fgravity
Felectric

Gme m p
ke
2
 4.4  10 41
so that the gravitation force is very much weaker.
Example: Assuming the orbit of the electron in the previous example is a circle, find the
speed of the electron.
Solution: The electrostatic force on the electron always point toward the proton, and acts
as the centripetal force for the circular motion. Therefore
Felectric
v 
me v 2

r
Felectricr
 2.18  10 6 m / s
me
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The proton experiences the same force Felectric directed toward the electron. However, its
acceleration is negligible compared with that of the electron, because the proton mass is
much larger.
When more than two point charges are present, to calculate the force on one of the
charges, it is necessary to add up the forces due to the rest of the charges. The addition of
forces obeys the law of vector addition.
Example: Three point charges qa  3C, qb  2C, qc  1C lie on a straight line.
The distance between q a and q b is 2m. Find the force on q c if
(1) it is closer to q b and at 1m from it.
(2) It is midway between q a and q b
Solution: (1) The configuration of charges is as shown:
a
3μC
b
c
2μC
1μC
2m
1m
The colored arrows represent the forces on q c due to q a (blue)and q b (red). We see that
the former is a repulsion and the latter is an attraction. These forces are
a on c: Fac  9.0  10 9 
3  10 6  1  10 6
 3  10 3 N
2
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directed to the right
b on c: Fbc  9.0  10 9 
2  10 6  1  10 6
 18  10 3 N
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directed to the left
The net force on q c is therefore 15  10 3 N directed to the left.
(2) The charge configuration is
c
a
b
3μC
1m
1μC
2μC
2m
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The forces on q c due to q a (blue) and q b (red) are
3  10 6  1  10 6
Fac  9.0  10 
 27  10 3 N
2
1
to the right
2  10 6  1  10 6
 18  10 3 N
2
1
to the right
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Fbc  9.0  10 9 
The net force is therefore 45  10 3 N to the right.
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3. Electric Field
Any collection of charges is called a charge distribution. A charge distribution can be
discrete, consisting of a number of point charges, or it can be continuous, occupying a
portion of a surface (2-D) or a volume (3-D) of space.
A point charge introduced into the vicinity of a charge distribution is called a test charge.
Let the charge it carries be q 0 . At any position the test charge is placed, it will experience

a force F due to the charges in the distribution, and this force is proportional to q 0
according to Coulomb’s law. The quantity

 F
E
q0
is therefore independent of the charge q 0 , depending only on the charge distribution and
the location of the point where the test charge is place. It is called the electric field at that
location. It is equal to the force acting on a test charge of  1C . The unit of electric field
is N/C (newton per coulomb). We can also write


F  q0 E
which tells us how to calculate the force on a point charge at a location where the electric
field is known. In particular, the direction of the force is the same as the direction of the
electric field if the charge is positive, and is opposite if the charge is negative.
Electric Field due to a point charge.
The simplest charge distribution is a point charge q . It creates an electric field
everywhere in space. At a point at a distance r from the charge, the magnitude of the
electric field is, from Coulomb’s law,

q
E k 2
R

The direction of E is radially outward from the point charge if it is positive, and inward
toward the point charge if it is negative. The following figure displays the electric field
vectors in a number of locations surrounding a positive and a negative charge
respectively.
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Electric Lines of Force
These are imaginary lines with assigned arrows drawn in space with the property that at
any point on the line, the direction of the electric field is tangent to the line and agrees
with the direction of the arrow. A number of lines of force are drawn for the positive and
point charge as shown:
For a charge distribution consisting of more than one point charge or an arbitrary
continuous distribution, the lines of force are curves. Customarily, we draw equally
spaced lines of force emanating from a positive point charge or converging onto a
negative point charge. The number of lines coming out or going into a point charge is
proportional to the magnitude of the charge. The lines of force have the following
properties:
1. Two different lines cannot intersect each other. (Because if they do, there would be
two directions for the electric field at the point of intersection.)
2. The lines are not closed. They either come out of positive charges, terminate at
negative charges, or extend to infinity.
3. Regions where the lines are more crowded together have stronger electric field.
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Electric field due to a number of point charges
The electric field due to a discrete point charge distribution can be calculated as the
vector sum of the electric field due to the individual charges.
One of the simplest discrete distribution is an electric dipole, consisting of a pair of point
charges of equal in magnitude but opposite in sign. As an example, consider a dipole
made up of charges  5nC and  5nC separated by a distance of 2m. Let’s find the
electric field at the location A, which is at 1m from the negative charge as shown:
 5nC
 5nC
A
The electric field at A due to the  5nC charge (blue) is
9.0  10 9 
5  10 9
 5 N / C to the right (blue arrow)
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The electric field at A due to the  5nC charge (red) is
5  10 9
9
9.0  10 
 45 N / C to the left (red arrow)
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Therefore the electric field of the dipole at the point A is 40N/C and points to the left, and
is indicated as a black arrow.
Next, we’ll find the electric field at the point B as shown:
y
B
θ
x
2
5m m
θ
5nC
1
m
5nC
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The electric fields due to the blue and red charges at B are indicated as blue and red
arrows. Since the distance between B and either one of the charge is 12  2 2  5 m,
The fields have the same magnitude of
9.0  10 9 
5  10 9
 5
2
 9N / C
To add the blue and red vectors, we introduce x and y axes at the point B as shown. We
have:
sum of y components of the two vectors =0;
sum of x components is 9 cos  9 cos  18 cos , where the angle θ can be calculated
from the triangle on the left as
cos 
1
5
 0.447
Therefore the electric field at the point B has magnitude 18  0.447  8.0N / C . Its
direction is indicated by the black arrow.
The following figure shows the field lines from a dipole:
For two equal point charges of the same sign, the pattern of field lines is as shown:
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Example: Four point charges lie on the corners of a square with side of length a as
shown. Find the electric field at the center of the square.
A
B
+q
-q
O
-q
-q
D
C
Solution: At the center O, the electric fields due to the charges at the corners B and D
cancel out because they are equal but in the opposite directions. On the other hand, the
fields due to A and C are equal but in the same direction. The length AO, denoted by r ,
can be found from Pythagoras theorem:
r 2  r 2  a2
r 
a
2
The magnitude of the electric field at O is therefore
2k
q 4kq
 2
r2
a
and points toward the corner C.
Example: Three point charges are at the corners of a square of side 1m as shown. Find
the electric field at the unoccupied corner.
A
B
1m
+4nC
-4nC
n
C
y
x
-4nC
-q
D
C
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Solution: In the figure above, the directions of the electric fields due to the three charges
at the point D are colored. Their magnitudes are
4  10 9
 36 N / C
12
E A  EC  9.0  10 9 
E B  9.0  10 
9
4  10 9
 2
2
 18 N / C since the length BD is
2 from Pythagoras theorem.
Choosing x and y axes at the point D as shown, we have, for the components of the net
electric field,
E x  18 cos 45  36  48.7
E y  18 sin 45  36  23.3

2
The magnitude of the electric field is E  48.7 2   23.3  54.0 N / C
The angle it makes with the x-axis is   tan 1
 23.3
 25.6 
48.7
Electric field of an infinite charge sheet
Consider an infinite plane sheet carrying a uniform 2-D distribution of charge. Let the
surface charge density be  . The unit of  is C/m2. The pattern of field lines are
shown below for positively charged and negatively charged sheets respectively.
Positive sheet
Negative sheet
At any point on either side of the sheet, the magnitude of the electric field is


E 
2 0
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For a pair of sheets bearing surface charge densities of equal magnitude but opposite
signs, the magnitude of the electric field between the sheets is
 
E 
0
being twice as much as that due to a single sheet. Elsewhere the electric field is zero. This
is because between the sheets, the contributions from the two sheets add, but they cancel
elsewhere as shown in the figure( positive sheet and its electric field is in blue, negative
sheet and its electric field in red)
In practice, sheet charges are finite in extent. The formulas for infinite sheets are
applicable at locations close to the surface compared with the extent. The configuration
of parallel sheet charges of opposite signs can be created by connecting two metallic
plates to the two terminals of a battery. This constitutes a parallel plate capacitor. Note
that at the ends of a parallel plate capacitor, the field lines are curved. Well inside the
plates, the electric field is uniform in space, meaning that it is independent of the
location.
Example: (Motion of electron in region of uniform electric field) An electron travelling
horizontally at 5.0  10 7 m / s enters into the space between two parallel plates where the
electric field is equal to 2.5  10 4 N / C and points downward. How large is the upward
deflection of the electron after travelling 0.2m in the region?
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Solution:
y
x
E
The force on the electron is upward and has magnitude F  eE
F
eE
 y  component of acceleration a y 

me me
x  component of acceleration a x  0
If v denotes the horizontal velocity when the electron enters the field region, the time to
x
travel a horizontal distance x is t 
v
During this time, the vertical distance travelled by the electron is
1
1 eE  x 
eEx 2
1.6  10 19  2.5  10 4  0.2 2
y  ayt 2 

 0.035m
  
2
2
2 me  v 
2me v 2
2  9.1  10 31  5.0  10 7
2


Shielding
Electrostatics deals with situations in which charges are not moving. In this case, the
electric field inside a conductor is zero, because otherwise the free electrons will move.
Excess charges in a conductor therefore reside on the surface in electrostatics. Since the
electric field arises from the surface charges and other charges outside the conductor, the
various contributions must cancel out inside the conductor. If we hollow out a volume
inside the conductor, the surface charges and the charges outside the conductor are not
disturbed. Therefore the electric field in the hollowing remains zero. This is obtained
even if the hollowing is complete so that only a thin shell of the conductor remains. We
thus arrive at a situation in which no matter how large the electric field is outside a
conductor shell, there is no electric field inside the shell. This works even if the shell is
replaced by a cage made with metallic wires. The phenomenon is known as shielding.
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+
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E=
0
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+
For this reason, it is quite safe to sit in your car during thunderstorm.
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4. Electric Potential
Definition of Electric Potential Consider first the simple case of a region of space where


the electric field E is uniform. In this region, a test charge q 0 experiences a force q 0 E .
The displacement of the charge in moving from the initial position Pi to the final position

Pf is the vector pointing from Pi to Pf , and is denoted by  r .
Δr
Pf
θ
Pi
E
The work done by the electric force on the charge is


W  q0 E  r  cos 


where  is the angle between the vector E and  r . Defining the dot product between
two vectors A and B as

   


A  B  A  B  cos angle between A and B

we can write
 
W  q0 E  r .
Note that in this formula q 0 can be either positive or negative.
In the general case where the electric field is not uniform, and an arbitrary path is
followed in moving the charge from Pi to Pf , the work done by the electric force along
the path can be calculated by dividing the path into small segments of displacements,
using the above formula with the local electric field for the work done in the small
segment, and adding up the contributions:
 
W  q0  E  r along the
path
from Pi
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to Pj
Pf
Δr
E
Pi
It can be shown that the electric force is conservative, which means that the work is
independent of the path, and depends only on the two end points. Therefore, we can
associate with each position of the test charge a quantity called its electric potential
energy U in such a way that
Ui U f  W
According to this definition, in going from a place with higher potential energy to one
with lower potential energy U i  U f  , positive work is done by the electric field on the
test charge. Conversely, in going from a point with lower to one with higher potential
energy, the work done is negative. This is similar to the definition of gravitational
potential energy U g  mgh : in going from a higher position to a lower position on the
earth’s surface, the gravitational attraction on a mass m does positive work.
We now define the electric potential V at any point by
V 
U
q0
so that U  q0V
and observe that
 
Vi  V f   E  r
along any path from Pi to Pf
The right side of this equation can be regarded as the work done by the electric force on a
unit positive test charge. Since only differences in potentials are involved in calculating
work, the value of the electric potential is determined only after we have chosen certain
point to be a reference point and call the electric potential zero there.
Just as the electric field, the electric potential at a point does not depend on the test
charge, and depends only on the point itself in relation to the charge distribution. The unit
of electric potential is J/C(Joule/Coulomb), which is known as V (volt).
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Specialized to the case of a uniform electric field, or applied to an infinitesimal path, we
can write, in view of the customary meaning of the delta notation as final value minus the
initial values,
 
V  E  r
which is a fundamental relation between the electric field and the electric potential.
Electric potential in a region of uniform electric field
A region of uniform electric field exists between the oppositely charged plates of a
parallel plate capacitor as we have shown. Suppose the x-axis is chosen to be the
direction of the electric field, whose x-component we denote by E x and is positive. We
consider two points A,B so that the line AB is perpendicular to the x-axis.
E
A
B
Applying the fundamental relation,
VA  VB  0
because the displacement is perpendicular to the electric field. This shows that the
potential on all points of a plane perpendicular to the electric field have the same electric
potential. A surface on which all points have the same electric potential is called an
equipotential surface. For a uniform electric field, we see that the equipotential surfaces
are planes perpendicular to the field lines.
Consider a point C “downwind” of the electric field from the point A and at a distance of
x from it.
E
A
C
Then for the displacement from A to C, the increment of potential is V  VC  V A , so
that according to the fundamental relation between potential and electric field, we have
V A  VC  E x x
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which is a positive number. Thus, locations “downwind” of the electric field have smaller
electric potential. Also, electric field points in the direction from high to low electric
potentials.
The field lines (red) and equipotential surfaces (blue) inside a parallel plate capacitor are
as shown:
x
d-x
x
d
With the x-axis chosen as shown, and the electric potential chosen to be zero at the plate
on the right, the potential V at a point whose coordinate is x satisfies the relation
V  Ed  x 
where d is the distance between the plates. The graph of V against x is a straight line:
V
V0
x
d
The potential of the left plate is given by V0  Ed .
Electric Potential of a Point Charge
For a point charge q , zero potential is usually chosen to be at infinity. The potential V at
a point P at a distance r from the charge can be calculated from the work done by the
electric force on a unit positive test charge (or, by the electric field) for the displacement
from P to a point at infinity. As we move from P to infinity, the electric field changes,
and so the path has to be divided into many small parts and then summed. This process is
known as integration. It leads to the result
q
V k
r
q
r
P
V  0
22
Note that V has the same sign as q : a positive (negative) charge creates positive
(negative) potential in its environment.
Equipotential surfaces of a point charge are concentric spherical surfaces centered on the
charge. The field lines, going out radially from the point charge, are perpendicular to the
surfaces.
The potential due to a distribution of point charges can be obtained from the sum of the
potential due to the individual charges. Being a scalar, the potential is easier to calculate
than the electric field.
Example: For the dipole consisting of point charges of 5nC and  5nC 2m apart as
shown, find the electric potentials at the points A and B.
B
θ
2m
5m m
θ
A
-5nC
2m
1m
5nC
m
m
Solution: Since the distances between A and positive and negative charges are 3m and 1m
respectively,
5  10 9
 15V
Potential at A due to the positive charge is V L 9.0  10 9 
3
5nC
Potential at A due to the negative charge is VR  9.0  10 9 
23
(5  10 9 )
 45V
1
Potential at A is V  15  45  30V
The point B is at equal distance of
V  VL  VR  9.0  10 9 
5  10 9
5
5m from the two charges. The potential there is
 9.0  10 9 
(5  10 9 )
5
0
Equipotential surfaces and field lines of a dipole can be found in the text book.
Graphical Representation of Relation between Electric Potential and Electric Field
In general, field lines are always perpendicular (normal) to equipotential surfaces.

To prove this, we consider a point P on an equipotential surface and the electric field E
at the point. We take another point Q on the same surface but very close to P, and let

 r denote the displacement from P to Q. Applying the fundamental relation
 
V  E  r , we find
E
Q
P
 
E  r  0
because Q and P have the same potential. This means the angle between the vectors


E and  r is 90º. Since the point Q can be in any direction away from P while still on the
surface, we conclude that the direction of the field is normal to the equipotential surface
In a 1-D problem, the electric potential depends only on the x coordinate and the electric
 
field has only x-component. The fundamental relation V  E  r reduces to
V   E x x
and can be rewritten as
24
Ex  
V
x
Thus in a plot of V against x , the negative of the slope at any point gives the xcomponent of the electric field at that point, as we have seen in the case of the parallel
plate capacitor. This also holds true in the following two situations, although both are not
1-D problem.
For a point charge q at the origin of the x-axis, the potential at any point on the x-axis is
V x  
kq
x
Plots of V versus x and E x versus x are as shown:
V
Ex
x
x
For a dipole made up of point charges  q and  q at a distance 2a apart, with the xaxis chosen so that the coordinates of the left and right charges are  a and a
respectively, the potential of an arbitrary point on the x-axis with coordinate x is
xa
-a
a
xa
q
q
V x   k
k
xa
xa
The plots of V and Ex versus x are as shown:
Ex
V
x
x
25
In a 2-D problem, the potential depends on both x and y coordinates. The equipotential
surfaces are now curves in the x-y space. They are usually drawn so that neighboring
curves differ by equal amount of potential. The electric field at any point P is
perpendicular to the equipotential surface it is located at, and is in the direction pointing
from the higher to the lower potential. The magnitude of the electric field is obtained
from
 V
E 
r
where V is the potential difference between neighboring surfaces, and r is
perpendicular distance between the neighboring surfaces at the point P.
Decreasing V

r
P
Thus, regions where the equipotential surfaces are closer together correspond to strong
electric field.
Electric Potential of Conductors.
Since the electric field inside a conductor is zero in electrostatics, zero work is done by
the electric field in going from one point to the other inside the conductor. Therefore, any
two points in the conductor have the same electric potential. In fact, the whole conductor
is an equipotential body. In particular, the surface of the conductor is an equipotential
surface. The electric field line passing through any point on the surface is perpendicular
to the surface. When multiple conductors are present, each is an equipotential body but
the potentials are in general different.
26
5. Electric Potential II
Potential Difference and Work Done on a Charge
From the definition of electric potential, we deduce that when a charge q moves from an
initial location where the electric potential is Vi to a final location where the potential is
V f , the work done by the electric field is
Wi  f  qVi  V f

Thus, in going from a place of high to one of lower electric potential, the electric field
does positive work on a positive charge and negative work on a negative charge. When
positive work is done on an object, its kinetic energy increases, otherwise the kinetic
energy decreases, according to the work energy theorem.
Example:
A battery maintains a fixed potential difference between its two terminals because of the
chemicals it contains. The potential is higher at the positive terminal. If a wire connects
the terminals, the electric field inside the wire exerts a force on the electrons, causing
them to move toward the positive terminal. Thus, negative charges are moved from a
location of low potential (the negative terminal) to one of high potential (the positive
terminal). The work done by the electric field on these charges is positive, which should
cause increase of kinetic energy of the electrons. However, this increase does not take
place when we have a steady flow of electrons: their speed does not change in going from
one to the other terminals. This is because the electrons are not completely free. They
collide with the positive ions, thereby experience a force of resistance. As a result, heat is
evolved in the wire. The work done by the electric field is then equal to the heat energy
evolved.
Customarily, we replace the flow of negative charge by a flow of fictitious positive
charge in the opposite direction, and the direction of electric current is taken to be the
direction of motion of these fictitious positive charges. The work done by the electric
field on these fictitious positive charges is equal to the real work done on the electrons.
Suppose in a certain battery the potential difference between the terminals is kept at 12V,
and the heat evolves at a rate of 60W in the wire connecting the terminals. How much
charge has been transferred between the terminals in one hour’s of operation?
Solution:
Heat energy evolved = power ×time =60×3600=216,000 J
Work done by electric field in moving a charge q = qVi  V f   q  12
27
Work done by electric field = heat evolved in circuit
 q  12  216,000
 q  18,000C
The energy unit electron volt (eV)
Atomic particles carry charges in multiples of the electronic charge e  1.6  10 19 C . For
their motion, it is convenient to introduce a unit of energy, known as the electron volt.
The definition is
1eV
 1.6  10 19 J
Example: Find the work done in eV by the electric field on a helium nucleus moving
from a point where the electric potential is 200V to another point where the potential is
600V.
Solution: Using the fact that the charge of a helium nucleus is 2e , the work done by the
electric field is
W  2  1.6  10 19  200  600 J
 2  200  600eV  800eV
 2  1.6  10 19  200  600 / 1.6  10 19 eV
Example: Find the speed of a 500eV electron.
Solution: The kinetic energy of the electron is
K  500eV  500  1.6  10 19 J  8.0  10 17 J
From the relation
1
K  me v 2 and me  9.1  10 31 kg for the mass of an electron,
2
the speed is v 
2K

me
2  8.0  10 17
 1.3  10 7 m / s
31
9.1  10
Example What is the minimum work required to move an electron initially at rest at a
distance of 5.29  10 11 m from the nucleus of a hydrogen atom to a point very far away
from the nucleus?
Solution
Because the electron experiences an attractive force from the nucleus, an external force is
required to move the electron away from the nucleus. If Wext denotes the work done by
this force, from the work-energy theorem, the change of kinetic energy of the electron is
28
K  Wext  WE
where WE is the work done by the electric force, and is given by
WE  eVi  V f

The potential at the initial location is
e
1.6  10 19
9
Vi  k  9.0  10 
 27.2V
r
5.29  10 11
while at the final location which is very far away, V f  0 . Thus the work done by the
electric force is
WE  eVi  1.6  10 19  27.2 J  27.2 eV
and is negative.
Since the initial kinetic energy K i is zero, the work energy theorem yields
Wext  K  WE  K f  WE
This work is minimum when K f  0 . It value is therefore
Wext  WE  27.2 eV
In a hydrogen atom, the electron does not stay at rest at a distance r from the nucleus.
Instead it travels in a circle around the nucleus with r as the radius. In this case, the
initial kinetic energy is not zero. In fact, it can be shown that K i  13.6eV . Still with
K f  0 , the minimum external energy to remove the electron from the atom is
Wext   K i  WE  13.6  27.2  13.6 eV
This is known as the ionisation energy of the hydrogen atom.
Relation between electric force and potential
We now consider the problem of determining the motion of a test charge q 0 in the
vicinity of a charge distribution. We know that the force on the test charge is given by



F  q0 E , and E is in the direction of decreasing potential. Therefore, for a positive test
charge, the electric force is in the direction of decreasing potential, while for a negative
29
test charge, the force is in the direction of increasing potential. In other words, positive
charges seek lower potential and negative charges seek higher potential. We have also
shown that the potential energy of the test charge is given by
U  q0V
Therefore, for both positive and negative test charges, the force is in the direction of
decreasing potential energy, as required by the concept of potential energy.
Application of conservation of energy
The law of conservation of energy for the motion of a charged particle in the electric field
created by a charge distribution states that the sum of kinetic energy and electric potential
does not change. It is a consequence of the work energy theorem and the conservative
nature of the electric force. If the charge and mass of the particle are q and
m respectively, the law can be written in the form
1 2
1
mv f  qV f  mvi2  qVi
2
2
where the subscripts on the symbol v for speed refer to the initial and final stages of the
motion. An alternative form is
1 2 1 2
mv f  mvi  q Vi  V f
2
2

which explicitly shows that only difference in potentials enters.
Example A potential difference of 2kV is applied between the two plates of a parallel
plate capacitor. A proton is introduced at rest on the positive plate. Find the speed of the
proton when it reaches the negative plate.
Solution We use energy conservation equation in the form
K f  K i  qVi  V f

Since K i  0 , q  e for proton, and Vi  V f  2000V , we find
K f  1.6 10 19  2000  3.2 10 16 J
Using m  1.67  10 27 kg for the mass of a proton, the speed is
30
vf 
2K f
m

2  3.2  10 16
 6.2  10 5 m / s
 27
1.67  10
Example If an electron is introduced at the positive plate of the capacitor in the previous
problem with a velocity of 10 7 m / s directed toward the negative plate, what fraction of
the distance between the plates can the electron reach?
Solution
When the electron moves toward the negative plate, its kinetic energy is continuously
reduced while the potential energy qV  eV  increases. But the electric potential
V  decreases. Let V be the drop in potential from the positive plate to the location
where the electron stops. Applying energy conservation’
0  K i   eV
Since
Ki 
 
1
1
me vi2   9.1  10 31  10 7
2
2
2
 4.6  10 17 J
V
2kV
K
4.6  10 17
V  i 
 288V
e
1.6  10 19
x
d
Since the graph of potential against the distance from the positive plate is a straight line,
the drop in potential is proportional to the distance from the positive plate. Using the fact
that the total potential drop from the positive to the negative plate is 2000V, the fractional
penetration of the electron is
x
V

 0.14
d 2000
Example An alpha particle (another name for the helium nucleus, carrying a charge
 2e ) with kinetic energy of K  5MeV travels directly toward a bare nucleus of an atom
whose atomic number is Z (The charge of the nucleus is therefore Ze ). Find the distance
of closest approach if the bare nucleus is gold. ( Z  79)
Solution: The potential created by the bare nucleus at a distance r is
V k
Ze
r
31
Let rmin denote the distance of closest approach. Applying energy conservation between
this location ( K f  0 ) and the initial location which is infinitely far
away Vi  0, K i  K  ,
 Ze
0  2e k
 rmin
 rmin

  K  0


2kZe2 2  9.0  10 9  79  1.6  10 19


K
5  10 6  1.6  10 19
which is approaching the size of the nucleus.
32

2
 4.6  10 14 m
6. Capacitance
Definition of Capacitance
A capacitor is a device to store electric charge. It also stores electric energy as we shall
demonstrate.
Conceptually, a capacitor is formed when we have two metallic conductors carrying
equal and opposite variable charges  Q and  Q respectively. These charges create an
electric field and electric potential in the environment, with the conductors forming
equipotential bodies. We denote the potential difference between the conductors by V .
This difference is defined as the potential of the positively charged conductor minus that
of the negatively charged bodies, and its value is therefore positive. From Coulomb’s
law, we expect V to be proportional to Q . The more the charge carried by the
conductors, the more the potential difference between them. Therefore the quantity
C
Q
V
is independent of the charge Q carried by the conductors. It is known as the capacitance
of the two conductors, and depends only on such geometrical factors as the sizes, shapes,
and distances of the conductors. The unit of capacitance is F (farad), which is equal to
coulomb per volt (C/V)
Rearranging the above equation to read
Q  CV
it follows that a large capacitance means large charge storage with modest potential
difference. This is important as a large potential difference translates into a large electric
field. When the electric field is too large, the medium surrounding the conductors can
become conducting, and the capacitor discharges. The minimum electric field for
discharge to happen is called the breakdown electric field. In air, the breakdown field is
about one MV/m.
Parallel Plate Capacitor
This simplest capacitor is formed by two parallel conductor plates so close to each other
that they can be considered as infinite plates as far as the calculation of the electric field
between them is concerned. A formula for its capacitance can be obtained by calculating
the potential difference V when the plates carry charges Q and  Q . Because of
attraction, these charges are concentrated on the inner sides of the metallic plates facing
each other, forming sheet charges. If A denote the area of each side of the plate, the
surface charge densities of the sheets are   where
33
d

Q
A
A
A
V
The uniform electric field between the plates then has the magnitude
E

Q

0 0 A
and points from the positive to the negative plates. Letting d be the distance between the
plates, the potential difference is given by
V  Ed
Substituting the equation for E in terms of Q , we find
V 
Qd
0 A
from which it follows from the definition of capacitance that
C  0
A
d
and depends on geometrical factors only.
Usually, the plates are charged by connecting them to the terminals of a dc power supply,
such as a battery. The potential difference is fixed by the power supply.
Example: Find the capacitance of a parallel plate capacitors made from circular plates of
radius 10cm at a distance of 1mm apart in air. How much charge is stored when the plates
are charged to a potential difference of 200V? How large is the electric field?
Solution
d  1mm  10 3 m
2
A  r 2    0.1  0.0314m 2
34
C   0
A
0.0314
 8.85  10 12 
 2.78  10 10 F
3
d
10
Q  CV  2.78  10 10  200  5.56  10 8 C
E
V 200
 3  2.00  10 5 V / m
d 10
The energy of a capacitor
To build up the charge on a capacitor to its final value Q starting from no charge, one
can imagine a process of continually extracting positive charge from the negative plate
and depositing it on the positive plate. Work is done by the external force during this
process. The total work done in reaching the final configuration can be considered to be
stored as electrical energy.
To calculate the work done, suppose at one time during the build up, the capacitor carries
the charge q , which is less than Q . At this time, the potential difference is denoted by v ,
which is less than the final potential difference V .
When a small amount of charge  q is moved from the negative to the positive plates,
work done by external force Wext  work done by electric force  qv
The total work ,which is equated to the energy stored, and is denoted by U is then
U  Wext   vq
q
, the graph of v against q is a straight line as shown in the
C
figure. The summation above can be carried out by noticing that Wext is equal to the area
underneath the graph. This area is that of a right angle triangle with sides equal to Q and
V.
v
From the relation v 
V
Therefore,
U 
1
QV
2
Δ
q
35
q
Q
By using the relation Q  CV , the energy can be written also in the forms
U
1
CV 2
2
or
U
Q2
2C
Example: How much energy is stored in the parallel plate capacitor in the previous
example.
Solution:
U
1
1
CV 2   2.78  10 10  200 2  5.56  10 6 J
2
2
Energy stored in an electric field
In the case of parallel plate capacitor, the stored energy can be rewritten as follows:
1
1 0 A
Ed 2  1  0 E 2  Ad 
U  CV 2 
2
2 d
2
The factor Ad is equal to the volume of the space between the plates, which is only
region with nonzero electric field. Therefore we can take the point of view that energy is
stored in the space where there is an electric field, and that
electric field energy density 
1
0E2
2
Dielectrics
A dielectric is a material made up of molecules that are permanent electric dipoles. When
it is inserted between the plates of a parallel plate capacitor, the electric field causes the
dipoles to be aligned, with the negative ends pointing to the positive plate. This results in
a negative surface charge on the face of the dielectric that is in contact with the positive
plate, and a positive surface charge on the face in contact with the negative plate. Inside
36
+
+
+
+
+
+
+
+
+
-
+
-
+
-
+
-
+
-
+
-
  induced

the dielectric, the charge is zero.
We say that a surface charge is induced on the dielectric as a result of polarization
induced by the electric field created by the charges on the plates. Let  induced denote the
surface charge density on the dielectric. Together with the charge  on the plates, a pair
of sheet charges with charge density
      induced
are formed. This is reduced from the value  without dielectric. Such an effect is known
as screening. The ratio


is a property of the dielectric called the dielectric constant. It is always greater than one.

The electric field inside the capacitor is now due to these sheets charges. Its value is
 

,
 0  0
and is less than the value E0    0 without dielectric.
E

Q
d
d
 0
 0 A
where we have used   Q A . Thus, the capacitance is
V  Ed 
C
Q

A
37