Physics
Lecture 4
 Electrostatics
 Electric flux and Gauss’s law
 Electrical energy
 potential difference and
electric potential
 potential energy of charged
conductors
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapters 15-16
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Lightning Review
Last lecture:
1. Properties of the electric field, field lines
2. Conductors in electrostatic equilibrium
 Electric field is zero everywhere within the conductor.



Any excess charge field on an isolated conductor resides on its surface.
The electric field just outside a charged conductor is perpendicular to the
conductor’s surface.
On an irregular shaped conductor, the charge tends to accumulate at
locations where the radius of curvature of the surface is smallest.
Review Problem: Would life be different if the electron were positively
charged and the proton were negatively charged? Does the choice of
signs have any bearing on physical and chemical interactions?
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15.10 Electric Flux and Gauss’s Law
A convenient technique was introduced by Karl F. Gauss
(1777-1855) to calculate electric fields.
Requires symmetric charge distributions.
Technique based on the notion of electrical flux.
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15.10 Electric Flux
To introduce the notion of flux, consider
a situation where the electric field is
uniform in magnitude and direction.
Consider also that the field lines cross a
surface of area A which is
perpendicular to the field.
The number of field lines per unit of
area is constant.
The flux, F, is defined as the product of
the field magnitude by the area crossed
by the field lines.
F  EA
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Area=A
E
4
15.10 Electric Flux
Units: Nm2/C in SI units.
Find the electric flux through the area A = 2 m2, which is
perpendicular to an electric field E=22 N/C
F  EA
Answer: F = 44 Nm2/C.
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15.10 Electric Flux
If the surface is not perpendicular to the field, the
expression of the field becomes:
F  EA cosq
Where q is the angle between the field and a normal to
the surface.
N
q
q
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15.10 Electric Flux
Remark:
When an area is constructed such that a closed surface
is formed, we shall adopt the convention that the flux
lines passing into the interior of the volume are negative
and those passing out of the interior of the volume are
positive.
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Example:
Question:
Calculate the flux of a constant E field (along x) through a
cube of side “L”.
y
1
2
E
x
z
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Question:
Calculate the flux of a constant E field (along x) through a cube of side “L”.
Reasoning:
 Dealing with a composite, closed surface.
 Sum of the fluxes through all surfaces.
 Flux of field going in is negative
 Flux of field going out is positive.
 E is parallel to all surfaces except surfaces labeled 1 and 2.
 So only those surface contribute to the flux.
Solution:
F1   EA1 cos q1   EL2
F 2  EA2 cos q 2  EL2
F net   EL2  EL2  0
y
1
2
E
x
z
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15.10 Gauss’s Law
The net flux passing through a closed surface
surrounding a charge Q is proportional to the magnitude
of Q:
Fnet   EA cosq  Q
In free space, the constant of proportionality is 1/eo
where eo is called the permittivity of of free space.
1
1
12
2
2
eo 


8.85

10
C
N

m
4 ke 4 8.99 109 Nm2 / C 2

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
10
15.10 Gauss’s Law
The net flux passing through any closed surface is equal
to the net charge inside the surface divided by eo.
F net   EA cos q 
Q
eo
Can be used to compute electric fields. Example: point
charge
Q
Q
2
F net   EA cos q  4 r E  E 
 ke 2
2
4e 0 r
r
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16.0 Introduction
The Coulomb force is a conservative force
A potential energy function can be defined for any
conservative force, including Coulomb force
The notions of potential and potential energy are
important for practical problem solving
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16.1 Potential difference and electric potential
The electrostatic force is
conservative
As in mechanics, work is
E
B
A
d
W  Fd cos
Work done on the positive charge
by moving it from A to B
W  Fd cos   qEd
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Potential energy of electrostatic field
The work done by a conservative force equals the
negative of the change in potential energy, DPE
DPE  W  qEd
This equation


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is valid only for the case of a uniform electric field
allows to introduce the concept of electric potential
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Electric potential
The potential difference between points A and B, VB-VA,
is defined as the change in potential energy (final minus
initial value) of a charge, q, moved from A to B, divided
by the charge
DPE
DV  VB  VA 
q
Electric potential is a scalar quantity
Electric potential difference is a measure of electric
energy per unit charge
Potential is often referred to as “voltage”
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Electric potential - units
Electric potential difference is the work done to move a
charge from a point A to a point B divided by the
magnitude of the charge. Thus the SI units of electric
potential
1V  1 J C
In other words, 1 J of work is required to move a 1 C of
charge between two points that are at potential
difference of 1 V
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Electric potential - notes
Units of electric field (N/C) can be expressed in terms of the
units of potential (as volts per meter)
1 N C  1V m
Because the positive tends to move in the direction of the
electric field, work must be done on the charge to move it in
the direction, opposite the field. Thus,


A positive charge gains electric potential energy when it is moved in
a direction opposite the electric field
A negative charge looses electrical potential energy when it moves
in the direction opposite the electric field
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Analogy between electric and gravitational fields
The same kinetic-potential energy theorem works here
A
A
d
E
q
B
d
g
m
B
If a positive charge is released from A, it accelerates in
the direction of electric field, i.e. gains kinetic energy
If a negative charge is released from A, it accelerates in
the direction opposite the electric field
KEi  PEi  KE f  PE f
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Example: motion of an electron
What is the speed of an electron accelerated from rest across a
potential difference of 100V? What is the speed of a proton
accelerated under the same conditions?
Given:
DV=100 V
me = 9.1110-31 kg
mp = 1.6710-27 kg
|e| = 1.6010-19 C
Vab
Observations:
1. given potential energy
difference, one can find the
kinetic energy difference
2. kinetic energy is related to
speed
KEi  PEi  KE f  PE f
KE f  KEi  KE f  DPE  qDV
Find:
ve=?
vp=?
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1 2
2 qDV
mv f  qDV  v f 
2
m
ve  5.9 106 m , v p  1.3 105 m
s
s
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16.2 Electric potential and potential energy
due to point charges
Electric circuits: point of zero potential is defined by
grounding some point in the circuit
Electric potential due to a point charge at a point in
space: point of zero potential is taken at an infinite
distance from the charge
With this choice, a potential can be found as
q
V  ke
r
Note: the potential depends only on charge of an object,
q, and a distance from this object to a point in space, r.
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Superposition principle for potentials
If more than one point charge is present, their electric
potential can be found by applying superposition
principle
The total electric potential at some point P due to several
point charges is the algebraic sum of the electric
potentials due to the individual charges.
Remember that potentials are scalar quantities!
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Potential energy of a system of point
charges
Consider a system of two particles
If V1 is the electric potential due to charge q1 at a point P,
then work required to bring the charge q2 from infinity to P
without acceleration is q2V1. If a distance between P and
q1 is r, then by definition
q2
P
q1
r
A
q1q2
PE  q2V1  ke
r
Potential energy is positive if charges are of the same
sign and vice versa.
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Mini-quiz: potential energy of an ion
Three ions, Na+, Na+, and Cl-, located such, that they
form corners of an equilateral triangle of side 2 nm in
water. What is the electric potential energy of one of the
Na+ ions?
Cl?
qNa qCl
qNa qNa
qNa
PE  ke
 ke
 ke
 qCl  qNa 
r
r
r
but : qCl  qNa !
Na+
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Na+
qNa
PE  ke
 qNa  qNa   0
r
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16.3 Potentials and charged conductors
Recall that work is opposite of the change in potential
energy,
W  PE  q VB  VA 
No work is required to move a charge between two points
that are at the same potential. That is, W=0 if VB=VA
Recall:
1. all charge of the charged conductor is located on its surface
2. electric field, E, is always perpendicular to its surface, i.e. no work is
done if charges are moved along the surface
Thus: potential is constant everywhere on the surface of a
charged conductor in equilibrium
… but that’s not all!
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Because the electric field in zero inside the conductor,
no work is required to move charges between any two
points, i.e.
W  q VB  VA   0
If work is zero, any two points inside the conductor have
the same potential, i.e. potential is constant everywhere
inside a conductor
Finally, since one of the points can be arbitrarily close to
the surface of the conductor, the electric potential is
constant everywhere inside a conductor and equal to its
value at the surface!
Note that the potential inside a conductor is not necessarily zero,
even though the interior electric field is always zero!
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The electron volt
A unit of energy commonly used in atomic, nuclear and
particle physics is electron volt (eV)
The electron volt is defined as the energy that electron
(or proton) gains when accelerating through a potential
difference of 1 V
Relation to SI:
Vab=1 V
1 eV = 1.6010-19 C·V = 1.6010-19 J
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Problem-solving strategy
Remember that potential is a scalar quantity
Superposition principle is an algebraic sum of potentials due
to a system of charges
Signs are important
Just in mechanics, only changes in electric potential are
significant, hence, the point you choose for zero electric
potential is arbitrary.
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Example : ionization energy of the electron
in a hydrogen atom
In the Bohr model of a hydrogen atom, the electron, if it is in the
ground state, orbits the proton at a distance of r = 5.2910-11 m. Find
the ionization energy of the atom, i.e. the energy required to remove
the electron from the atom.
Note that the Bohr model, the idea of electrons as tiny balls orbiting the nucleus, is not
a very good model of the atom. A better picture is one in which the electron is spread
out around the nucleus in a cloud of varying density; however, the Bohr model does
give the right answer for the ionization energy
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In the Bohr model of a hydrogen atom, the electron, if it is in the ground state,
orbits the proton at a distance of r = 5.29 x 10-11 m. Find the ionization energy,
i.e. the energy required to remove the electron from the atom.
Given:
r = 5.292 x 10-11 m
me = 9.1110-31 kg
mp = 1.6710-27 kg
|e| = 1.6010-19 C
Find:
E=?
The ionization energy equals to the total energy of the
electron-proton system,
E  PE  KE
e2
v2
, KE  m
with PE  ke
r
2
The velocity of e can be found by analyzing the force
on the electron. This force is the Coulomb force;
because the electron travels in a circular orbit, the
acceleration will be the centripetal acceleration:
mac  Fc
or
v2
e2
m  ke 2 ,
r
r
or
e2
v  ke
,
mr
2
Thus, total energy is
e2 m  ke e2 
e2
18
E   ke  


k


2.18

10
J  -13.6 eV

e
r 2  mr 
2r
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16.4 Equipotential surfaces
They are defined as a surface in space on which the potential
is the same for every point (surfaces of constant voltage)
The electric field at every point of an equipotential surface is
perpendicular to the surface
convenient to represent by drawing
equipotential lines
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