Lecture 3
o Aim of the lecture
 Gauss’ Law:
 Flux
 More mathematical descriptions
 Generalised field distributions
o Main learning outcomes
 familiarity with
 Electric fields, potentials
 Coulomb’s and Gauss’ Laws
 Calculation of fields and forces
Reminder:
o Objects have a property called ‘charge’
Basic property like mass
o Charge is quantised – comes in units of 1.6 x 10-19C
Charge comes in two types, which are called
Positive
Negative
o The total charge in any closed system cannot be changed
o Charges interact with each other, causing
Repulsive force if charges are the same (++ or --)
Attractive forces if charges are opposite (+-)
+
+
F
F
-
Reminder:
o The forces are equal and in the opposite direction
o The form of the force law is: F = keq1q2/r2
o The nature of the interaction between charge is described using
 An electric field
 On a diagram the field is represented by lines
 emerge from a source (positive charge)
 end in a sink (negative charge)
 The density of field lines represents the field strength
 but caution needed – this is fully accurate in 3-D, but
 in a drawing in 2-D the line density cannot always be exact
 this is only a problem drawing NOT a problem with the theory
Exact, density of
field lines falls off
like 1/r2
Only a representation,
Density is falling like 1/r -wrong
Reminder:
o The interaction between charges can be described
 as an interaction of one of the charges with the field of the other
 The force law is: F = q1E
 where E is the electric field the charge q1 is in.
 eg E = keq2/r2 where r is the distance from q2
o The electric field is conservative so
 there is an energy associated with position in the field
 just like gravity
o Therefore there is a potential associated with position in the field
 The potential is called the electric potential,
 measured in volts
 Equipotentials are drawn on diagrams,
 these are accurate representations of 3-D system
Reminder: Gauss’ Law
o Idea behind Gauss’ Law already introduced
 The field lines come from a charge,
 so no extra lines appear away from it
 and none disappear
 the number of lines is a constant
 total number of lines through any closed surface
surrounding a charge must be constant
For a single point charge, the number of
lines passing through a sphere surrounding it
cannot depend on the radius of the sphere.
Or in fact on the shape of the surrounding surface
That’s Gauss’ Law
Reminder:
The correct expressions are vectors:
F = qE where F is the force on charge q in field E
E = ke q/r2 r is the field created by a point charge q
Where r is a vector pointing away from + charges
F the potential is a scalar field, there is no direction.
Units:
F is in Newtons
E is in Volts/metre (or Newtons/Coulomb)
F is in Volts (or Joules/Coulomb)
Note F is also often denoted Fe or V
To define Gauss’ Law more precisely (and accurately!)
need concept of a Flux
o E is a vector field
o Another example of a vector field is
o V where this is the velocity vector
for airflow at any point in a volume
Consider an open door with
air blowing in and out from
both sides
oThe net volume of air flowing through the door
 is the sum of all these vectors
 it is called the flux
 in this case an air volume flux
 it is a vector sum
 it depends on the direction as well as the magnitude
o At every point in the door
there is a net velocity
 magnitude
 and a direction, ie
 V – a vector field
oThe net volume of air flowing through the door
 is the sum of all these vectors
 it is called the flux
 in this case an air volume flux
 it is a vector sum
 it depends on the direction as well as the magnitude
Air volume/second = ∫∫dxdy V.a
where a is a unit vector perpendicular to the door
The quantity ∫∫dxdyV.a is the volume flux of air, we usually write:
F=
∫s V.da
where ∫s means integration over the surface
For Gauss’ Law we will use
F=
∫ V.da
∫
This symbol means an integral over a closed surface
da
means that the integral is being performed as a
vector dot product with the local surface direction
F is a volume flux
it is a measure of how much air is flowing
through the door.
Gauss’ Law
To state it properly, need to define
Electric Flux, dF through a small area dA as
dF = E.dA
where the direction of A
is perpendicular to the surface dA
So
dF =
E.dA
And
Gauss’ Law is:
q = e0
∫ E.dA
Where q is the charge inside the surface
dF =
E.dA
So the total flux through a surface is F = ∫sdF = ∫E.dA
but
q = e0
∫ E.dA
Where q is the charge inside the surface, so
Evaluate F for
A dipole configuration
Through a sphere
F
F == 0q/e
- q/e
00
q = e0
∫ E.dA
Gauss’ Law allows cute proofs
eg All excess charge on a conductor is on its surface
o If there was an electric field
inside A then free charges would move.
oThis means flux through surface
is everywhere zero
o because A can be moved arbitrarily
close to the conductor surface.
o No flux means no charge inside
Concentric spheres
Around a wire
Planar non-conducting sheet
Electric Potential
Potential Energy associated with Electric Field
Consider gravity
Potential Energy, U
U = mgh
Remember where this comes from:
F = Gmemt/r2
The work done, WD is force times distance
WD = ∫Fdr
re+h
= ∫dr Gmtme/r2
re
= Gmt(1/(re+h)-1/re)me
≈ mt{Gme/re}h
= mtgh
The formula arises because the strength of the gravitational field is
approximately constant for ‘small’ changes in height above earth
The equipotential surfaces for
the gravitational field above the
earth are spheres, which are
just lines in this 2-D picture
Potential Energy, U
U = mgh
18J/kg
16J/kg
14J/kg
12J/kg
The equipotential surfaces for
the gravitational field above the
earth are spheres, which are
just lines in this 2-D picture
Each line can be given a value = gh
10J/kg
8J/kg
6J/kg
4J/kg
2J/kg
0
Potential Energy, U
U = mgh
These are just the height multiplied by a constant
These are the values of the gravitational potential
When you multiply the potential by the mass you get
the potential energy
+ Voltage = V0
Between two metal plates
which have a voltage V across them,
There are equipotential lines,
each of which has a voltage
+
A charge has a potential energy
U = qV
just like gravity, except that the mass is replaced by q
and gh is replaced by the voltage
Voltage = 0
+ Voltage = V0
So the potential Energy,U
U = q {V0/D} h
D
+
h
Voltage = 0
compare with gravity
U = m {Gme/re} h