Transcript Mini-lecture on Gauss`s law

Gauss’s Law
What does
it mean?
E

dA


Dominic Berry
University of Waterloo
q
0
How do
we use it?
Griffith University
8 February, 2011
What does it mean?

First we need the concept of flux.
Area A
What does it mean?

First we need the concept of flux.
Electric
field E
Area A
What does it mean?

First we need the concept of flux.
Electric
field E
Area A
Flux is just electric field times area
  EA
What does it mean?

First we need the concept of flux.
E
If electric field does not pass through the area, flux is zero.
0
What does it mean?

First we need the concept of flux.
A
In general we use a normal vector to the plane, A.
What does it mean?

First we need the concept of flux.
E
A


For more general angles the flux varies as cos.
  EA cos  E  A
What does it mean?

First we need the concept of flux.
cos  1
E
E
E
A
  0
A


A cos  0
  90
0
  EA
For more general angles the flux varies as cos.
  EA cos  E  A
What does it mean?

The total flux through a closed surface.
What does it mean?

The total flux through a closed surface.

We use the convention that
the normal always points
outward.
What does it mean?

The total flux through a closed surface.
E

We use the convention that
the normal always points
outward.
What does it mean?

The total flux through a closed surface.
E


We use the convention that
the normal always points
outward.
For the four sides,
EA0
What does it mean?

The total flux through a closed surface.
E



We use the convention that
the normal always points
outward.
For the four sides,
EA0
For the top,
E  A  EA
What does it mean?

The total flux through a closed surface.
E




We use the convention that
the normal always points
outward.
For the four sides,
EA0
For the top,
E  A  EA
For the bottom,
E  A   EA
What does it mean?

The total flux through a closed surface.
E





We use the convention that
the normal always points
outward.
For the four sides,
EA0
For the top,
E  A  EA
For the bottom,
E  A   EA
The total flux is
  EA  EA  0
What does it mean?

What does the integral mean?
E


E

dA

The circle indicates an
integral over the closed
surface.
What does it mean?

What does the integral mean?
E



E

dA

The circle indicates an
integral over the closed
surface.
In practice we will not have
to evaluate the interval.
What does it mean?

What does the integral mean?
E




E

dA

The circle indicates an
integral over the closed
surface.
In practice we will not have
to evaluate the interval.
We break the surface up
into sections where the flux
is easy to calculate.
What does it mean?

What does the integral mean?
E


dA


In principle sum over
infinitesimal elements dA.
E

dA

The circle indicates an
integral over the closed
surface.
In practice we will not have
to evaluate the interval.
We break the surface up
into sections where the flux
is easy to calculate.
What does it mean?

The full Gauss’s law.
E
E

dA



q
0
The left side is the total flux
out through the surface.
What does it mean?

The full Gauss’s law.
E
E

dA



+q

q
0
The left side is the total flux
out through the surface.
The right side is
proportional to the charge,
q, inside the surface.
What does it mean?

The full Gauss’s law.
E
E

dA



+q


q
0
The left side is the total flux
out through the surface.
The right side is
proportional to the charge,
q, inside the surface.
The constant, 0, is the
usual vacuum permittivity.
How do we use it?

For example, consider a charge +q.
+q
r
How do we use it?


For example, consider a charge +q.
We choose a spherical surface.
+q
r
How do we use it?



For example, consider a charge +q.
We choose a spherical surface.
By spherical symmetry the electric E
field must be directed radially
outwards.
E
+q
E
r
E
E
How do we use it?




For example, consider a charge +q.
We choose a spherical surface.
By spherical symmetry the electric E
field must be directed radially
outwards.
The magnitude of the electric field
must be constant on the surface.
E
+q
E
r
E
E
How do we use it?





For example, consider a charge +q.
We choose a spherical surface.
By spherical symmetry the electric E
field must be directed radially
outwards.
The magnitude of the electric field
must be constant on the surface.
The flux is just EA.
E
+q
E
r
E
E
How do we use it?






For example, consider a charge +q.
We choose a spherical surface.
By spherical symmetry the electric E
field must be directed radially
outwards.
The magnitude of the electric field
must be constant on the surface.
The flux is just EA.
Gauss’s law gives
q
EA 
0
E
+q
E
r
E
E
How do we use it?






For example, consider a charge +q.
We choose a spherical surface.
By spherical symmetry the electric E
field must be directed radially
outwards.
The magnitude of the electric field
must be constant on the surface.
The flux is just EA.
Gauss’s law gives
q
EA 
0
E 4 r 2 
+q
E
r
q
0
E
E
E
How do we use it?






For example, consider a charge +q.
We choose a spherical surface.
By spherical symmetry the electric E
field must be directed radially
outwards.
The magnitude of the electric field
must be constant on the surface.
The flux is just EA.
Gauss’s law gives
q
EA 
0
E 4 r 2 
E
q
4 0 r
2
+q
E
r
q
0
E
E
E
How do we use it?






Consider a shell of charge +q.
We choose a spherical surface.
By spherical symmetry the electric
field must be directed radially
outwards.
The magnitude of the electric field
must be constant on the surface.
The flux is just EA.
Gauss’s law gives
q
EA 
E
E
+q
E
0
E 4 r 2 
E
r
q
0
q
4 0 r
2
E
E
How do we use it?






Consider a shell of charge +q.
We choose a spherical surface.
By spherical symmetry the electric
field must be directed radially
outwards.
The magnitude of the electric field
must be constant on the surface.
The flux is just EA.
Gauss’s law gives
0
EA 
E
E
+q
E
0
r
E 4 r 2  0
E
E0
E
How do we use it?
General procedure:


1.
2.
Choose a surface corresponding
to the symmetry of the problem.
E
E
Break the surface up into
subsurfaces where the electric
field is either
constant and parallel to the
normal, or
perpendicular to the normal.

Evaluate the total flux using the
electric field as a free parameter.

Solve Gauss’s law for E.
+q
E
r
E
http://www.dominicberry.org/presentations/gauss.ppt
E