gauss physics science general chemistry 2

Electrochemistry is the branch of chemistry that studies the relationship between electricity and chemical reactions. A galvanic cell (or voltaic cell) is an electrochemical cell that converts chemical energy into electrical energy through a spontaneous redox reaction. In this experiment, we will use a lemon or potato as an electrolyte to create a simple battery.

The lemon or potato contains natural acids (citric acid or phosphoric acid) that allow the movement of ions. When a zinc nail (anode) and a copper coin (cathode) are inserted into the lemon or potato, they act as electrodes. The chemical reaction between the metals and the electrolyte generates a small electric current, which can be used to power an LED or a buzzer.

1. Gauss’s Law
0
q
EdA

 



4. What does it mean?
First we need the concept of flux.
Area A

5. What does it mean?
First we need the concept of flux.
E
Area A
Electric
field

6. What does it mean?
First we need the concept of flux.
E
Area A
Electric
field
Flux is just electric field times areaEA

7. What does it mean?
First we need the concept of flux.
If electric field does not pass through the area, flux is zero.
0
E


8. What does it mean?
First we need the concept of flux.
In general we use a normal vector to the plane, .
A

A


9. What does it mean?
First we need the concept of flux.


For more general angles the flux varies as cos.
cosEA E A  

A

E


10. What does it mean?
First we need the concept of flux.


90 
E

0 
E

A

A

A

For more general angles the flux varies as cos.
cosEA E A  

E

EA 0
cos 1
cos 0

11. What does it mean?
The total flux through a closed surface.

12. What does it mean?
The total flux through a closed surface.
The convention that the
normal always points
outward.

13. What does it mean?
The total flux through a closed surface.
E
 Electric Flux- the product
of electric field and the
area of the surface
projected in a place
perpendicular to the field,
can be defined by
considering an area in
question.

14. What does it mean?
The total flux through a closed surface.
E
 The Greek letter phi (Φ) is
the symbol for electric flux in
physics:
The formula for electric flux is
ΦE = EA, where:
ΦE: Represents electric flux
E: Represents the magnitude
of the electric field
A: Represents the surface area

15. What does it mean?
The total flux through a closed surface.
E

The SI unit for
electric flux is
newton-meters
squared per
coulomb
(N · m2 /C). 

16. 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,
For the top,
0E A 

E A EA 


17. 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,
For the top,
For the bottom,
0E A 

E A EA 

E A EA 


18. 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,
For the top,
For the bottom,
The total flux is
0E A 

E A EA 

E A EA 

0EA EA  

19. What does it mean?
What does the integral mean?
E

The circle indicates an
integral over the closed
surface.
EdA 



20. What does it mean?
What does the integral mean?
E

EdA 


The circle indicates an
integral over the closed
surface.
In practice we will not have
to evaluate the interval.

21. What does it mean?
What does the integral mean?
E

EdA 


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.

22. What does it mean?
What does the integral mean?
E

EdA 


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.
dA

In principle sum over
infinitesimal elements .dA


23. What does it mean?
The full Gauss’s law.
E

The left side is the total flux
out through the surface.
0
q
EdA

 



24. What does it mean?
The full Gauss’s law.
E

The left side is the total flux
out through the surface.
The right side is
proportional to the charge,
q, inside the surface.
0
q
EdA

 


+q

25. What does it mean?
The full Gauss’s law.
E

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.
0
q
EdA

 


+q

26. How do we use it?
For example, consider a charge q.
+q
r

27. How do we use it?
For example, consider a charge q.
We choose a spherical surface.
+q
r

28. How do we use it?
For example, consider a charge q.
We choose a spherical surface.
By spherical symmetry the electric
field must be directed radially
outwards.
E

E

E

E

E

+q
r

29. How do we use it?
For example, consider a 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.
E

E

E

E

E

+q
r

30. How do we use it?
For example, consider a 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.
E

E

E

E

E

+q
r

31. How do we use it?
For example, consider a 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
E

E

E

E

E

0
q
EA


+q
r

32. How do we use it?
For example, consider a 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
E

E

E

E

E

0
q
EA


2
0
4
q
E r


+q
r

33. How do we use it?
For example, consider a 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
E

E

E

E

E

0
q
EA


2
0
4
q
E r


2
0
4
q
E
r

+q
r

34. How do we use it?
E

E

E

E

E

0
q
EA


2
0
4
q
E r


2
0
4
q
E
r

+q
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
r

35. How do we use it?
E

E

E

E

E

0
0
EA


2
4 0E r
0E
+q
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
r

36. How do we use it?
General procedure:
Choose a surface corresponding
to the symmetry of the problem.
Break the surface up into
subsurfaces where the electric
field is either
1.constant and parallel to the
normal, or
2.perpendicular to the normal.
Evaluate the total flux using the
electric field as a free parameter.
Solve Gauss’s law for E.
E

E

E

E

E

r
+q
http://www.dominicberry.org/presentations/gauss.ppt