Transcript PHYS_2326_012709

Steps to Applying Gauss’ Law
To find the E field produced by a charge distribution at a point of
distance r from the center
1. Decide which type of symmetry best complements the
problem
2. Draw a Gaussian surface (mathematical not real)
reflecting the symmetry you chose around the charge
distribution at a distance of r from the center
3. Using Gauss’s law obtain the magnitude of E
Gauss’s Law
E 


 E d A
 qi
0
Applications of the Gauss’s Law
Remember – electric field lines must start and must end on charges!
If no charge is enclosed within Gaussian surface – flux is zero!
Electric flux is proportional to the algebraic number of lines leaving
the surface, outgoing lines have positive sign, incoming - negative
Examples of certain field configurations
Remember, Gauss’s law is equivalent to Coulomb’s law
However, you can employ it for certain symmetries to solve the reverse problem
– find charge configuration from known E-field distribution.
Field within the conductor – zero
(free charges screen the external field)
Any excess charge resides on the
surface


E d A0
S
Field of a charged conducting sphere
Field of a thin, uniformly charged conducting wire
Field outside the wire can only point
radially outward, and, therefore, may
only depend on the distance from the wire


Q
 E d A  0

E
2 r 0
- linear density of charge
Field of the uniformly charged sphere
Uniform charge within a sphere of radius r
r
q  Q 
a
3
'
E

r
3 0

Q - total charge
Q
- volume density of charge
V
Field of the infinitely large conducting plate
s
Q
A
s- uniform surface charge density
s
E
2 0
Charged Isolated Conductors
• In a charged isolated conductor all the charge moves to the
surface
• The E field inside a conductor must be 0 otherwise a
current would be set up
• The charges do not necessarily distribute themselves
uniformly, they distribute themselves so the net force on
each other is 0.
• This means the surface charge density varies over a
nonspherical conductor
Charged Isolated Conductors cont
• On a conducting surface
s
E
o
• If there were a cavity in the isolated conductor, no charges
would be on the surface of the cavity, they would stay on
the surface of the conductor
Charge on solid conductor resides on surface.
Charge in cavity makes a equal but opposite charge reside on
inner surface of conductor.
Properties of a Conductor in Electrostatic Equilibrium
1. The E field is zero everywhere inside the conductor
2. If an isolated conductor carries a charge, the charge resides on its
surface
3. The electric field just outside a charged conductor is
perpendicular to the surface and has the magnitude given above
4. On an irregularly shaped conductor, the surface charge density is
greatest at locations where the radius of curvature of the surface
is smallest
Charges on Conductors
Field within conductor
E=0
Experimental Testing of the Gauss’s Law
Earnshaw’s theorem
A point charge cannot be in stable equilibrium in electrostatic field of
other charges
(except right on top of another charge – e.g. in the middle of a
distributed charge)
Stable equilibrium with other
constraints
Atom – system of charges with only Coulombic forces in play.
According to Earhshaw’s theorem, charges in atom must move
However, planetary model of atom doesn’t work
Only quantum mechanics explains the existence of an atom
Electric Potential Energy
Concepts of work, potential energy
and conservation of energy
For a conservative force, work can always
be expressed in terms of potential energy difference
b

Wa b   F d l  U  (U b  U a )
a
Energy Theorem
For conservative forces in play,
total energy of the system is conserved
Ka  U a  Kb  U b
Wa b  Fd  q0 Ed
U  q0 Ey
Wa b  U  q0 E ( ya  yb )
Potential energy Uincreases
as the test charge q0 moves in the direction opposite to

the electric force F  q0 E : it decreases as it moves in the same direction as the force
acting on the charge
Electric Potential Energy of Two Point Charges
b
Wa b

rb
qq
  F d l   ke 20 cos  dl
r
a
r
a
1 1
Wa b  ke qq0   
 ra rb 
qq0
U  ke
r
Electric potential energy of two point charges
Example: Conservation of energy with electric forces
A positron moves away from an a – particle
m p  9.1  1031 kg

0
a-particle
positron
ma
7000m p
qa  2e
r0  1010 m
V0  3  106 m / s
What is the speed at the distance r  2r0  2 1010 m ?
What is the speed at infinity?
Suppose, we have an electron instead of positron. What kind of motion we would expect?
Conservation of energy principle
K0  U 0  K1  U1
Electric Potential Energy of the System of Charges
Potential energy of a test charge q0
in the presence of other charges
U
q0
qi

4 0 i r i
Potential energy of the system of charges
(energy required to assembly them together)
U

qi q j
4 0 i  j r ij
Potential energy difference can be equivalently described as a work done by external force
required to move charges into the certain geometry (closer or farther apart). External
force now is opposite to the electrostatic force


Wa b  (U b  U a )   Fext d l