1 Gauss Law Electric Flux and applications.ppt

Gauss Law Electric Flux and applications

1. 1/17/07 184 Lecture 6 1
PHY 184PHY 184
Spring 2007
Lecture 6
Title: Gauss’ Law

2. 1/17/07 184 Lecture 6 2
AnnouncementsAnnouncements
Homework Set 2 is due Tuesday at 8:00 am.
We will have clicker quizzes for extra credit
starting next week (the clicker registration closes
January 19).
Homework Set 3 will open Thursday morning.
Honors option work in the SLC will start next week
•Honors students sign up after class for time slots.

3. 1/17/07 184 Lecture 6 3
OutlineOutline
/1/ Review
/2/ Electric Flux
/3/ Gauss’ Law

4. 1/17/07 184 Lecture 6 4
Review - Point ChargeReview - Point Charge
► The electric field created by
a point charge, as a function
of position r,
r
r
q
rE ˆ
4
1
)(
2
0


► The force exerted by an electric
field on a point charge q located at
position x
)(xEqF


… direction tangent to the field line through x

5. 1/17/07 184 Lecture 6 5
Review – Electric DipoleReview – Electric Dipole
2 equal but opposite
charges, -q and +q
Dipole moment
(direction: - to +)
On the axis, far from the
dipole,
dqp


3
02r
p
E



6. 1/17/07 184 Lecture 6 6
Force and Torque on an Electric DipoleForce and Torque on an Electric Dipole
Assume the 2 charges (-q and +q) are
connected together with a constant
distance d, and put the dipole in a
uniform electric field E.
Net force = 0
Torque about the center:


sin
sinsin
2
sin
2
armmoment Farmmoment Ftorque
21
pEτ
qdE
d
qE
d
qEτ



Torque
vector
Ep



7. 1/17/07 184 Lecture 6 7
Gauss’ Law

8. 1/17/07 184 Lecture 6 8
ObjectiveObjective
So far, we have considered point charges. But how can we treat
more complicated distributions, e.g., the field of a charged wire, a
charged sphere or a charged ring?
Two methods
Method #1: Divide the distribution into infinitesimal elements dE
and integrate to get the full electric field.
Method #2: If there is some special symmetry of the
distribution, use Gauss’ Law to derive the field.

Gauss’ Law
The flux of electric field through
a closed surface is proportional
to the charge enclosed by the
surface.

9. 1/17/07 184 Lecture 6 9
Gauss’ Law
The flux of electric field through
a closed surface is proportional
to the charge enclosed by the
surface.

10. 1/17/07 184 Lecture 6 10
Electric FluxElectric Flux
Let’s imagine that we put a ring with area A
perpendicular to a stream of water flowing
with velocity v
The product of area times velocity, Av, gives
the volume of water passing through the ring
per unit time
•The units are m
3
/s
If we tilt the ring at an angle , then the
projected area is
A cos, and the volume of water per unit time
flowing through the ring is Av cos .
Av


11. 1/17/07 184 Lecture 6 11
Electric Flux (2)Electric Flux (2)
We call the amount of water flowing through the ring
the “flux of water”
We can make an analogy with electric field lines from
a constant electric field and flowing water
We call the density of electric field lines through an
area A the electric flux given by
cosFlux Av


E

A
cos Flux Elecric EA

12. 1/17/07 184 Lecture 6 12
Math PrimerMath Primer
Surfaces and Normal Vectors Surfaces and Normal Vectors
For a given surface, we define the normal unit
vector n,n, which points normal to the surface and
has length 1.
Electric Flux

13. 1/17/07 184 Lecture 6 13
Math Primer - Gaussian Surface Math Primer - Gaussian Surface
A closed surface, enclosing
all or part of a charge
distribution, is called a
Gaussian Surface.Gaussian Surface.
Example: Consider the flux
through the surface on the
right. Divide surface into
small squares of area A.
Flux through surface:

14. 1/17/07 184 Lecture 6 14
Electric Flux (3)Electric Flux (3)
In the general case where the electric field is not
constant everywhere
We define the electric flux through a closed surface
in terms of an integral over the closed surface
)(rE



S
AdE


15. 1/17/07 184 Lecture 6 15
Gauss’ LawGauss’ Law
Gauss’ Law (named for German mathematician and
scientist Johann Carl Friedrich Gauss, 1777 - 1855)
states
•(q = net charge enclosed by S).
If we add the definition of the electric flux we get
another expression for Gauss’ Law
Gauss’ Law : the electric field flux through S is
proportional to the net charge enclosed by S.

0q


S
qAdE

0


16. 1/17/07 184 Lecture 6 16
Theorem: Gauss’ Law and Coulomb’s LawTheorem: Gauss’ Law and Coulomb’s Law
are equivalent.are equivalent.
Let’s derive Coulomb’s Law from Gauss’ Law.
We start with a point charge q.
We construct a spherical surface with radius r
surrounding this charge.
•This is our “Gaussian surface”“Gaussian surface”
q
r
 = E A

17. 1/17/07 184 Lecture 6 17
Theorem (2)Theorem (2)
The electric field from a point
charge is radial, and thus is
perpendicular to the Gaussian
surface everywhere.
/1/ The electric field direction is
parallel to the normal vector for any
point.
/2/ The magnitude of the electric field
is the same at every point on the
Gaussian surface.
  
 EAdAEdAEAdE

/1/ /2/  = E A

18. 1/17/07 184 Lecture 6 18
Theorem (3)Theorem (3)
Now apply Gauss’ Law
2
04
1
r
q
E


Q. E.
D.
2
0
r4π AArea
EAΦ whereqΦε



19. 1/17/07 184 Lecture 6 19
ShieldingShielding
An interesting application of Gauss’ Law:
The electric field inside a charged conductor is
zero.
Think about it physically…
•The conduction electrons will move in response to any
electric field.
•Thus the excess charge will move to the surface of the
conductor.
•So for any Gaussian surfaceGaussian surface inside the conductor --
encloses no charge! – the flux is 0. This implies that the
electric field is zero inside the conductor.

20. 1/17/07 184 Lecture 6 20
Shielding IllustrationShielding Illustration
Start with a hollow
conductor.
Add charge to the
conductor.
The charge will move to
the outer surface
We can define a
Gaussian surface that
encloses zero charge
•Flux is 0
•Ergo - No electric field!

21. 1/17/07 184 Lecture 6 21
Cavities in ConductorsCavities in Conductors
a) Isolated Copper block with
net charge. The electric field
inside is 0.
b) Charged copper block with
cavity. Put a Gaussian surface
around the cavity. The E field
inside a conductor is 0. That
means that there is no flux
through the surface and
consequently, the surface
does not enclose a net charge.
There is no net charge on the
walls of the cavity

22. 1/17/07 184 Lecture 6 22
Shielding DemonstrationShielding Demonstration
We will demonstrate shielding in two ways
We will place Styrofoam peanuts in a container on a
Van de Graaff generator
•In a metal cup
•In a plastic cup
We will place a student in a wire cage and try to fry
him with large sparks from a Van de Graaff
generator
•Note that the shielding effect does not require a solid
conductor
•A wire mesh will also work, as long as you don’t get too close to
the open areas

23. 1/17/07 184 Lecture 6 23
Lightning Strikes a CarLightning Strikes a Car
The crash-test dummy is safe,
but the right front tire didn’t
make it …
High Voltage Laboratory,
Technical University Berlin, Germany