Transcript PHY 184 lecture 6 - Home Page | MSU Department of Physics
1/17/07
PHY 184
Spring 2007 Lecture 6
Title: Gauss’ Law
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Announcements
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.
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Outline
/1/ Review /2/ Electric Flux /3/ Gauss’ Law 1/17/07
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Review - Point Charge
► The electric field created by a point charge, as a function of position r,
E
(
r
) 1 4 0
q r
2
r
ˆ ► x The force exerted by an electric field on a point charge q located at position
F
q E
(
x
) … direction tangent to the field line through x 1/17/07
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Review – Electric Dipole
2 equal but opposite charges, -q and +q Dipole moment (direction: - to +)
q d
On the axis, far from the dipole,
E
2
p
0
r
3 1/17/07
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Force 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:
τ τ
torque
qE
pE
d
F 1 moment sin 2 sin
qE d
2 arm sin F 2 moment arm
qdE
sin Torque vector
E
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Gauss’ Law
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Objective
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.
1/17/07 Gauss’ Law The flux of electric field through a closed surface is proportional to the charge enclosed by the surface.
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1/17/07 Gauss’ Law The flux of electric field through a closed surface is proportional to the charge enclosed by the surface.
184 Lecture 6
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Electric Flux
Let’s imagine that we put a ring with area with velocity
v A
perpendicular to a stream of water flowing
v A
The product of area times velocity, per unit time • The units are m 3 /s
Av
, gives the volume of water passing through the ring If we tilt the ring at an angle projected area is
A
cos flowing through the ring is
Av
, then the , and the volume of water per unit time cos
.
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Electric Flux (2)
We call the amount of water flowing through the ring the “
flux
of water”
Flux
Av
cos
We can make an analogy with electric field lines from a constant electric field and flowing water
E
A
We call the density of electric field lines through an area
A
the electric flux given by
Elecric Flux
EA
cos
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Math Primer Surfaces and Normal Vectors
For a given surface, we define the normal unit vector
n,
which points normal to the surface and has length 1. Electric Flux
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Math Primer - Gaussian Surface
A closed surface, enclosing all or part of a charge distribution, is called a
Gaussian Surface.
Example: Consider the flux through the surface on the right. Divide surface into small squares of area A.
Flux through surface: 1/17/07
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Electric Flux (3)
constant everywhere
E
(
r
) We define the electric flux through a closed surface in terms of an integral over the closed surface
S E
d
A
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Gauss’ Law
Gauss’ Law states (named for German mathematician and scientist Johann Carl Friedrich Gauss, 1777 - 1855) 0
q
• (q = net charge enclosed by S).
If we add the definition of the electric flux we get 0
S E
d A
q
Gauss’ Law : the electric field flux through S is proportional to the net charge enclosed by S.
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Theorem: Gauss’ Law and Coulomb’s Law 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” r q
= E A
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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.
E
d A
/1/ /2/
E dA
E
dA
EA
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= E A
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Theorem (3)
1/17/07 Now apply Gauss’ Law
ε
0
Φ
q where Φ
EA Area A
4π r
2
E
1 4 0
q r
2
Q. E. D.
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Shielding
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 surface inside the conductor - encloses no charge! – the flux is 0. This implies that the electric field is zero inside the conductor.
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Shielding 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!
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Cavities 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 1/17/07
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Shielding 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 1/17/07
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Lightning 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 1/17/07
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