Download PHY 184 lecture 6

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
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Review - Point Charge
► The electric field created by a point charge,
as a function of position r,
 
E (r ) 
q ˆ
r
2
40 r
1
► The force exerted by an electric field on a point charge
q located at position x

 
F  qE (x )
… direction tangent to the field line through x
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Review – Electric Dipole
 2 equal but opposite
charges, -q and +q
 Dipole moment
(direction: - to +)


p  qd
 On the axis, far from the
dipole,
E
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p
20r 3
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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:
torque  F1  moment arm  F2  moment arm
d
d
τ  qE sin   qE sin   qdE sin 
2
2
τ  pE sin 
Torque vector
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 
  p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.
Gauss’ Law
The flux of electric field through a closed surface is
proportional to the charge enclosed by the
surface.
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Gauss’ Law
The flux of electric field through a closed surface is
proportional to the charge enclosed by the
surface.
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Electric Flux
 Let’s imagine that we put a ring with area A
perpendicular to a stream of water flowing
with velocity v
v
A
 The product of area times velocity, Av, gives
the volume of water passing through the ring
per unit time
• The units are m3/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 .
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
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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:
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Electric Flux (3)
 In the general case where the electric field is not
constant everywhere
 
E (r )
 We define the electric flux through a closed surface
in terms of an integral over the closed surface
 
   E  dA
S
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Gauss’ Law
 Gauss’ Law (named for German mathematician and
scientist Johann Carl Friedrich Gauss, 1777 - 1855)
states
0  q
• (q = net charge enclosed by S).
 If we add the definition of the electric flux we get
another expression for Gauss’ Law
 
 0  E  dA  q
S
 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”
q
=EA
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r
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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.
/1/
/2/
 
 E  dA   E dA  E dA  EA
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=EA
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Theorem (3)
Now apply Gauss’ Law
ε0 Φ  q
where
Area A  4π r
Φ  EA
2
q
E
2
40 r
1
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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
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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
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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
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