Transcript slides - Phenix at Vanderbilt

PHYS117B: Lecture 4
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Last lecture: We used
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Coulomb’s law
Principle of superposition
To find the electric field of continuous charge
distributions
Today: I’m going to teach you the easy way!
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The key word is
SYMMETRY
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We will:
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Recognize and use symmetry to determine
the shape of electric fields
Calculate electric flux through a surface
Use Gauss’s law to calculate the electric field
of symmetric charge distributions
Use Gauss’s law to understand the properties
of conductors in electrostatic equilibrium
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Suppose we know only 2 things about
electric fields:
1.
2.
An electric field points away from + charges
and towards negative charges
An electric field exerts a force on a charged
particle
What can we deduce for the electric field of an
infinitely long charged cylinder ?
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The charge distribution has cylindrical
symmetry
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What does this mean ?
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There is a group of geometrical transformations
that do not cause any physical change
Let’s try it (I have a cylinder here)
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Translate,
rotate, reflect
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If you can’t tell that the charge
distribution (the cylinder) was
transformed geometrically, then the
electric field should not change either !
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The symmetry of the electric field MUST
match the symmetry of the charge
distribution!
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Can the electric field of the cylinder look
like this ?
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Can the electric field of the cylinder look
like this ?
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What about this situation?
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What about this situation ?
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Three basic symmetries: planar, cylindrical,
spherical
We will make heavy
use of these three
basic symmetries.
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We learned how to use symmetry to
determine the direction of the electric
field, if we know the geometry of the
charge distribution.
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Can we reverse the problem ? Can we use
the symmetry of the electric field
configuration to determine the geometry of
the charge distribution that causes this field ?
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The mystery box
Take a test charge, move it around, measure the force on it,
figure out the field configuration => deduce the symmetry of the
charge distribution inside
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Imagine a situation like this:
No field around the box, or
same “amount of field” going in and out
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Or you can have:
Same box, same geometry of the field, but
“more field” going out of the box
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We need some way to measure “how
much” field goes in or out of the box
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To measure the volume of water that passes
through a loop per unit time, we use FLUX: the
dot product of the velocity vector and the area
vector gives volume/time
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We can define electric field flux:
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If the field is non-uniform:
This is trouble, I need to do an integral
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The surface is NOT flat : ooh bother !
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Relax ! We are going to deal with two
EASY situations: the field is UNIFORM
in both.
The field is EVERYWHERE tangent
the surface: FLUX = 0 !
The field is EVERYWHERE
perpendicular to the surface:
FLUX = EA
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What good is the electric field flux ?
Gauss’s law gives a relation between the
electric field flux through a closed surface
and the charge that is enclosed in that
surface.
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From symmetry: determine the shape (direction in every point
in space) of the electric field
From Gauss’s law: determine the magnitude of E
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What surface are we talking about ?
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This is NOT a physical surface
It is an IMAGINARY surface
If we want to make life simple, we have to figure out
what type of surface to choose, so that the integral
in Gauss’s law is EASY to do
We need to choose a surface that has the same
symmetry as the charge distribution
Then the field will be either tangent to the surface or
perpendicular to the surface => no angles to deal
with in the dot product!
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Let’s do it for an infinitely long wire with
uniform linear charge density l
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Last time we used Coulomb’s law: this is the
HARD way
Today: or sweet simplicity – we’ll use
Gauss’s law to get the same result !
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Field of a line of charge: use Coulomb’s
law, superposition and symmetry !
line of charge: length 2a
and linear charge density l
E
2al k
x x2  a2
xˆ
For an infinite line of charge
i.e. x<< a
2l k
l
E
xˆ 
xˆ
x
2 x 0
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See now how to do it using symmetry and
Gauss’s law
Qencl  l L
   top   bot   wall
  0  0  Awall E
  2 rLE
l
E
rˆ
2 r 0
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