Let’s do another experiment
Expt: see 1/r2 dependence, see vectorial nature, see
proportionality to charge
Lecture 2
Electric Field and Gauss’s Law
How can we write this mathematically..
F =k
q1 q2
r2
What is the meaning of
k?
.. How can we picture this?
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A Force Field....
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The Electroscope
http://www.engr.uky.edu/~gedney/courses/ee468/expmnt/escope-graph.gif
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http://upload.wikimedia.org/wikipedia/commons/b/b5/Electroscope.png
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What’s going on?
Is the electric field real?
Is it real? Does it matter?
How does one charge know that the other is there? Where does
the force come from? Does it go to infinity?
Why not just imagine that charged particles interact with each
other without the need for a mediating field?
The conventional solution is to say:
in this case how do they feel each other?
(1) that there is an electric field surrounding every charged
particle, and
you can do this. If you take this approach and there is more
than one charge then one calculates the individual pairs of
forces and then sums them vectorially for the resultant. This
can be a nightmare. You need to look at Ex. 21.3,21.4
before next lecture
(2) that a charged particle within a field ‘feels’ the field and
experiences a force
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An Electric Field....
We get a real ‘small’
charge of magnitude q,
and measure the force at each point
in the field
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An Electric Field....
How do we measure the field?
Electric Field Units: Newtons/Coulomb
Ẽ(x, y, z) = F̃ (x, y, z)/qtest
This is a vector field
i.e. at every point there is a direction
We draw lines tangent to the
force, and the density of lines
naturally ends up proportional to
strength of the field
lines never cross
Why is this?
lines never terminate except on a charge
i.e. at a point (x,y,z) there is a field (Ex,Ey,Ez)
or could give each component separately
i.e. Ex =Ex(x,y,z) & Ey = Ey(x,y,z) ...
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Picturing Field
Measuring fields - The Electroscope
http://www.engr.uky.edu/~gedney/courses/ee468/expmnt/escope-graph.gif
http://upload.wikimedia.org/wikipedia/commons/b/b5/Electroscope.png
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An Electric Field....
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Electric Fields by Calculation
qsource qtest
1
where k =
∼ 9 x 109
2
r
4π"0
qsource qtest
r̂ where r̂ = !r/|r|
F̃ = k
r2
qtest
qsource
Electric Field Units: Newtons/Coulomb
F =k
Ẽ(x, y, z) = F̃ (x, y, z)/qtest
This is a vector field
i.e. at every point there is a direction
!r
i.e. at a point (x,y,z) there is a field (Ex,Ey,Ez)
Ẽ = F̃ /qtest
or could give each component separately
i.e. Ex =Ex(x,y,z) & Ey = Ey(x,y,z) ...
so
Ẽ = k
qsource
r̂
r2
Defn: Field is force per unit test charge, think of gravity...
(mass makes gravity, gravity acts on mass)
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Vector Addition of Forces
All
positive charges
Dipole Field (pls look at Ex. 21.10-21.12)
Force adds
vectorially,
thus so does
field
q = 2.0 !C
F = 0.29 N
0.3m
q = 4.0 !C
0.4m
!
!
E+ = k q+/r2
2
E- = k q-/r2
FTotal = 0.46 N
Ex+ = -k |q+|/r2 cos !
Ex- = -k |q-|/r2 cos !
ExTOT = - k (|q-|+ |q+|)/r2 cos !
EyTOT = 0
0.3m
E+ = -k |q+|/r2
q-
-
q = 2.0 !C
ExTOT = - k (|q-|+ |q+
1
|)/r2
E- = -k
+
|q-|/r2
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Visualizing the field
q+
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Lots of charges
It is all very well to calculate the fields like this but isn’t there an
easier way?
If the charge distribution is symmetric in some way then it is
possible to use an entirely different approach to get to exactly
the same thing i.e. it doesn’t give you any more information
than the Coulomb Law and Electric Field, but it can make things
a lot easier to see.
Not useful in unsymmetric situations
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Flux lines
Does this feel right?
Expt 1: play with polystyrene ball with outgoing needles
Expt 2: what about two positive charge models?
What did we learn?
that if I make a box of any size around the charges then the
number of needles penetrating the surface is the same
Positive Charge ! Outward Flux
that the number of penetrating needles is proportional to the
charge inside the box.
if the line penetrates the box going outwards lets count as
positive and if going inward then we will call negative
Twice the Box ! The same Flux
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Twice the Charge ! Twice the Flux
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Enclosed Dipole
No enclosed charge - but adjacent charge
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Gauss’s Law: part 1
Field Lines and Electric Field
! =k q
E
r2
ΦE ∝ qenclosed
but Gauss said even more...
r1
but we will need to wait till next time to find this out
2r1
E1
=4
E2
A1
1
=
but also
A2
4
If (I) lines are continuous,
& (II) Coulomb’s Law is right,
& (III) lines are spherically symmetric from a point charge
then Areal density of flux lines is proportional to Electric Field
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In this lecture and the next
We have covered sections 21.3-21.6 and then 22.0-22.3 in the
textbook: this all part of the examinable material of the course
In the next lecture we will see how to use Gauss’s law for some
interesting shapes, and to realize some important things.
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