Scott Hughes
3 February 2005
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2005
Lecture 2:
Electric fields & flux; Gauss’s law
2.1
The electric field
Last time, we learned that Coulomb’s law plus the principle of superposition allowed us to
write down fairly simple formulas for the force that a “swarm” of
N
charges exerts on a test
charge
Q
:
~
F
=
N
X
i
=1
Q q
i
ˆ
r
i
r
2
i
.
We found another simple formula for the force that a blob of charged material with charge
density
ρ
exerts on a test charge:
~
F
=
Z
Q ρ dV
ˆ
r
i
r
2
i
.
And, of course, similar formulas exist if the charge is smeared out over a surface with surface
density
σ
, or over a line with line density
λ
.
In all of these cases, the force ends up proportional to the test charge
Q
. We might as
well factor it out. Doing so, we define the
electric field
:
~
E
≡
~
F
Q
=
N
X
i
=1
q
i
ˆ
r
i
r
2
i
(
N
Point charges)
=
Z
ρ dV
ˆ
r
i
r
2
i
(Charge continuum)
.
Given an electric field, we can figure out how much force will be exerted on some point
charge
q
in the obvious way:
~
F
=
q
~
E
.
In cgs units, the electric field is measured in units of dynes/esu, or esu/cm
2
, or stat
volts/cm — all of these choices are equivalent. (We’ll learn about statvolts in about a week.)
In SI units, first of all, we must multiply all of the above formulae by 1
/
4
π²
0
. The units of
electric field are then given in Newtons/Coulomb, or Coulomb/meter
2
, or Volts/meter.
Let’s look at a couple of examples.
2.1.1
Example: Electric field at center of charged ring
Take a ring of radius
R
with uniform charge per unit length
λ
=
Q/
2
πR
.
What is the
electric field right at its center? Well, if we don’t think about it too carefully, we want to do
something like this:
~
E
=
Z
λ
ˆ
r
r
2
ds ,
11
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where the integral will be take around the ring;
ds
will be a path length element on the ring,
r
is the radius of the ring, and ˆ
r
is a unit vector pointing from each length element on the
ring to the ring’s center ...
But wait a minute! You should quickly notice that for each contribution due to a little
length element on the ring, there is an equal contribution from the ring’s opposite side
pointing in the opposite direction
. Each contribution to the
~
E
field at the center of the ring
is precisely canceled by its opposite contribution:
So the field at the center of the ring is zero.
2.1.2
Example: Electric field on ring’s axis
Suppose we move along the ring’s axis to some coordinate
z
above the disk’s center. The
field does not cancel in this case, at least not entirely: Considering the contribution of pieces
of the ring, we see that there is a component in the radial direction, and a component along
z
. The radial component cancels by the same symmetry argument as we used above. The
z
component, however, definitely does not cancel:
Radius R
Height above ring z
cancels in radial direction, does not
cancel in z direction.
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 Spring '10
 ScottHughes
 Electric charge, charge density

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