Scott Hughes
10 February 2004
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2004
Lecture 3:
Electric field energy.
Potential; Gradient.
Gauss’s law revisited; divergence.
3.1
Energy in the field
Suppose we take our spherical shell of radius
r
, with charge per unit area
σ
=
q/
4
πr
2
, and
squeeze it. How much work does it take to do this squeezing? To answer this, we need to
know how much pressure is being exerted by the shell’s electric field on itself. Pressure is
force/area;
σ
is charge/area; force is charge times electric field. Hence, our intuitive guess is
that the pressure should be
P
guess
=
(electric field)(charge/area)
=
4
πσ
2
.
We have used the fact that the electric field just at the surface of the shell is 4
πσ
to write
down this equation. This answer turns out to be almost right — we just have to divide by
two:
P
= 2
πσ
2
.
Purcell gives a careful derivation of where this 1
/
2 comes from. Physically, it arises because
we need to
average
the field at
r
over its discontinuity — the field is 0 at radius
r

²
, it is
4
πσ
at
r
+
²
.
Now that we know the pressure it is easy to compute the work it takes to squeeze:
dW
=
F dr
=
P
4
πr
2
dr
=
2
πσ
2
dV .
On the last line, we’ve used the fact that when we compress the shell by
dr
, we have squeezed
out a volume
dV
= 4
πr
2
dr
. Now, using
E
= 4
πσ
, we can write this formula in terms of the
electric field:
dW
=
E
2
8
π
dV .
The quantity
E
2
/
8
π
is the energy density of the electric field itself. (In SI units, we would
have found
²
0
E
2
/
2.) It shows up in this work calculation because when we squeeze the shell
we are
creating
new electric field: the shell from
r

dr
to
r
suddenly contains field, whereas
it did not before. We had to put
dW
= (energy density)
dV
amount of work into the system
to make that field.
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3.2
Electric potential difference
I move a charge
q
around in an electric field
~
E
. The work done
by me
as I move this charge
from position
~a
to
~
b
is given by
U
ba
=
Z
~
b
~a
~
F
me
·
d~s
=

Z
~
b
~a
~
F
q
·
d~s
=

q
Z
~
b
~a
~
E
·
d~s .
~
F
me
is the force that I exert. By Newton’s 3rd, this is equal and opposite to the force
F
q
that
the electric field exerts on the charge. This is why we flip the sign going from line 1 to line
2. Make sure you understand this logic — that sign flip is an important part of stuff we’re
about to define! On the last line, we’ve expressed this in terms of the electric field. Recall
that one of the things we learned about electric fields is that they generate
conservative
forces. This means that this integral must be independent of the path we take from
~a
to
~
b
.
We’ll return to this point later.
Since the work we do is proportional to the charge, we might as well divide it out. Doing
so, we define the
electric potential difference
between
~a
and
~
b
:
φ
ba
=

Z
~
b
~a
~
E
·
d~s .
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 Spring '08
 Hughes
 Vector Calculus, Electrostatics, Energy, Mass, Magnetic Field, Electric charge, Vector field

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