of notation, we write this as dUe∝E^2 dv, or dUe=aE^2 dv, where
ais a constant of proportionality. Note that we never really made
use of any of the details of the geometry of figure a, so the reasoning
is of general validity. In other words, not only is dUe=aE^2 dvthe
function that works in this particular case, but there is every reason
to believe that it would work in other cases as well.
It now remains only to finda. Since the constant must be the
same in all situations, we only need to find one example in which
we can compute the field and the energy, and then we can deter-
minea. The situation shown in figure a is just about the easiest
example to analyze. We let the square capacitor plates be uni-
formly covered with charge densities +σand−σ, and we writeb
for the lengths of their sides. Lethbe the gap between the plates
after they have been separated. We chooseh b, so that the
field experienced by the negative plate due to the positive plate
isE+= 2πkσ. The charge of the negative plate is−σb^2 , so the
magnitude of the force attracting it back toward the positive plate
is (force) = (charge)(field) = 2πkσ^2 b^2. The amount of work done
in separating the plates is (work) = (force)(distance) = 2πkσ^2 b^2 h.
This is the amount of energy that has been stored in the field be-
tween the two plates, Ue = 2πkσ^2 b^2 h = 2πkσ^2 v, wherev is the
volume of the region between the plates.
We want to equate this toUe=aE^2 v. (We can writeUeand
vrather than dUeand dv, since the field is constant in the region
between the plates.) The field between the plates has contributions
from both plates,E=E++E−= 4πkσ. (We only used half this
value in the computation of the work done on the moving plate,
since the moving plate can’t make a force on itself. Mathematically,
each plate is in a region where its own field is reversing directions,
so we can think of its own contribution to the field as being zero
within itself.) We then haveaE^2 v=a· 16 π^2 k^2 σ^2 ·v, and setting
this equal toUe= 2πkσ^2 vfrom the result of the work computation,
we finda= 1/ 8 πk. Our final result is as follows:
The electric energy possessed by an electric fieldEoccupying an
infinitesimal volume of space dvis given bydUe=1
8 πk
E^2 dv,whereE^2 =E·Eis the square of the magnitude of the electric field.
This is reminiscent of how waves behave: the energy content of
a wave is typically proportional to the square of its amplitude.Section 10.4 Energy in fields 605