a/Two oppositely charged
capacitor plates are pulled apart.
10.4 Energy in fields
10.4.1 Electric field energy
Fields possess energy, as argued on page 579, but how much
energy? The answer can be found using the following elegant ap-
proach. We assume that the electric energy contained in an infinites-
imal volume of space dvis given by dUe=f(E) dv, wherefis some
function, which we wish to determine, of the fieldE. It might seem
that we would have no easy way to determine the functionf, but
many of the functions we could cook up would violate the symme-
try of space. For instance, we could imaginef(E) =aEy, wherea
is some constant with the appropriate units. However, this would
violate the symmetry of space, because it would give theyaxis a
different status fromxandz. As discussed on page 216, if we wish
to calculate a scalar based on some vectors, the dot product is the
only way to do it that has the correct symmetry properties. If all
we have is one vector,E, then the only scalar we can form isE·E,
which is the square of the magnitude of the electric field vector.
In principle, the energy function we are seeking could be propor-
tional toE·E, or to any function computed from it, such as√
E·E
or (E·E)^7. On physical grounds, however, the only possibility that
works isE·E. Suppose, for instance, that we pull apart two oppo-
sitely charged capacitor plates, as shown in figure a. We are doing
work by pulling them apart against the force of their electrical at-
traction, and this quantity of mechanical work equals the increase
in electrical energy,Ue. Using our previous approach to energy, we
would have thought ofUeas a quantity which depended on the dis-
tance of the positive and negative charges from each other, but now
we’re going to imagineUeas being stored within the electric field
that exists in the space between and around the charges. When the
plates are touching, their fields cancel everywhere, and there is zero
electrical energy. When they are separated, there is still approxi-
mately zero field on the outside, but the field between the plates is
nonzero, and holds some energy.
Now suppose we carry out the whole process, but with the plates
carrying double their previous charges. Since Coulomb’s law in-
volves the productq 1 q 2 of two charges, we have quadrupled the force
between any given pair of charged particles, and the total attractive
force is therefore also four times greater than before. This means
that the work done in separating the plates is four times greater, and
so is the energyUestored in the field. The field, however, has merely
been doubled at any given location: the electric fieldE+due to the
positively charged plate is doubled, and similarly for the contribu-
tionE−from the negative one, so the total electric fieldE++E−
is also doubled. Thus doubling the field results in an electrical en-
ergy which is four times greater, i.e., the energy density must be
proportional to the square of the field, dUe∝(E·E) dv. For ease604 Chapter 10 Fields