Simple Nature - Light and Matter

10.4 Energy In Fieldsa / Two oppositely chargedcapacitor plates are pulled apart.10.4.1 Electric field energyFields possess energy, as argued on page 559, but how muchenergy? The answer can be found using the following elegant approach.We assume that the electric energy contained in an infinitesimalvolume of space dv is given by dU e = f(E) dv, where f is somefunction, which we wish to determine, of the field E. It might seemthat we would have no easy way to determine the function f, butmany of the functions we could cook up would violate the symmetryof space. For instance, we could imagine f(E) = aE y , where ais some constant with the appropriate units. However, this wouldviolate the symmetry of space, because it would give the y axis adifferent status from x and z. As discussed on page 212, if we wishto calculate a scalar based on some vectors, the dot product is theonly way to do it that has the correct symmetry properties. If allwe have is one vector, E, then the only scalar we can form is E · E,which is the square of the magnitude of the electric field vector.In principle, the energy function we are seeking could be proportional√ to E · E, or to any function computed from it, such asE · E or (E · E) 7 . On physical grounds, however, the only possibilitythat works is E · E. Suppose, for instance, that we pull aparttwo oppositely charged capacitor plates, as shown in figure a. Weare doing work by pulling them apart against the force of their electricalattraction, and this quantity of mechanical work equals theincrease in electrical energy, U e . Using our previous approach toenergy, we would have thought of U e as a quantity which dependedon the distance of the positive and negative charges from each other,but now we’re going to imagine U e as being stored within the electricfield that exists in the space between and around the charges.When the plates are touching, their fields cancel everywhere, andthere is zero electrical energy. When they are separated, there is stillapproximately zero field on the outside, but the field between theplates is nonzero, and holds some energy. Now suppose we carryout the whole process, but with the plates carrying double theirprevious charges. Since Coulomb’s law involves the product q 1 q 2 oftwo charges, we have quadrupled the force between any given pairof charged particles, and the total attractive force is therefore alsofour times greater than before. This means that the work done inseparating the plates is four times greater, and so is the energy U estored in the field. The field, however, has merely been doubled atany given location: the electric field E + due to the positively chargedplate is doubled, and similarly for the contribution E − from the negativeone, so the total electric field E + + E − is also doubled. Thusdoubling the field results in an electrical energy which is four timesgreater, i.e., the energy density must be proportional to the squareof the field, dU e ∝ (E · E) dv. For ease of notation, we write this582 Chapter 10 Fields