self-check D
We can think of the quantity dUe/dvas theenergy densitydue to the
electric field, i.e., the number of joules per cubic meter needed in order
to create that field. (a) How does this quantity depend on the compo-
nents of the field vector,Ex,Ey, andEz? (b) Suppose we have a field
withEx 6 = 0,Ey=0, andEz=0. What would happen to the energy density
if we reversed the sign ofEx? .Answer, p. 1059
A numerical example example 17
.A capacitor has plates whose areas are 10−^4 m^2 , separated by
a gap of 10−^5 m. A 1.5-volt battery is connected across it. How
much energy is sucked out of the battery and stored in the elec-
tric field between the plates? (A real capacitor typically has an
insulating material between the plates whose molecules interact
electrically with the charge in the plates. For this example, we’ll
assume that there is just a vacuum in between the plates. The
plates are also typically rolled up rather than flat.)
.To connect this with our previous calculations, we need to find
the charge density on the plates in terms of the voltage we were
given. Our previous examples were based on the assumption that
the gap between the plates was small compared to the size of the
plates. Is this valid here? Well, if the plates were square, then
the area of 10−^4 m^2 would imply that their sides were 10−^2 m in
length. This is indeed very large compared to the gap of 10−^5
m, so this assumption appears to be valid (unless, perhaps, the
plates have some very strange, long and skinny shape).
Based on this assumption, the field is relatively uniform in the
whole volume between the plates, so we can use a single symbol,
E, to represent its magnitude, and the relationE = dV/dx is
equivalent toE=∆V/∆x= (1.5 V)/(gap) = 1.5× 105 V/m.
Since the field is uniform, we can dispense with the calculus, and
replace dUe= (1/ 8 πk)E^2 dvwithUe= (1/ 8 πk)E^2 v. The volume
equals the area multiplied by the gap, so we haveUe= (1/ 8 πk)E^2 (area)(gap)=1
8 π× 9 × 109 N·m^2 /C^2(1.5× 105 V/m)^2 (10−^4 m^2 )(10−^5 m)= 1× 10 −^10 Jself-check E
Show that the units in the preceding example really do work out to be
joules. .Answer, p. 1059
Whykis on the bottom example 18
It may also seem strange that the constantkis in the denomina-
tor of the equation dUe= (1/ 8 πk)E^2 dv. The Coulomb constantk
tells us how strong electric forces are, so shouldn’t it be on top?606 Chapter 10 Fields