Digital Signal Processing Reference
In-Depth Information
The identity can be rewritten in terms of the electrostatic vector potential, sub-
stituting
ψ
=
and
a
=
ε E
:
ε
2
W
e
=
(
∇·
E
−
E
·∇
) dV
V
E
=−∇
, the equation can be simplified further:
Since
ε
2
(
∇·
E
+
E
·
E)dV
W
e
=
(2-72)
V
The divergence theorem of vector calculus states (Appendix A) that
(
∇·
F)dV
S
F
·
=
d
s
V
allowing further simplification of (2-72):
E
·
E
dV
ε
2
ε
2
E
·
n ds
+
W
e
=
(2-73)
S
V
where
n
is a unit vector normal to the surface.
Equation (2-73) describes the total energy in an electric field that is induced
by a volume of charges. If we expand the volume of integration to include
all space, any contribution of (2-73) outside the charge distribution will con-
tribute nothing to the work done since there are no charges in that space.
Also note that if the volume of integration is chosen to be infinity, the sur-
face integral disappears. To understand why, remember that the surface integral
sums all the contributions evaluated at the surface. Since
∝
1/
r
[equation
1/
r
2
[equation (2-60)], and
ds
∝
r
2
, the limit of
E
·
nds
is pro-
portional to (1/
r
)(1/
r
2
)
r
2
, whose limit goes to zero when
r
is infinity and the
surface integral disappears. The volume integral includes contributions over the
entire volume, not only on the surface. Subsequently, equation (2-73) can be
reduced to (2-74), which is the work done by accumulating charges to create an
electric field:
(2-62)],
E
∝
ε
2
all space
(E
2
) dV
W
e
=
joules
(2-74)
This leads to the definition of the
volume energy density
, which expresses the
stored energy in a charge distribution in terms of the electric field:
ε
2
E
2
joules
/
m
3
w
e
=
(2-75)
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