Digital Signal Processing Reference
In-Depth Information
Also note that (2-65) can be derived using vector identities, which is significantly
simpler but does not provide any intuition. The alternative derivation is shown
here because we employ a similar technique when studying magnetostatics in the
next section.
For an electrostatic field, Ampere's law is reduced to
∇×
E
=
0 because the
field does not vary with time. For any differentiable scalar function, the following
vector identity holds true (from Appendix A):
∇×∇
ψ
=
0
E
must be derivable from the gradient of a scalar
function. Since (2-61) shows a relationship between the electrostatic potential and
the electric field, a leap of logic says that the scalar function in the vector identity
must be the electrostatic potential:
∇×
E
=
Therefore, since
0,
∇×
E
=∇×
(
−∇
)
=
0
(2-66)
Equation (2-65), known as the
electrostatic scalar potential
, is used often when
solving electrostatic problems such as transmission-line impedance or calculating
the effective dielectric constant of a microstrip, as we demonstrate in Chapter 3.
2.4.2 Energy in an Electric Field
To calculate the energy stored in an electric field, it is necessary to begin with
a stationary charge (
q
1
) in free space that is infinitely far way from any other
charges. It takes no work to move the first charge into position (
W
1
0) because
there are no other nearby charges to provide electrostatic repulsion. Then we
calculate how much work it takes to bring another charge (
q
2
) into the vicinity
of the first charge using (2-61) and (2-62):
=
q
1
q
2
4
πε
0
r
12
W
2
=
q
2
12
=
(2-67)
If we bring another charge,
q
3
, into the vicinity of
q
1
and
q
2
, the work is calculated
as
q
1
r
13
+
1
4
πε
0
q
2
r
23
W
3
=
q
3
(
13
+
23
)
=
q
3
Thus, the total work done to bring the three charges together is
q
1
r
13
+
q
1
q
2
4
πε
0
r
12
+
q
3
4
πε
0
q
2
r
23
W
tot
=
W
1
+
W
2
+
W
3
=
0
+
q
1
q
2
r
12
1
4
πε
0
q
1
q
3
r
13
q
2
q
3
r
23
=
+
+
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