Biomedical Engineering Reference
In-Depth Information
Q
Q
∫
2
. in
r
θθφ
d d
=
2
4
πε
r
ε
S
0
0
In other words, the area integral of the electric field over any closed surface
is equal to the net charge enclosed in the surface divided by the permittivity of
space. Gauss' law is useful for calculating the electrical fields when they originate
from charge distributions of sufficient symmetry to apply it. Gauss' law permits the
evaluation of the electric field in many situations by forming a symmetric Gaussian
surface surrounding a charge distribution and evaluating the electric flux through
that surface. Since charge is enclosed in a 3D space, a normal practice is to express
charge either by unit surface area called surface charge density (
σ
ch
with units C/
ρ
ch
with units C/cm
3
) and then
integrate accordingly. When charge enclosed is electric charge density of
cm
2
) or by unit volume basis called charge density (
ρ
ch
, then
total charge,
∫
Q
=
ρ
•
dV
(3.35)
ch
V
EXAMPLE 3.10
Consider an infinitely long plate with surface charge density of 2 C/m
2
in a medium that
has a permittivity of 25
×
10
−
12
C
2
/Nm
2
. What
is the electrical field strength?
Solution: Because the plate is infinite and symmetrical, the only direction the
E
field
can go is perpendicular to the plate. So,
∫
E A AE
=
2
S
2
2
[]
Q
⎡
C/cm
⎤
*
A
⎡
cm
⎤
2
A C
=
σ
=
⎣
⎦
⎣
⎦
ch
ch
From (3.34),
σ
ε
2
AE
ch
A
=
Hence
2
2 C/cm
⎡
⎤
σ
⎣
⎦
[
]
E
ch
=4
10
10
N/C
==
×
2
−
12
2
2
ε
22510 [C/Nm]
××
and
σ
E
ch
n
ˆ
=
2
ε
where
n
is a unit vector perpendicular to the plate.
