ECE106 Lecture Notes - Lecture 24: Intelligence Quotient, Electric Field, Dielectric
6 Sep 2018
School
Department
Course
Professor
156
·
·
∫
∫
E
⊥
ds
∴
E
˙
·
d
˙
s
=
0
There is no flux on the top and bottom! Now Gauss law states,
I
∫
Eds
+
∫
E
˙
d
˙
s
=
Q
enc
s0
E
˙
d
˙
s
=
Q
enc
s0
cylindrical
surface
topor
bottom
The second integral in the above equation is zero so there is no need to worry about it. What is
Q
enc
?
∴
cylindrical
surface
Q
enc
=
ρL
ρL
Eds
=
s0
E ds
=
ρL
s0
ρL
E
2
πrL
=
E
=
s0
ρ
2πs0r
or
E
˙
=
ρ
r
ˆ
2
πs
0
r
(4.10)
Check it with the calculation done previously for a continuous charge distribution. Which one
is easier? Can we do this method if the line charge was of finite length ”
L
”? Will symmetry hold in
that case?
Note on Gauss Law
: As mentioned earlier, laws of force are also helpful in visualizing the
field patterns (electric, magnetic or any other field).
1.
flux (Φ
e
) of the electric field
→
”flow” of the field.
2.
flux
densit
y
(
D
˙
)
(flux
or
”flo
w”
p
er
unit
area).
3.
flux
densit
y
(
D
˙
)
is
related
to
the
electric
field
in
tensit
y
(
E
˙
)
at
the
p
oin
t:
D
˙
=
s
E
˙
(4.11)
in free-space,
s
=
s
0
; in dielectric medium
s
=
s
0
s
r
.
F
or
example,
a
fluid
flo
w
c
haracterized
b
y
a
constan
t
flo
w
v
ector
˙
v
A
˙
