Download Notes 10 3318 Gauss`s Law I

ECE 3318
Applied Electricity and Magnetism
Spring 2016
Prof. David R. Jackson
ECE Dept.
Notes 10
1
Gauss’s Law
A charge q is inside a surface.
z
   D  nˆ dS
S (closed surface)
S
 q
 
 Nf

nˆ (outward normal)
q
x
y
E
Assume q produces Nf flux lines

 N S

NS  # flux lines that go through S
From the picture:
NS = Nf (all flux lines go through S)
Hence
 q
2
Gauss’s Law (cont.)
The charge q is now outside the surface
NS = 0
(All flux lines enter and
then leave the surface.)
q
S
Hence
   D  nˆ dS  0
S
3
Gauss’s Law (cont.)
To summarize both cases, we have Gauss’s law:
Carl Friedrich Gauss
 D  nˆ dS  Q
encl
S
n̂  outward normal
By superposition, this law must be true
for arbitrary charges.
(his signature)
4
Example
2q
-q
q
S3
S1
S2
 D  nˆ dS  q
 D  nˆ dS  0
 D  nˆ dS  2q
S1
S2
S3
Note: E  0 on S2 !
Note: All of the charges contribute to the electric field in space.
5
Using Gauss’s Law
Gauss’s law can be used to obtain the electric field
for certain problems.
The problems must be highly symmetrical.
 The problem must reduce to one unknown field component
(in one of the three coordinate systems).
Note: When Gauss’s law works, it is usually easier to use than Coulomb’s law
(i.e., the superposition formula).
6
Choice of Gaussian Surface
 D  nˆ dS  Q
encl
S
Rule 1: S must be a closed surface
Rule 2: S should go through the observation point (usually called r).
Guideline: Pick S to be  to E as much as possible
LHS 
E
 D  nˆ dS
n̂
S
SE
 nˆ D
(This simplifies the dot product calculation.)
S
7
Example
 D  nˆ dS  Q
Find E
z
encl
q
S
r
S
Assume D  rˆ Dr
r
y
(only an r component)
  D rˆ   rˆ dS  q
q
r
S
x
D
nˆ  rˆ
r
dS  q
S
Point charge
Assume Dr  Dr  r 
Then
Dr  dS  q
or
(only a function of r)
Dr  4 r 2   q
S
8
Example (cont.)
We then have
LHS 
2
ˆ
D

n
dS

D
4

r

r 

S
RHS  Qencl  q
so
Dr  4 r 2   q
Hence
q
Dr 
4 r 2
q 

2


D  rˆ 
C/m
2  
 4 r 
 q 
E  rˆ 
2 
 4 0r 
V/m
9
Note About Spherical Coordinates
Note: In spherical coordinates, the LHS is always:
LHS  Dr  4 r 2 
Assumption:
v  r, ,   f  r  (a function of r only, not  and  )
D  rˆ Dr  r  (in the rˆ direction, and a function of r only, not  and  )
LHS 
 D  nˆ dS 
 D  rˆ dS 
2
D
dS

D
4

r


r
 r
S
S
S
10
Example
z
s= s0
Find E everywhere
y
Case a) r < a
a
LHS = RHS
x
Hollow shell of
uniform surface charge
Dr  4 r 2   Qencl  0
so
s0
r
a
Dr  0
r
Hence
E  0 [V/m]
11
Example (cont.)
Case b) r > a
LHS = RHS
r
a
Dr  4 r 2   Qencl   s 0 4 a 2
4 a 2  s 0
 Dr 
4 r 2
Q
 Dr 
4 r 2
r
s0
Hence
E  rˆ
Q
4 0 r
2
[V/m]
The electric field outside the sphere of surface charge is the same as
from a point charge at the origin.
12
Example (cont.)
z
Summary
s = s0
y
r<a
E  0 [V/m]
a
x
r>a
E  rˆ
Q
4 0 r
2
[V/m]
Note: A similar result holds for the force due to gravity
from a shell of material mass.
13
z
Example (cont.)
s = s0
y
Note: The electric field is
discontinuous as we cross the
boundary of a surface charge density.
a
x
Er
Q/(40
Q
a2)
4 0 r 2
r
a
14
Example
z
v = v0
Find E(r) everywhere
y
a
Case a) r < a
x
Solid sphere of uniform volume charge
 D  nˆ dS  Q
encl
S
r
 Dr  4 r 2   Qencl
r
V
a
v0
Qencl    v  r  dV
V
Gaussian surface S
15
Example (cont.)
Calculate RHS:
Qencl    v 0 dV
r
r
V
V
  v 0  dV
a
V
v0
Gaussian surface S
4 3
 v 0   r 
3

LHS = RHS
4 3
Dr  4 r    v 0   r 
3

2
16
Example (cont.)
z
Hence, we have
v = v0
1 
Dr  v 0  r 
3 
r
y
a
x
The vector electric field is then:
ra
 r 
E  rˆ v 0 
 [V/m]
 3 0 
17
Example (cont.)
 D  nˆ dS  Q
Case b) r > a
encl
S
 Dr  4 r 2   Qencl
r
a
r
Qencl
v = v0
4 3
 v0   a 
3

Gaussian surface S
so
4

Dr  4 r 2    v 0   a 3 
3

 a3 
 Dr   v 0  2 
 3r 
Hence, we have
 v 0 a3 
E  rˆ 
V/m
2  
 3 0r 
18
Example (cont.)
We can write this as:
z
  v 0 a 3    4 / 3 
E  rˆ 
2 

3

r
 0    4 / 3 
v = v0



r
y
Hence
a
E  rˆ
Q
4 0 r
2
x
ra
where
4

Q  v 0   a3 
3

The electric field outside the sphere
of charge is the same as from a
point charge at the origin.
19
z
Example (cont.)
Summary
v = v0
y
a
 r 
E  rˆ v 0 
 [V/m] (r < a)
 3 0 
v 0 a 3
Q
ˆ
E  rˆ

r
3 0 r 2
4 0 r 2
x
 V/m
(r > a)
Er
Note: The electric field is continuous
as we cross the boundary of a
volume charge density.
v0 a/ (30)
r
a
20