Transcript Gauss`s Law

Gauss’s Law
Electric Flux
•Electric flux is the amount of electric field going across a surface
•It is defined in terms of a direction, or normal unit vector,
perpendicular to the surface
•For a constant electric field, and a flat surface, it is easy to calculate
•Denoted by E
 E  AE cos 
 E  AE  nˆ
•Units of Nm2/C
•When the surface is flat, and the fields are constant, you
can just use multiplication to get the flux
•When the surface is curved, or the fields are not constant,
you have to perform an integration
 E   E  nˆ dA
E  0
n̂
E  A E
n̂
Electric Flux For a Cylinder
A point charge q is at the center of a cylinder of radius a and height 2b.
What is the electric flux out of (a) each end and (b) the lateral surface?
top
r  s 2  b2
ke q
E 2
r
ke qb
n̂
E
dA  2 sds
n̂
•Consider a ring of radius s and thickness
ds
a
a
2 ke qb
2 ke qbsds
2 ke qb
 2 ke q 

 E   E  nˆ dA  
3/2
2
2
2
2
b2  a 2
b s 0
0 b  s 
lateral surface
 E   E  nˆ dA 
r  z a
b

2 ke qa dz
2
b  a  z
2

E  nˆ 
2
2 3/2

E
s
ke qb
ke q
E  nˆ  2 cos   3 
2
2 3/2
r
r
b  s 
2

ke qa
r3
2 ke qz
a2  z 2
dA  2 adz
b

b
4 ke qb
a 2  b2
b
z
r
a
r
b
q
b
a
Total Flux Out of Various Shapes
n̂
E
A point charge q is at the “center” of a (a) sphere (b) joined hemispheres
(c) cylinder (d) cube. What is the total electric flux out of the shape?
ke q
E  nˆ  2
b
a
 E   E1   E 2   E 3
a
ke q
 E   E  nˆ dA
 E1  EA  2 2 a 2
q
ke q
a
 2 4 a 2
 E1  2 ke q
a
n̂ q
a
 E  4 ke q
E2  0
n̂
E
 E 3  2 ke q
E n̂
 E  2 Ecap   Elat
E

2 ke qb  4 ke qb
 E  4 ke q
 2  2 ke q 


2
2
b a 
b2  a 2

q
 E  4 ke q
Gauss’s Law
•No matter what shape you use, the total electric flux out of a region
containing a point charge q is 4keq = q/0.
Why is this true?
•Electric flux is just measuring how many field
lines come out of a given region
q
•No matter how you distort the shape, the field lines
come out somewhere
 E  4 ke q
E  q 0
•If you have multiple charges inside the region their effects add
•However, charges outside the region do not contribute
E   q1  q2  q3   0
q4
q3
E 
q1
q2
qin
0
Using Gauss’s Law
Gauss’s Law can be used to solve three types of problems:
1. Finding the total charge in a region when you know the electric field outside
that region
2. Finding the total flux out of a region when the charge is known
a) It can also be used to find the flux out of one side in symmetrical
problems
b) In such cases, you must first argue from symmetry that the flux is
identical through each side
3. Finding the electrical field in highly symmetrical situations
a) One must first use reason to find the direction of the electric field
everywhere
b) Then draw a Gaussian surface over which the electric field is constant
c) Use this surface to find the electric field using Gauss’s Law
d) Works generally only for spherical, cylindrical, or planar-type problems
Sample Problem
A very long box has the shape of a regular pentagonal prism.
Inscribed in the box is a sphere of radius R with surface charge
density . What is the electric flux out of one lateral side of the box?
End view
Perspective
view
•The flux out of the end caps is negligible
•Because it is a regular pentagon, the flux from
the other five sides must be the same
5E ,side  4 R2  0
 E  qin  0
5 E ,side   A  0
E ,side  4 R2 5 0
Using Gauss’s Law to find E-field
A sphere of radius a has uniform charge density  throughout. What
is the direction and magnitude of the electric field everywhere?
•Clearly, all directions are created equal in this problem
•Certainly the electric field will point away from the sphere at all points
•The electric field must depend only on the distance
•Draw a sphere of radius r around this charge
qin   0 E
•Now use Gauss’s Law with this sphere
V    0 AE  nˆ
4
3
 a    0 4 r E
3
2
 a3
E  rˆ
3 0 r 2
r
E
Note: We implicitly assumed r > a
when we calculated the enclosed charge.
E  rˆE
a
Using Gauss’s Law to find E-field (2)
A sphere of radius a has uniform charge density  throughout. What
is the direction and magnitude of the electric field everywhere?
•When computing the flux for a
Gaussian surface, only include
the electric charges inside the
surface
qin   0 E
E  rˆE
V    0 AE  nˆ
4
3
 r 3    0 4 r 2 E
E  rˆ  r 3 0
rˆ  a 3 0 r
E
 rˆ  r 3 0
3
2
r
for r  a,
for r  a.
E
a
Electric Field From a Line Charge
What is the electric field from an infinite line with linear charge density ?
n̂
E
E  rˆE
n̂
L
r
•Electric field must point away from the line charge, and depends only on distance
•Add a cylindrical Gaussian surface with radius r and length L
•Use Gauss’s Law q   
in
0
E
•The ends of the cylinder don’t contribute
•On the sides, the electric field and the normal are parallel
E
 L   0  E  nˆ da   0 EA   0 E  2 rL 

2 0 r
 rˆ
E
2 0 r
Electric Field From a Plane Charge
What is the electric field from an infinite plane with surface charge density ?
E
n̂
n̂
n̂
•Electric field must point away from the surface, and depends only on
E  kˆ E
distance d from the surface
•Add a box shaped Gaussian surface of size 2d  L  W
•Use Gauss’s Law
qin   0 E
•The sides don’t contribute
•On the top and bottom, the electric field and the normal are parallel
 kˆ
E

 LW   0  E  nˆ da   0 EA   0 E  2LW 
E
2 0
2 0
Conductors and Gauss’s Law
•Conductors are materials where charges are free to flow in response to electric
forces
•The charges flow until the electric field is neutralized in the conductor
Inside a conductor, E = 0
•Draw any Gaussian surface inside the conductor
qin   0 E   0  E  nˆ dA  0
In the interior of a conductor,
there is no charge
The charge all flows to the surface
Electric Field at Surface of a Conductor
•Because charge accumulates on the surface of a conductor,
there can be electric field just outside the conductor
•Will be perpendicular to surface
n̂
•We can calculate it from Gauss’s Law
•Draw a small box that slightly penetrates the surface
•The lateral sides are small and have no flux through
them
•The bottom side is inside the conductor and has no electric field
•The top side has area A and has flux through it
 E  E  nˆ A  EA
•The charge inside the box is due to the surface charge 
•We can use Gauss’s Law to relate these qin   A   0  E   0 EA

E  nˆ
0
Sample problem
An infinitely long hollow neutral conducting cylinder has inner
radius a and outer radius b. Along its axis is an infinite line charge
with linear charge density . Find the electric field everywhere.
end-on
view
b
a
perspective
view
•Use cylindrical Gaussian surfaces when needed in each region
•For the innermost region (r < a), the total charge comes entirely from the line charge
 rˆ
•The computation is identical to before
E
2 0 r
•For the region inside the conductor, the electric field is always zero E  0
•For the region outside the conductor (r > b), the electric field can be calculated like
before
 rˆ
•The conductor, since it is neutral, doesn’t contribute E 
2 0 r
Where does the charge go?
How can the electric field appear, then disappear, then reappear?
 rˆ

E   2 0 r
 0

if r  a or r  b
if a  r  b
+
+
–
+
–
–
+
+
–
–
–
•The positive charge at the center attracts negative
– –
+
charges from the conductor, which move towards it
+
•This leaves behind positive charges, which repel
+
each other and migrate to the surface
end-on view
•In general, a hollow conductor masks the distribution of the charge
+
inside it, only remembering the total charge
+
+
•Consider a sphere with an irregular cavity in it
––
cutaway
–
–
view
q +
ke q
+
–
Eout  2 rˆ
–
r
+
+