Transcript lecture3

4. Gauss’s law
4.1 Electric flux
•Uniform electric field
N

area
A

E
- number of electric field lines
Perpendicular
to the area
•Not uniform electric field
~N
 E  EAcos
Definition:

E
A  A
N~E
N~A
N ~ cos
E  EA cos
E   E   EA cos
Units:
 E   Nm 2
C  Vm
Example: Flux through open surface
Compare the electric flux through two squares of areas A1 and A2 (A2>A1)
placed in a region with a uniform electric field E:
(Side view)
θ
A1 A2 cosθ
A2
A. Φ1 < Φ2
θ
E
B. Φ1 = Φ2
C. Φ1 > Φ2
A2
The number of lines going through them is the same.
The effective area for the large square is A2cosθ = A1
•If you reverse the direction of A, you reverse the sign of the flux
•For an open surface, choose any direction
• For a closed surface it is conventional to take the area vector pointing outwards
Ф > 0  lines going out
Ф < 0  lines coming in
Ф = 0  no lines or a balance between incoming and outgoing lines
Example 1: Flux through closed surface
A cubic box is placed in a region of uniform electric field as shown in figure 1.
If the cube is tilted, as shown in figure 2, the net electric flux through it:
E
E
A. Increases
B. Decreases
C. Stays the same
The net flux is zero in both cases!
Every line that comes in, goes out.
Example 2: Two spheres have radii R and 2R.
Their centers lie on a positive charge.
Compare the electric flux through the two surfaces:
A. ΦR < Φ2R
B. ΦR = Φ2R
C. ΦR > Φ2R
The number of lines going through
both surfaces is the same.
The area at 2R is 4 times larger, but
the electric field is 4 times smaller!
2R
+
R
4.2 Gauss’s law - generalization of Coulomb’s law
A  4r 2
cos  1
Q
Ek 2
r
Gauss’s law:
Example:
Q
 E   EA cos   E  A  EA  k 2 4r 2
r
 E  4kQ  Q /  0
k
1
4 0
4.3 Applications of Gauss’s law
•Shell theorem:
r
Q
R
~
r
~
q
q
Qq
F k 2
r
Q
Ek 2
r
rR
~
F 0
~
E 0
~
r R
•Cylindrical symmetry

E
A  2rh
cos  1
Q /  0  E   EA cos  EA  E2rh

E

2 0 rh 2 0 r
Q
Q

h
Examples (applications of Gauss’s law):
Electric field created by
Point charge, charged sphere
or spherical shell
(Spherical symmetry)
E
Long cylinder or tube
(Cylindrical symmetry)

E
2 0 r
Q

L
Conducting plane surface, and
two conducting parallel plates
(Planar symmetry)

E
0
Q

A
One conducting plate
(Planar symmetry)

E
2 0


F  QE
Q
4 0 r 2
Planar symmetry
a) Conducting surface
b) Conducted plate
Q
E=0
+
+
+
+
+
  EA  Q /  0
E
1 Q
0 A

E
0
+Q
Q/2+Q/2=Q
E
E
+
+
+
+
+
+
+
+
+
+
c) Two conducting plates
E
-Q
+
+
E=0
E
+
+
+
-
  E2A  Q / 0
E
1 Q
0 2A

E
2 0
  
E  E1  E2
E

0
E=0
Example: A solid metal sphere of radius 3.00 m carries a total charge of 3.5μC.
What is the magnitude of the electric field at a distance from the sphere’s
center of (a) 0.15 m, (b) 2.90 m, (c) 3.10 m, and (d) 6.00 m? (e) How would the
answers differ if the sphere were a thin shell?
(a) and (b) Inside a solid metal sphere the electric field is 0
(c) Outside a solid metal sphere the electric field is the same as if all the
charge were concentrated at the center as a point charge:
E k
Q
r
2

 8.988  10 N  m C
9
2
2


(d) Same reasoning as in part (c):
E k
Q
r
2

 8.988  10 N  m C
9
2
2

3.50  106 C
 3.10 m 

2
  3.27 10
3.50  106 C
 6.00 m 
2
3
  8.74 10
(e) The answers would be no different for a thin metal shell
2
N C
N C
Example: Find the magnitude of the electric field produced by a uniformly
charged sphere with radius R and total charge Q .
 E   EA cos  E A  EA  E4r
For r > R:
For r < R:
E  Q 0
E
Vencl
r3
qQ
Q 3
V
R
Q r3
E  q 0 
 0 R3
R
Q
4 0 r 2
Qr
E
4 0 R 3
E
~
r
2
1
r2
r
R
Example: Find the magnitude of the electric field produced by a uniformly
charged cylinder with radius R length L and total charge Q .
 E   EA cos  EA  EA  E2rL
E  Q 0
For r > R:
E
~
R
R

E

2 0 rL 2 0 r
Q
Vencl
r2
qQ
Q 2
V
R
Q r2
E  q 0 
0 R2
For r < R:
r
Qr
r
E

2
2 0 R L 2 0 R 2
1
r
r
A charged isolated conductor
E0
   EA cos  Q  0
E0
E 00Q0
V   Ed ||
•The static electric field inside a conductor is zero – if it were not, the charges
would move.
•The net charge on a conductor in equilibrium is on its surface, because the
electric field inside the conductor is zero.
Spherical conducting shell:
Gauss’s surface
inside shell
inside conductor
Charge inside the surface
(+Q)
(+Q) + (-Q)=0
otside shell (+Q) + (-Q) + (+Q) = +Q
Example: Compare the magnitude of the electric field
at point P before and after the shell is removed.
Hint: Use Gauss’s law to find EP.
σin
A. Ebefore < Eafter
σout
B. Ebefore = Eafter
C. Ebefore > Eafter
Q
In both cases, the symmetry is the same.
We will use the same Gaussian surface:
a sphere that contains point P.
So the flux will look just the same:
P
  E 2R 2
And the enclosed charge is also the same: Q. Therefore,
  E 4R 2  Q  0
Q
E
4R 2 0