Gauss’s Law (II)
• Examples: charged spherical shell, infinite plane,
long straight wire
Review:
For a closed surface S:
Net outward flux through S 
net charge enclosed by S


Calculating E for symmetric charge distributions
1) Use symmetry to sketch the behaviour of E
(possible only for very symmetric distributions)
2) Pick an imaginary “Gaussian surface” S
3) Calculate the flux through S, in terms of an
unknown |E|
4) Calculate the charge enclosed by S from geometry
5) Relate 3) and 4) by Gauss’s Law, solve for |E|
Example: Uniformly-Charged Thin Sheet
charge

unit area
E
is uniform
+
+
+
+
+
+
+
+
+
+
+
Gaussian
surface
Find E due to an infinite charged sheet
Solution
Example: Infinite Line Charge (Long thin Wire)
r
+
+
+
+
+
+
+
L
charge

 constant
unit length
Find E at distance r from the wire
+
+
Solution
Conductors (in Electrostatic Equilibrium)

1) E  0 inside a conductor.

2) E  surface just outside a conductor.
3) Any net charge on a conductor resides on the
surface only.
Example: Hollow Thin Spherical Shell
Total charge
+
Q
+
+
+
r
a
+
r
+
+
+
P2
+
Find E outside at P1 : (r > a):
Find E inside at P2 :
P1
+
(r < a):
+
+
+
Solution
Consequences if E was not zero:

1) If E  0 , charges would move!!!

2) If E||  0 , charges would move!!!
E
Ell
+
E=0 inside

3) E  0 inside, so any Gaussian surface inside
the conductor encloses zero net charge.
+
+
+
+
+
+
+ +
+ + + +
+ +


+
+ +
+
+
+
+

+
+
+
+
+
+ +
  0 So net charge =0
Since E = 0 everywhere .
+
-
-
+
conductor
-

E 0
-
+
+
Surface Charge Density on a Conductor
E
dA1
+
+
+
+
dA2
E=0
+
inside
What is E at the surface of the conductor?
charge
 
unit area
Solution
Example:
• Metal ball of radius R1 has charge +Q
Metal shell of radius R2,and R3 has total charge -3Q )
• Find: charges on each surface. (and E)
R3
R1
-3Q
+Q
R2
Solution
Quick Recap
Conductor:

E

Infinite Sheet of charge:
At surface of conductor

E
2 
Why is the lower value half of the upper?