1. Physics 212
Lecture 4
Gauss’ Law

2. Gauss’ Law In general, the integral to calculate flux is difficult.
But in cases with symmetry (spheres, cylinders, etc.) we can find a closed surface where,
is constant everywhere on the surface.
For such cases, 19

3. E field from an infinite line of charge with charge density 
Example
E field from an infinite line of charge with charge density  r L
Choose cylinder of radius r, length L, centered on the line of charge.
By symmetry, E field points radially outward ( for positive  )
and so

4. Examples with Symmetry
Spherical
Cylindrical
Planar 21

5. Checkpoint 4 Recall that for an infinite plane
In which case is E at point P the biggest?
A) A B) B C) the same
Recall that for an infinite plane
with charge density  (Coul/m2),
E = /20 27

6. Superposition !
NET - + + +
Case A
Case B 34

7. Checkpoint 1 The E field from a charged cube
is not constant on any of these surfaces.
Gauss’s law is true, but it does not help us to calculate E for this particular problem.
(D) The field cannot be calculated using Gauss’ Law
(E) None of the above
How to calculate the field? Go back and find E by superposition. Add the contributions to E
from each infinitesimal charge dq in the cube: 23

8. Conductors = charges free to move
Claim: E = 0 inside any conductor at equilibrium
Why ?
Charges in a conductor will move if E  0. They redistribute themselves until E = 0 and everything comes to equlibrium.
Claim: Any excess charge in a conductor is on the surface (in equilibrium).
Careful here… inside means part of conductor, does NOT include cavity of a conductor!
Why?
Take Gaussian surface (dashes) to be just
inside conductor surface.
E = 0
E = 0 everywhere inside conductor.
Use Gauss’ Law: 06

9.  Charges reside on surfaces of conductors.
E = 0 inside conductors
 Charges reside on surfaces of conductors.
Induced Charges
Begin with a neutral conductor:
Q = 0
No charge inside or on the surface
Now bring a positive charge +Q near
the conductor.
+ Q - +
This induces (-) charge on one end and (+) charge on the other.
The total charge on the conductor is
still zero.
The induced charge is on the surface of the conductor. 09

10. Charge in Cavity of Conductor
A particle with charge +Q is placed in the center of an uncharged conducting hollow sphere. How much charge will be induced on the inner and outer surfaces of the sphere?
A) inner = –Q, outer = +Q
B) inner = –Q/2 , outer = +Q/2
C) inner = 0, outer = 0
D) inner = +Q/2, outer = -Q/2
E) inner = +Q, outer = -Q
Write E = 0 in conductor
Draw induced charge
Qouter Q
Qinner 10

11. Checkpoint 3
Draw gaussian surfaces and field 26

12. What is direction of field OUTSIDE the red sphere?
Checkpoint 3
What is direction of field OUTSIDE the red sphere?
Using a Gaussian surface that enclosed both conducting spheres, the net enclosed charge will be zero. Therefore, the field outside will be zero. 29

13. Checkpoint 2 31

14. Calculation Point charge +3Q at center of neutraly r2
neutral
conductor
Point charge +3Q at center of neutral
conducting shell of inner radius r1 and
outer radius r2.
What is E everywhere?
+3Q r1 x A B C D
Magnitude of E is function of r.
Magnitude of E is function of (r-r1).
Magnitude of E is function of (r-r2).
None of the above. A B C D
Direction of E is along
None of the above 36

15. Calculation Use Gauss’ Law with a spherical surfacey
Calculation r2
neutral
conductor
E is a function of r . Direction of is along
+3Q r1 x
Use Gauss’ Law with a spherical surface
Centered on the origin to determine E(r).
Do calculation
r < r1 A B C
r1 < r < r2 A B C
r > r2 40

16. What is the induced surface charge density at r1 ?
neutral
conductor r2
r < r1
r > r2
+3Q r1
r1 < r < r2
What is the induced surface charge density at r1 ?
Gauss’ Law: A B C r1 r2
+3Q
Similarly: 44

17. Now suppose we give the conductor a charge of -Qy
Now suppose we give the conductor a charge of -Q -Q
What is E everywhere?
What are charge distributions at r1 and r2? r2
conductor
r1 < r < r2 r1 r2
+3Q
-3Q +
+2Q
+3Q r1 x A B C
r < r1
r > r2 A B C 46