1. Applied Electricity and Magnetism
ECE 3318
Applied Electricity and Magnetism
Spring 2017
Prof. David R. Jackson
ECE Dept.
Notes 10

2. NS  # flux lines that go through S
Gauss’s Law
A charge q is inside a surface.
S (closed surface)
NS  # flux lines that go through S
From the picture:
NS = Nf (all flux lines go through S)
Assume q produces Nf flux lines
Hence

3. Gauss’s Law (cont.) NS = 0 The charge q is now outside the surface
(All flux lines enter and then leave the surface.)
Hence

4. Gauss’s Law (cont.) To summarize both cases, we have Gauss’s law:
Carl Friedrich Gauss
By superposition, this law must be true for arbitrary charges.
(his signature)

5. Note: All of the charges contribute to the electric field in space.
Example
Note: E  0 on S2 !
Note: All of the charges contribute to the electric field in space.

6. 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).

7. Choice of Gaussian Surface
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 E S
(This simplifies the dot product calculation.)

8. Example
Point charge
Find E

9. Example (cont.) S Assume Assume Then or (only an r component)
(only a function of r)
Then or

10. Example (cont.)
We then have so
Hence

11. Example (cont.)
Summary

12. Note About Spherical Coordinates
Note: In spherical coordinates, the LHS is always:
Assumption:

13. Hollow shell of uniform surface charge density
Example
Hollow shell of uniform surface charge density
Find E everywhere

14. Example (cont.)
Case a) r < a
LHS = RHS so
Hence

15. Example (cont.) Case b) r > a LHS = RHS Hence
The electric field outside the sphere of surface charge is the same as from a point charge at the origin.

16. Example (cont.) Summary Note:
A similar result holds for the force due to gravity from a shell of material mass.

17. Example (cont.)
Note:
The electric field is discontinuous as we cross the boundary of a surface charge density.

18. Solid sphere of uniform volume charge density
Example
Solid sphere of uniform volume charge density
Find E(r) everywhere

19. Example (cont.)
Case a) r < a
Gaussian surface S

20. Example (cont.)
Calculate RHS:
Gaussian surface S
LHS = RHS

21. Example (cont.)
Hence, we have
The vector electric field is then:

22. Example (cont.)
Case b) r > a
Gaussian surface S so
Hence, we have

23. Example (cont.) We can write this as: Hence where
The electric field outside the sphere of charge is the same as from a point charge at the origin.

24. Example (cont.) Summary Note:
The electric field is continuous as we cross the boundary of a volume charge density.