1. Notes 10 ECE 3318 Applied Electricity and Magnetism Gauss’s Law I
Spring 2019
Prof. David R. Jackson
Dept. of ECE
Notes 10
Gauss’s Law I
Notes prepared by the EM Group University of Houston

2. NS  # flux lines that go through S
Gauss’s Law
A charge q is inside a closed 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 in a simple way.
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 the same:
Assumption:
Spherical Gaussian surface
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.