1. Using Symmetry to Solve really complex vector calculus. . .
Gauss’ lAW
Using Symmetry to Solve really complex vector calculus. . .
a.k.a “You don’t know the divergence theorem yet or Greene’s Theorem, so I have to wing it because you’re not in vector calc.”
By Mr. Mac

2. Learning Targets:
How to determine the amount of charge within an imaginary, closed surface by examining the electric field on the surface.
What is meant by flux and how you can calculate it for mass, volume, and electrical flow.
How to use Gauss’s Law (Integral Form) to calculate the electric field as a function of (r) for a symmetric distribution of charges. (plane, cylinder, sphere).

3. A charge hidden inside a box can be probed with a test charge qo to measure Force/E field outside the box.

4. The volume (V) flow rate (dV/dt) of fluid through the wire rectangle (a) is:
v x A when the area of the rectangle A is perpendicular to the velocity vector v.
This flow rate (“Volume Flux”) (b) is vA cos f when the rectangle is tilted at an angle f. (Dot Product)
Fv=v ● A Units (m3/s)
For mass, Density (r) xVolume=Mass
Fm=rv ● A Units (kg/s)

5. (a) The electric flux through the surface = E●A.
(b) When the area vector makes an angle f with the vector E, the area projected onto a plane oriented perpendicular to the flow is A perp.   = A cos f. 
The flux is zero when f = 90o because the rectangle lies in a plane parallel to the flow and no E -Field Lines flow through the Area.
Imagine E-Field is real! (like water from a sprinkler)
What about other shapes made of flat surfaces?

6. Electric FLUX through a Gaussian sphere centered on a point charge +q.
FE = ò E . dA = ò E dA cos f = ò  E dA = E ò dA
(Why can we take out E?) = E (4p R2) = (1/4p eo) q /R2) (4p R2) = q / eo. So the electric flux  FE = q / eo. 

7. Gauss Law (Integral Form)
Now we can write Gauss's Law: FE = ò E . dA = ò |E||dA|cos f =qenc./eo

8. Flux FE from a point charge q.
The projection of an element of area dA of a sphere of radius R UP onto a concentric sphere of radius 2R. The projection multiplies each linear dimension by 2, so the area element on the larger sphere is 4 dA. The same number of lines of flux pass thru each area element.
Flux FE from a point charge q.

9. Flux through an irregular surface.
The projection of the area element dA onto the spherical surface is dA cos f.
Flux through an irregular surface.

10. Spherical Gaussian surfaces around (a) positive and (b) negative point charge.

11. Gauss’s Law can be used to calculate the magnitude of the E field vector:
Symmetry was able to help us in last example!
If we can mimic the shape of the charge distribution with a closed surface (Gaussian Surface) so that the E-Field is constant at each surface and has a constant angle, f, then the messy math and calculus simplifies.
You could still use this law to find the E-Field for any shape! But symmetry would be messy!
Each surface would then have a different angle and E-Field strength. (You’ll have to wait until vector calc to solve these)
To be honest, very often you have to use a computer.
(How could you write Area in vector form for a box?)
How could you write E-Field in vector form for a sphere?)
C J. F. Becker

12. Use the following recipe for Gauss’s Law problems: 1
Use the following recipe for Gauss’s Law problems: 1. Carefully draw a figure - location of all charges, direction of electric field vectors E

13. Use the following recipe for Gauss’s Law problems: 1
Use the following recipe for Gauss’s Law problems: 1. Carefully draw a figure - location of all charges, direction of electric field vectors E 2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field. 

14. Use the following recipe for Gauss’s Law problems: 1
Use the following recipe for Gauss’s Law problems: 1. Carefully draw a figure - location of all charges, direction of electric field vectors E 2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E o dA

15. Use the following recipe for Gauss’s Law problems: 1
Use the following recipe for Gauss’s Law problems: 1. Carefully draw a figure - location of all charges, direction of electric field vectors E 2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E o dA 4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the |E| out of the integral.

16. Use the following recipe for Gauss’s Law problems: 1
Use the following recipe for Gauss’s Law problems: 1. Carefully draw a figure - location of all charges, direction of electric field vectors E 2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E . dA 4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the |E| out of the integral. 5. Determine the value of Qencl from your figure and insert it into Gauss's equation. (This is where those charge densities come into play- (l,s,r)

17. Use the following recipe for Gauss’s Law problems: 1
Use the following recipe for Gauss’s Law problems: 1. Carefully draw a figure - location of all charges, and direction of electric field vectors E 2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E o dA 4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the |E| out of the integral. 5. Determine the value of Qencl from your figure and insert it into Gauss's equation. 6. Solve the equation for the magnitude of E.
C J. F. Becker

18. Gaussian surface
Under electrostatic conditions, any excess charge resides entirely on the surface of a solid conductor.

19. Electric field = zero (electrostatic) inside a solid conducting sphere
Under electrostatic conditions the electric field inside a solid conducting sphere is zero. Outside the sphere the electric field drops off as 1 / r2, as though all the excess charge on the sphere were concentrated at its center.
Electric field = zero (electrostatic) inside a solid conducting sphere

20. A coaxial cylindrical Gaussian surface is used to find the electric field outside an infinitely long charged wire.

21. A. a uniformly charged sphere of radius R
Q22.9
For which of the following charge distributions would Gauss’s law not be useful for calculating the electric field?
A. a uniformly charged sphere of radius R
B. a spherical shell of radius R with charge uniformly distributed over its surface
C. a right circular cylinder of radius R and height h with charge uniformly distributed over its surface
D. an infinitely long circular cylinder of radius R with charge uniformly distributed over its surface
E. Gauss’s law would be useful for finding the electric field in all of these cases.
Answer: C

22. A cylindrical Gaussian surface is used to find the electric field of an infinite plane sheet of charge.

23. Electric field between two (large) oppositely charged parallel plates.
(Ignoring
edge effects)
Electric field between two (large) oppositely charged parallel plates.

24. “Volume charge density“: r = charge / unit volume (C/m^3)
is used to characterize the charge distribution.
The electric field of a uniformly charged INSULATING sphere.

25. +q
The solution of this problem lies in the fact that the electric field inside a conductor is zero and if we place our Gaussian surface inside the conductor (where the field is zero), the charge enclosed must be zero (+ q – q) = 0.
Find the electric charge q on surface of a hole in the charged conductor.

26. A Gaussian surface drawn inside the conducting material of which the box is made must have zero electric field on it (field inside a con-ductor is zero).  If the Gaussian surface has zero field on it, the charge enclosed must be zero per Gauss's Law. E
The E field inside a conducting box (a “Faraday cage”) in an electric field.