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Gauss’s Law
PH 203
Professor Lee Carkner
Lecture 5
Gauss
= q/0 = ∫EdA
Note that: Flux only depends on net q internal to surface
For a uniform surface and uniform q, E is the
same everywhere on surface so ∫EdA = EA
Cylinder E field is always radially outward
We want to find E a distance r away
To solve Gauss’s Law: = q/0 = ∫EdA = EA q is h
Solve for E
a)
Plane
We can again capture the flux with a cylindrical Gaussian surface
Useful for large sheet or point close to sheet
b)
Spherical Shell
Consider a spherical shell of charge of radius r and total charge q
c)
d)
Surface within Sphere What if we have total charge q,
uniformly distributed with a radius R?
e)
What if surface is inside R?
If we apply r3/R3 to the point charge formula we get,
E = (q/40R3)r
Conductors and Charge
The charges in the conductor are free to move and so will react to each other
Like charges will want to get as far away from each other as possible
No charge inside conductor
Charge Distribution How does charge distribute itself over a surface?
e.g., a sphere
No component parallel to surface, or else the charges would move
Excess charge there may spark into the air
Conductors and External
The positive charges will go to the surface “upfield” and the negative will go to the surface “downfield”
The field inside the conductor is zero The charges in the conductor cancel out the external
field A conductor shields the region inside of it
Conducting Ring
Charges
Pushed
To
Surface
No E Field
InsideField Lines
Perpendicular
to Surface
Faraday Cage
If we make the conductor hollow we can sit inside it an be unaffected by external fields
Your car is a Faraday cage and is thus a good place to be in a thunderstorm
Next Time
Read 24.1-24.6 Problems: Ch 23, P: 24, 36, 45, Ch 24, P:
2, 4
A uniform electric field of magnitude 1 N/C is pointing in the positive y direction. If the cube has sides of 1 meter, what is the flux through sides A, B, C?
A) 1, 0, 1B) 0, 0, 1C) 1, 0, 0D) 0, 0, 0E) 1, 1, 1
A
B
C
Consider three Gaussian surfaces. Surface 1 encloses a charge of +q, surface 2 encloses a charge of –q and surface 3 encloses both charges. Rank the 3 surfaces according to the flux, greatest first.
A) 1, 2, 3
B) 1, 3, 2
C) 2, 1, 3
D) 2, 3, 1
E) 3, 2, 1
+q -q
1 2
3
Rank the following Gaussian surfaces by the amount of flux that passes through them, greatest first (q is at the center of each).
A) 1, 2, 3
B) 1, 3, 2
C) 2, 1, 3
D) 3, 2, 1
E) All tie
q q
1
2
q
3
Rank the following Gaussian surfaces by the strength of the field at the surface at the point direction below q (where the numbers are).
A) 1, 2, 3
B) 1, 3, 2
C) 2, 1, 3
D) 3, 2, 1
E) All tie
q q
1
2
q
3