# 4 Gauss s Law by hx6s897

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```									AP Physics C
Mrs. Coyle
Electric Flux, F:
The number of electric (flux) field lines which pass
through a given cross-sectional area A.
Which has the largest
electric flux? Does flux depend on area?

Which has the largest
electric flux? Does flux depend on
direction?

The area vector A is perpendicular
to the surface A and has a
magnitude equal to the area A.
Electric Flux

F= E dA =  EdA cos q

 Flux is a dot product and therefore a scalar
 Units:
N 2
[F ]=      m
C
[E]=     N
C
[A]= area m2
q: angle formed between E
and the normal to the surface.
When E is perpendicular to the loop?

F= EA

Why?
What if you have a closed surface?
For a closed surface:
Gauss’s Law: the total electric flux
through a closed surface is
proportional to the enclosed
charge.
qin
F=        surface
E dA =
0
Permittivity of free space:
ε0 = 8.8542 x 10-12 C2 / (N m2)
Remember:
Coulomb’s Law constant, k

1
k=           = 9 x10 Nm / C
9   2      2

4 0
Note: the dot product

qin
F=      surface
EdA cos q =
0
Notes:
 The “Gaussian” surface can have any shape. You choose
the surface.
 The flux is positive if more electric field lines go out of
the surface and negative if more go in.
 The flux from a particular charge qin is the same
regardless of the shape or size of the surface.
 If the surface contains no charge, the flux through it is
zero. That means that every field line that enters the
surface will also exit the surface.
More Notes on Gauss’s Law
 The normal area vector is taken to point from the
inside of the closed surface out.

 Gauss’s Law is also known as Maxwell’s first equation.

 Gauss’s Law is second way used to find the electric
field. What is the first way you learned?
Problem 1 The Electric Field due to
a point charge. Prove E=kq/r2
What do you think would be
a different Gaussian surface?
Problem 2: Show E= σ /(2 ε0) for a
rectangular metal slab positively
charged with a uniform surface
charge density σ.
Problem 3: Field Due to a
Spherically Symmetric Charge
Distribution of an Insulator
a)Outside the sphere( r>a)
qin
F E =  E  dA =  EdA =
εo
EA = E 4πr 2
Q          Q
E=           = ke 2
4πεo r 2
r
Problem 3: Field Due to a Spherically
Symmetric Charge Distribution of an Insulator
b)Inside the sphere( r<a)
 qin < Q and since the
volume charge density is
uniform:
Q       qin            r3
=         qin = Q 3
4 a 3
4 r 3
a
qin
F E =  E  dA =  EdA =
εo
qin        Q
E=          = ke 3 r
4πεo r 2
a
Problem 3: Field Due to a
Spherically Symmetric Charge
Distribution (Insulator)
Outside the sphere
Q
E = ke 2
r

Inside the sphere
Q
E = ke 3 r
a
Problem 4: Field Due to a Thin
Spherical Shell
 For r > a, the enclosed charge is Q and E = keQ / r2
 For r < a, the charge inside the surface is 0 and E = 0
Problem 5: Field due to a Line of
Charge
Problem 5: Field Due to a Line
of Charge
End view

 The flux through the
ends of the gaussian
cylinder is 0. Why?
Problem 5: Field Due to a Line
of Charge
qin
F E =  E  dA =  EdA =
εo
λ
E  2πr  =
εo
λ          λ
E=        = 2ke
2πεo r       r

```
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