First Pillar: Gauss’ Law

1. There are 4 pillars that
make up the foundation
of Electricity &
Magnetism.
Gauss’Law(Electricity)
Gauss’Law(Magnetism)
Faraday’slawofinduction
Ampere’sLaw
Electricity & Magnetism
We’ll study each of
these in varying
degrees.
Maxwell’s
Equations

2. Karl Fredrick Gauss (1777-1855)
He was a contemporary of Charles Coulomb (1736-1806)
Instead of finding the field from a single charge, Gauss found the field
from a bunch of charges (charge distribution).
Why is Gauss’ Law important?
Specific
Coulomb’s Law finds a
field/charge from point charges.
+Q
General
Gauss’ Law finds a field/charge
from any charged object.
+Q
+Q
-Q
+Q
First Pillar: Gauss’ Law

3. The electric field coming through a certain area is
proportional to the charge enclosed.
Gaussian Surface
 An imaginary surface around a charge distribution
(group of charges) arbitrarily chosen for its symmetry (so
the Electric Field coming through the imaginary surface is
fairly constant through all areas of the surface).
Examples:
Point Charge Wire Strange Shape Parallel Plates
(Capacitor)
+Q
-Q -Q -Q -Q -Q +Q
+Q
-Q
+Q + + + + +
- - - - -
Sphere:
Surface Area = 4πr2
Cylinder:
Surface Area = 2πrh Strange Surface:
Calculus
Box:
Surface Area = L x W
What is Gauss’ Law?

4. Quick reminders on Electric Field Lines
1. More field lines = stronger field.
2. Field lines always come out of the surface perpendicularly.
3. Out of +, into ‒
(show the direction a + charge will move)
The electric field coming through a certain area is
proportional to the charge enclosed.
2
2. Electric Fields
3. Charge enclosed
The field is proportional to the charge inside the Gaussian Surface.
More Field Lines = Stronger Field = Stronger Charge Inside.
3
What is Gauss’ Law?

5. Gauss’ Law:
∫EdA α Q
The electric field coming through a certain area is
proportional to the charge enclosed.
How do we make this an equation? – Add a constant!
∫EdA = cQ
c = 1/εo  remember this?!?
Permitivity Constant
εo = 8.85x10-12
Nm2
/C  k = 1/4πεo = 8.99 x 109
Nm2
/C2
o
Q
EdA
ε
=∫
How much field through a certain area
Rename this to be Electric Flux (ΦE)  how much field comes through a
certain area.

6. The electric field coming through a certain area is
proportional to the charge enclosed.
o
E
Q
EdA
ε
==Φ ∫
And finally…
ΦE = Electric Flux (Field through an Area)
E = Electric Field
A = Area
q = charge in object (inside Gaussian surface)
εo = permittivity constant (8.85x 10-12
)
Gauss’ Law Summary

7. Sample Problem 1
A Van de Graaff machine with a radius of 0.25 m
has been charged up. What is the electric field
0.1 m away from the center of the sphere?
Hint: Where are all the charges? On the outside.
So how much charge is in the center? None.
0
0
===Φ
o
E EA
ε
Since all the charge is on the surface, it proves there is no field
inside a conducting surface!

8. Sample Problem 2
Find the electric field around a point charge,
Q.
Remember the area of a sphere (Gaussian
Surface in this case) is 4πr2
.
2
2
2
4
1
kforgettdon'
4
)4(
r
kQ
E
r
Q
E
Q
rE
Q
EA
oo
o
o
=
=→=
=
=
πεπε
ε
π
ε
What does this look like?
Coulomb’s electric field for point charges!
+Q

9. Sample Problem 3
A solid sphere of radius R = 40 cm has a total positive
charge of 26 μC uniformly distributed throughout its
volume. Calculate the magnitude of the electric field at
the following distances from the center of the sphere.
(a) 0 cm (b) 30 cm (c) 60 cm
0
0
===Φ
==Φ
o
E
o
E
EA
Q
EA
ε
ε
0
0
===Φ
==Φ
o
E
o
E
EA
Q
EA
ε
ε
Inside sphere Q = 0 Still inside sphere
∴Q still = 0
CNxE
x
x
E
Q
rE
Q
EA
o
o
E
/1049.6
1085.8
1026
)6.04(
)4(
5
12
6
2
2
=
=
=
==Φ
−
−
π
ε
π
ε

10. Sample Problem 3
A solid sphere of radius R = 40 cm has a total positive
charge of 26 μC uniformly distributed throughout its
volume. Calculate the magnitude of the electric field at
the following distances from the center of the sphere.
(a) 0 cm (b) 30 cm (c) 60 cm
0
0
===Φ
==Φ
o
E
o
E
EA
Q
EA
ε
ε
0
0
===Φ
==Φ
o
E
o
E
EA
Q
EA
ε
ε
Inside sphere Q = 0 Still inside sphere
∴Q still = 0
CNxE
x
x
E
Q
rE
Q
EA
o
o
E
/1049.6
1085.8
1026
)6.04(
)4(
5
12
6
2
2
=
=
=
==Φ
−
−
π
ε
π
ε