There are 4 pillars that make up the foundation

of Electricity & Magnetism.

Gau

ss’

Law

(E

lect

rici

ty)

Gau

ss’

Law

(M

agne

tism

)

Far

aday

’s la

w o

f in

duct

ion

Am

pere

’s L

aw

Electricity & Magnetism

We’ll study each of these in varying

degrees.

Maxwell’s Equations

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?SpecificCoulomb’s Law finds a field/charge from point charges.

+Q

GeneralGauss’ Law finds a field/charge from any charged object.

+Q

+Q

-Q

+Q

First Pillar: Gauss’ Law

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).

Exa

mpl

es:

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?

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 enclosedThe field is proportional to the charge inside the Gaussian Surface.

More Field Lines = Stronger Field = Stronger Charge Inside.

3

What is Gauss’ Law?

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 = cQc = 1/εo remember this?!?

Permitivity Constantεo = 8.85x10-12 Nm2/C k = 1/4πεo = 8.99 x 109 Nm2/C2

o

QEdA

How much field through a certain area

Rename this to be Electric Flux (ΦE) how much field comes through a certain area.

The electric field coming through a certain area is proportional to the charge enclosed.

oE

QEdA

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

Sample Problem 1A 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.

00

o

E EA

Since all the charge is on the surface, it proves there is no field inside a conducting surface!

Sample Problem 2Find 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 k forget t don'

4

)4(

r

kQE

r

QE

QrE

QEA

oo

o

o

What does this look like?

Coulomb’s electric field for point charges!

+Q

Sample Problem 3A 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

00

oE

oE

EA

QEA

00

oE

oE

EA

QEA

Inside sphere Q = 0 Still inside sphere Q still = 0

CNxE

x

xE

QrE

QEA

o

oE

/1049.6

1085.8

1026)6.04(

)4(

5

12

62

2