1

Chapter 22

2

Flux

Number of objects passing through a surface

3

Electric Flux,

is proportional to the number of electric field lines passing through a surface

Assumes that the surface is perpendicular to the lines If not, then we use a cosine of the angle

between them to get the components that are parallel

Mathematically:

AEAE

cos

4

Simple Cases

E

=EA

E

A

A

=0

=EAcos

E

A

E cos

5

From to

A represents a sum over a large a collection of objects

Integration is also a sum over a collection of infinitesimally small objects, in our case, small areas, dA

So

AdE

ytechnicallthendxdyrepresentsAdSince

AdE

,,

6

Gauss’s Law

The field lines emitted by a charge are proportional to the size of the charge.

Therefore, the electric field must be proportional to the size of the charge

In order to count the field lines, we must enclose the charges in some geometrical surface (one that we choose)

7

Mathematically

q0q

0enclosedq

AdE

Charge enclosed within bounding limits of this closed surface integral

8

Fluxes, Fluxes, Fluxes

9

3 Shapes

SphereCylinderPillbox

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Sphere

When to use: around spherical objects (duh!) and point charges Hey! What if an object is

not one of these objects? Closed surface integral

yields:

r is the radius of the geometrical object that you are creating

)4( 2rEAdE

11

Sphere Example

3

0

5

0

2

2

532

3

3

4)4(

)4(

3

4)

3

4)((

:

rA

E

rArE

rEAdE

rArArq

Vq

spheretheInside

enclosed

enclosed

What if you had a sphere of radius, b, which contained a material whose charge density depend on the radius, for example, =Ar2

where A is a constant with appropriate units?

2

5

0

5

0

2

2

532

3

3

4)4(

)4(

3

4)

3

4)((

:

r

bA

E

bArE

rEAdE

bAbAbq

Vq

spheretheOutside

enclosed

total

At r=b, both of these expressions should be equal

12

Cylinder

When to use: around cylindrical objects and line charges

Closed surface integral yields:

r is the radius of the geometrical object that you are creating and L is the length of the cylinder

)2( rLEAdE L

13

Cylinder Example

What if you had an infinitely long line of charge with a linear charge density, ?

rr

Eorr

E

LrLE

Lq

rLEAdE

enclosed

ˆ22

)2(

)2(

00

0

14

Pillbox

When to use: around flat surfaces and sheets of charge

Closed surface integral yields:

A is the area of the pillbox

EAAdE

15

Charge Isolated Conductor in Electrostatic Equilibrium

If excess charge is placed on an isolated conductor, the charge resides on the surface. Why? If there is an E-field inside the conductor then it would exert

forces on the free electrons which would then be in motion. This is NOT electrostatic.

Therefore, if there is no E-field inside, then, by Gauss’s Law, the charge enclosed inside must be zero If the charges are not on the outside, you are only left with the

surface A caveat to this is that E-field lines must be perpendicular to

the surface else free charges would move.

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Electric field on an infinitely large sheet of charge

A

qLet

+++++++++++++++++++++++++++++++++++++++++++++

AE

E

nEorE

AEA

So

Aq

EAAEEAAdE

enclosed

ˆ22

2

2))((

00

0

17

Electric field on a conducting sheet

A

qLet

+++++++++++++++++++++++++++++++++++++++++++++

AE

nEorE

AEA

So

Aq

EAEAAdE

enclosed

ˆ

0

00

0

So a conductor has 2x the electric field strength as the infinite sheet of charge

18

A differential view of Gauss’s Law

Recall the Divergence of a field of vectors vv

)(Div

Div=0

Div=+largeHow much the vector diverges around a given point

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Divergence Theorem (aka Gauss’s Thm or Green’s Thm)

Advdv Suspiciously like LHS of Gauss’s Law

Sum of the faucets in a volume = Sum of the water going thru the surface

A place of high divergence is like a faucet Bounded surface of some region

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Div(E)

0

0

0

E

so

dqandq

dE

qAdE

enclosedenclosed

enclosed

So how the E-field spreads out from a point depends on the amount of charge density at that point