Gauss

Electric flux

When electric field lines cut through an area, we say there is an electric flux.

To define it precisely we define an area vector for a planar area that is perpendicular

to the area and having a magnitude equal to the area.







ˆ

nA





E

A





A  An A

ˆ

Then for a uniform field the electric flux through a planar area is

 

  E A

If the field is not uniform or if the area is contorted, we have to find the flux of

many little pieces that are small enough to be treated as planar and over which

the field is uniform.



dA 

dA may change direction, but for a closed surface

 it is defined to point outward, normal to

E surface.



E may change direction or magnitude





dA

 



 

E  

area

E  dA

d E  E  dA

Exercise: Find the flux of a uniform field through a cylinder with axis parallel to

the field. 

dA 

dA











E

 

 sheath  0 E  dA  0

 

 Rcap  EA E  dA  EdA

 

 Lcap   EA E  dA   EdA



 E   sheath   Lcap   Rcap

 E  0  EA  EA  0

Flux due to a field created by a point charge outside the closed surface









Flux positive



Flux negative 2

Area increases as R

1

Field decreases as

R2

Magnitude of flux stays same









Charges outside a closed area

contribute a net zero flux

How can we get a net flux through the surface??

Only if charges are contained within a closed surface will there be a net flux

through it.



















No flux cancellations for a

charge inside a closed

surface



The more charge inside the greater the flux.

It is reasonable to expect direct proportion between flux and total charge inside.



 E  Qin

Gauss’s Law   Qin



Gaussian

E  dA 

0



Here the integration is done over a closed surface, a so called “Gaussian

surface”.

This surface is often not a “real” physical surface, but a mathematical

device introduced to apply the law.

You need to become proficient at drawing Gaussian surfaces that will

allow you to extract useful information about a charge distribution.



Simple stuff

2C

3C

What is the electric flux through the

2C

dotted surface?





Draw a Gaussian surface with zero flux.

5C

Is the field zero at such a surface?

4C

Conductors in electrostatics

Conductors have mobile charges, usually electrons, that can freely move

within the material should they experience a force.

What if a conductor is placed into a static electric field.

after

before 

E  

 







E0 













•Electrostatic equilibrium is quickly established; charges cease to move.



•Mobile charges rearrange themselves at the surface to cancel the original field.



•Field lines enter and leave the conductor at right angles to the surface, otherwise

charges at the surface would move.

Charge on a Conductor



A conductor can be neutral or it can have an excess charge.

In either case, Gauss’s law and the properties of a conductor say very

specific things about how the charge can be distributed over the conductor.

No charge can lie within these surfaces

Inside the conductor the field is zero.

Thus the flux through any surface we

could imagine drawing must also be

zero.



E0









Excess charge resides on surface of a conductor

Field near surface of a conductor

We can use Gauss’s law to determine the field near the surface of a conductor if

we take advantage of three related properties of the conductor:

1) Excess charge will be distributed on the surface. In general the

surface density σ will vary over the surface.

2) The field is zero inside the conductor.

3) The field exits (or enters) the surface at right angles to the surface,

i.e., parallel to the local area vector.



We can draw a teensy, tiny Gaussian cylinder with its axis parallel to the field at

the surface and with one cap inside the conductor.



Qin

 E  EA 

0



A 

EA  E

0



E Qin  A

0

Symmetry and Gauss’s Law

When a charge distribution has a high degree of symmetry, choosing a

Gaussian surface that respects the symmetry will allow you to solve for the

electric field.

Symbolically this works like:

  Qin

 E  dA 

Gaussian

0

It’s the LAW, obey it!!





Qin

 EdA 

Gaussian

0

A properly chosen surface will always have the

field and area element parallel. Parts of the area

my have perpendicular elements that do not

contribute.



Qin 

E  dA  

Gaussian 0

E will have a constant magnitude over a

properly chosen surface so it factors out of

integral.



Qin

E With E factored out of integral, it just evaluates

 0 AGaussian to the area of the Gaussian surface.

Spherical Symmetry

Examples: spherical conductors, non-conductors, concentric combinations of

these.

Example: Spherical non-conductor radius R carrying uniformly distributed Q.

rR rR





E 4 r 2

 Q



E 4 r 2

 

Qin

0 0

r

4 3

E

Q r

4  0 r 2 r 

E 4 r 2  

Q 3

 0 4  R3

Behaves like a E R 3

point charge Qr

E

E 4 0 R 3



E

4

r

R 2R



E 

dA





 

E dA





r



r



Cylindrical Symmetry E 

dA

r



R r





rR rR



E 2 Lr  



  R2L  L



  r2L 

0 E 2 Lr  

0

 R 2 

E  r

2 0 r 2 0 r E E

E

2 0

2

r

R 2R

Planar Symmetry

Examples: large non-conducting slabs; multiple slabs can be accommodated

with the superposition principle.

Conducting sheets can also be handled, but one needs to know the final charge

distribution on the surface; superposition of individual solutions will not usually

work. Why?



Symmetry dictates that the field point perpendicular to

 surface

 



  



  



  



 







 

Charge redistributes



 

  on the two

conductors

 



 

Example: infinite non-conducting sheet, thickness t, charge density ρ







E 

E

A





t  t 



At Side view

 2 EA 

E 0 

E

t 

E 

2 0 2 0

Inside the slab



E

t

x





E Side view





A2 x 

2 EA 

0

x  2x

E 

 0 2 0 t