Gauss’ Law. Class Objectives Introduce the idea of the Gauss’ law as another method to calculate...

Gauss’ Law

Class Objectives

• Introduce the idea of the Gauss’ law as another method to calculate the electric field.

• Understand that the previous method of calculating the electric field strength does not consider symmetry.

• Consider the different types of symmetry.

Class Objectives

• Introduce the idea of the Gaussian surface.

• Define the properties of Gaussian surfaces.

• Show how to choose a Gaussian surface.

• Show how Gaussian surfaces can be used to take advantage of symmetry.

Class Objectives

• Show that the results of Gauss’s law are the same as the standard results but quicker and easier!

Student Objectives

• Be able to use Gauss’ law to calculate the electric field for various objects.

• Be able to determine the type of symmetry in the problem and hence the type of Gaussian surface to be used.

Gauss’ Law

• Symmetry in problems arise naturally.

Gauss’ Law

• Symmetry in problems arise naturally.

• Gauss’ law is an alternate method to coulomb’s law where there is symmetry.

Gauss’ Law

• Outline of Gauss’ Law1. Central idea is a hypothetical closed

surface called a Gaussian surface.2. The surface can be chosen any shape to

complement the symmetry. Eg cylinder or sphere.

3. Gauss’ law relates the electric field at a point on the closed surface to the net charge enclosed by the surface.

Gauss’ Law

• The amount of charge enclosed is measured by the amount of flux passing through the surface.

Gauss’ Law

• Example of a spherical Gaussian closed surface.

Electric flux

• The electric flux is a measure of the number of field lines passing an area.

• The flux is represented by the symbol .

Electric flux

• The electric flux is a measure of the number of field lines passing an area.

• The flux is represented by the symbol .

• Assuming an arbitrary Gaussian surface in a field.

Electric flux

• If we divide the surface into small squares of area then the curvature of the surface can be ignored if the squares are taken small enough.

dA

Electric flux

• If we divide the surface into small squares of area then the curvature of the surface can be ignored if the squares are taken small enough.

• By convention we assign an area vector .

• The direction of which is perpendicular to the surface of the element directed away from its interior.

dA

Ad

Electric flux

• Because can be made arbitrarily small,

may be taken as constant for a given square.

• The flux for a Gaussian surface is given as

dA E

AdE

Electric flux

• Because can be made arbitrarily small,

may be taken as constant for a given square.

• The flux for a Gaussian surface is given as

• The circle indicates that the integral is to be taken over the entire closed surface.

dA E

AdE

Electric flux

• The follow diagrams gives three cases for the relative orientation of the field and dA. The angle measured is the smallest angle between the vectors.

Electric flux

• Note that all of the elements on the surface do not contribution to the flux.

Gauss’ Law

• Guass’ Law is often written as

• Where is the sum of the enclosed charges.

0encqAdE

encq

Gauss’ Law

Choosing a Gaussian surface

Gauss’ Law

• Choosing a Gaussian surface is not hard but subtle.

Gauss’ Law

• Choosing a Gaussian surface is not hard but subtle.

• A simple surface must be chosen to take advantage of symmetry.

• Consider the following rules as a guide.

Gauss’ Law

1. Choose the surface perpendicular to the field so the E and dA are parallel.

Gauss’ Law


2. Choose so that points on the surface are equal distance away from the charge so the E doesn’t vary.

Gauss’ Law


2. Choose so that points on the surface are equal distance away from the charge so the E doesn’t vary.

3. If this is not possible (1 and 2) choose a surface such that the dot product is zero. Ie. They are perpendicular.

Gauss’ Law and Coulomb’s Law

• Consider a point charge q.


• Consider a point charge q.

• For symmetry we use a spherical Gaussian surface.


• the chosen Gaussian surface holds true for the first two guidelines.


• the chosen Gaussian surface holds true for the first two guidelines.

1. E and dA are parallel.

2. Points on the surface are equal distance away from the charge so the E doesn’t vary.

dA

Gauss’ Law

• From Gauss’ Law

0encqAdE

Gauss’ Law


• Since the two vectors are parallel.

0encqAdE

0encqdAE

Gauss’ Law


• Since the two vectors are parallel.

• Also we note that the enclosed charge is simply q. So we can write that,

0encqAdE

0encqdAE

0q

dAE

Gauss’ Law

• The area of the closed surface is the area of a circle. So that

0

24

qrE

Gauss’ Law

• The area of the closed surface is the area of a circle. So that

• Therefore

0

24

qrE

204 r

qE

Conductor in an Electric Field


• For the cases to investigated, we will consider the cases where the conductor is in equilibrium (electrostatics).


• For the cases to investigated, we will consider the cases where the conductor is in equilibrium (electrostatics).

• For a conductor in equilibrium we have following conditions:


1. The charge exists entirely on the surface of conductor(no charge is found within the body of the conductor).



2. The electric field within the conductor is zero.



2. The electric field within the conductor is zero. The charge distributes itself so as to get as far from each other as possible.


3. The external electric field is perpendicular to the surface of the conductor.


3. The external electric field is perpendicular to the surface of the conductor. If not it would cause the charges to move along the surface of the conductor.


3. The external electric field is perpendicular to the surface of the conductor. If not it would cause the charges to move along the surface of the conductor.

Note: unless the conductor is spherical, the charge does not distribute itself uniformly.


• When a conductor is place in an external field, the mobile electrons experience a force pushing them in the opposite direction to the field

E

e


• Thus making the top of the conductor positive and bottom negative.

E

e_ _ _ _ _ _ _ _ _ _ _

+ + + + + + + + + +


• Thus making the top of the conductor positive and bottom negative. This sets up an internal electric which grows in strength as more electrons move.

E

e_ _ _ _ _ _ _ _ _ _ _

+ + + + + + + + + +

Eint

Electric field causedby pd between the topand bottom of theconductor


• Electrons move until the two fields have the same magnitude. Hence the net electric field E=0.

E

e_ _ _ _ _ _ _ _ _ _ _

+ + + + + + + + + +

Eint


• Electrons move until the two fields have the same magnitude. Hence the net electric field E=0. The magnitude of the charge on the top & bottom of the conductor are the same. E

e_ _ _ _ _ _ _ _ _ _ _

+ + + + + + + + + +

Eint

E

Eint


• Let us now consider the electric field just outside the surface of a conductor.


• Let us now consider the electric field just outside the surface of a conductor.

E


• Let us now consider the electric field just outside the surface of a conductor. We use a cylindrical Gaussian surface.

E

A


• Assume the conductor has charge per unit area and total charge Q with uniform charge distribution.


• Recall Gauss’ Law:

0encqAdE



• We note that

0encqAdE

Aqenc



• We note that

• We can therefore write that:

0encqAdE

Aqenc

0A

AdE


• Integrating as before we get that:

0A

EA



• Hence

0A

EA

0

E



• Hence

• The electric field.

0A

EA

0

E


Cylindrical rod


• Consider an infinitely long cylindrical rod with uniform linear charge density λ. What is the electric field a distance r from its axis.

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

r


• We choose a cylindrical Gaussian surface.

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

r

EE

h

• Writing Gauss’ law:

0encqAdE


• Hence,

0encqAdE

0encq

EA


• Hence,

0encqAdE

0encq

EA0

hEA


• Hence,

• The area enclosed is that of a cylinder.

0encqAdE

0encq

EA0

hEA

0

2 h

rhE


• Hence,

• The area enclosed is that of a cylinder.

0encqAdE

0encq

EA0

hEA

0

2 h

rhE r

E02

• The electric field due to a line of charge is

rE

02

Charged sphere

• We previously looked at the properties of a conductor.

• For a spherical conductor of charge +Q with uniform charge density, what is the electric field outside and inside the conductor?

+Q

• We know the charge resides on the surface.

• Consider a Gaussian surface inside the sphere.



• No charge is enclosed by the surface.



• No charge is enclosed by the surface.

• Therefore E = 0 as expected.

• Now consider a Gaussian surface outside the sphere.


• The total charge enclosed is Q.



• Therefore from Gauss’ Law,

0encqAdE

0Q

AdE



• Therefore from Gauss’ Law,

0encqAdE

0Q

AdE

0Q

EA 0

24

QrE

• Hence,

• As expected!!

204 r

QE

Uniform Charge non-conducting Sphere

• Consider a solid non-conducting sphere of radius R with uniform charge distribution ρ. What is the electric field outside and inside the sphere?

r

R

• The total charge is given by:

3

4 3rQ


• Case r > R (outside the sphere)

3

4 3rQ


• Case r > R (outside the sphere)• Gauss’ Law:

3

4 3rQ

0encqAdE


• Case r > R (outside the sphere)• Gauss’ Law:• The charge enclosed is the total charge.

3

4 3rQ

0encqAdE

0Q

AdE


• Case r > R (outside the sphere)• Gauss’ Law:• The charge enclosed is the total charge.

• E is constant on the surface, therefore

3

4 3rQ

0encqAdE

0Q

AdE

0Q

dAE

• Integrating over the surface we have, 0

24

QrE

• Integrating over the surface we have, 0

24

QrE

024 rQ

E

• Integrating over the surface we have,

• That is the sphere acts like a point charge (as expected).

0

24

QrE

024 rQ

E

• Case r < R (inside the sphere) r

R

• Case r < R (inside the sphere)• Again we use Gauss’ Law.• However the total charge is not enclosed.

r

R

• Case r < R (inside the sphere)• Again we use Gauss’ Law.• However the total charge is not enclosed.• The total charge enclosed is

r

R

3

4 3rQenc

• Case r < R (inside the sphere)• Again we use Gauss’ Law.• However the total charge is not enclosed.• The total charge enclosed is• Therefore,

r

R

3

4 3rQenc

0

3

3

4

r

AdE


r

R

3

4 3rQenc

0

3

3

4

r

AdE 0

32

34

4

r

rE


r

R

3

4 3rQenc

0

3

3

4

r

AdE 0

32

34

4

r

rE

03r

E

Gauss’ Law. Class Objectives Introduce the idea of the Gauss’ law as another method to calculate...

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