Chapter 21: Gauss

1
Chapter 21 Gausss Law

• Gausss law introduction

• The total electric flux through a closed surface
  is
• equal to the total (net) electric charge
  inside the surface
• divided by e0

• Gausss law is equivalent to Coulombs law

2

• Gausss law introduction (contd)

• Consider a distribution of charge

• Surround it with an imaginary surface that
  encloses the charge
• Look at the electric field at various points on
  this imaginary surface

• To find out charge distribution inside the
• imaginary surface, we need to measure
• electric fields especially on the surface

• To do that place a test charge of a known
• charge amount and measure the electric
• force

q
imaginary surface


test charge
3

• Charge and Electric Flux

• Electric fields by different charges

outward flux
outward flux



inward flux
-
-
inward flux
-
4

• Charge and Electric Flux

• Electric flux

outward flux
outward flux



When the distance to the surface doubled, the
area of the surface quadrupled and the electric
field becomes ¼.

5

• Charge and Electric Flux

• Definition of electric flux

For any point on a small area of a surface, take
the product of the average component of E
perpendicular to the surface and the area. Then
the sum of this quantity over the surface is
the net electric flux

• A qualitative statement of Gausss law

• Whether there is a net outward or inward
  electric flux through a closed
• surface depends on the sign of the enclosed
  charge.

• Charges outside the surface do not give a net
  electric flux through the
• surface.

• The net electric flux is directly proportional
  to the net amount of charge
• enclosed within the surface but is otherwise
  independent of the size of
• the closed surface.

6

• Calculating Electric Flux

• Analogy between electric flux and field of
  velocity vector

A good analogy between the electric flux and the
field of velocity vector in a flowing fluid can
be found.
A (area)
volume flow rate
velocity vector (flow speed)
A vector area that defines the plane of the
area, perpendicular to the plane
volume flow rate
f
7

• Calculating Electric Flux

• Analogy between electric flux and field of
  velocity vector

volume flow rate
A (area)
electric flux
Electric field vector
A vector area that defines the plane of the
area, perpendicular to the plane
E field is perpendicular to this plane
electric flux
f
8

• Calculating Electric Flux

• A small area element and flux

• Total flux for an area

• Example Electric flux through a disk

r 0.10 m
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• Calculating Electric Flux

• Example Electric flux through a cube

L
10

• Calculating Electric Flux

• Example Electric flux through a sphere

r0.20 m

q
q3.0 mC A4pr2
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• Gausss Law

• Preview

The total electric flux through any closed
surface (a surface enclosing a definite volume)
is proportional to the total (net) electric
charge inside the surface.

• Case 1 Field of a single positive charge q

A sphere with rR
at rR
rR

The flux is independent of the radius R of the
surface.
12

• Gausss Law

• Case 1 Field of a single positive charge q
  (contd)

Every field line that passes through the smaller
sphere also passes through the larger sphere
The total flux through each sphere is the same
rR
The same is true for any portion of its surface
such as dA

r2R
This is true for a surface of any shape or
size provided it is a closed surface enclosing
the charge
13

• Gausss Law

• Case 2 Field of a single positive charge
  (general surface)

surface perpendicular to

14

• Gausss Law

• Case 3 An closed surface without any charge
  inside

Electric field lines that go in come
out. Electric field lines can begin or end
inside a region of space only when there is
charge in that region.

• Gausss law

The total electric flux through a closed surface
is equal to the total (net) electric charge
inside the surface divided by e0
15

• Applications of Gausss Law

• Introduction

• The charge distribution the field

• The symmetry can simplify the procedure of
  application

• Electric field by a charge distribution on a
  conductor

• When excess charge is placed on a solid
  conductor and is at rest,
• it resides entirely on the surface, not in the
  interior of the material
• (excess charge charge other than the ions and
  free electrons that
• make up the material conductor

A Gaussian surface inside conductor
Charges on surface
Conductor
16

• Applications of Gausss Law

• Electric field by a charge distribution on a
  conductor (contd)

A Gaussian surface inside conductor
Charges on surface
Conductor
E at every point in the interior of a conducting
material is zero in an electrostatic situation
(all charges are at rest). If E were non-zero,
then the charges would move

• Draw a Gaussian surface inside of the conductor
• E0 everywhere on this surface (inside
  conductor)
• The net charge inside the surface is zero
• There can be no excess charge at any point
  within a solid conductor
• Any excess charge must reside on the conductors
  surface
• E on the surface is perpendicular to the surface

Gausss law
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• Applications of Gausss Law

• Example Field of a charged conducting sphere
  with q

Gaussian surface
E



Draw a Gaussian surface outside the sphere


R



r
2R
3R
R
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• Applications of Gausss Law

• Example Field of a line charge

Gaussian surface chosen according to symmetry
line charge density
19

• Applications of Gausss Law

• Example Field of an infinite plane sheet of
  charge

two end surfaces
area A

area A




Gaussian surface
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• Applications of Gausss Law

• Example Field between oppositely charged
  parallel
• conducting plane

plate 2
plate 1
Solution 1
-

b
-

c
-
a

outward flux
-

S1
S4
-

-

-

S2
S3
-

inward flux
-

Solution 2
At Point a
No electric flux on these surfaces
b
c
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• Applications of Gausss Law

• Example Field of a uniformly charged sphere

Gaussian surface








rR




R
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• Applications of Gausss Law

• Example Field of a hollow charged
• (uniformly on its surface) sphere

r0.300 m
R0.250 m
Gaussian surface
Hollow charged sphere
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• Charges on Conductors

• Case 1 charge on a solid conductor resides
  entirely on
• its outer surface in an
  electrostatic situation

The electric field at every point within a
conductor is zero and any excess charge on a
solid conductor is located entirely on its
surface.

• Case 2 charge on a conductor with a cavity

If there is no charge within the cavity, the
net charge on the surface of the cavity is zero.








Gauss surface


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• Charges on Conductors

• Case 3 charge on a conductor with a cavity and
  a charge q
• inside the cavity

• The conductor is uncharged and insulated from
• charge q.

-

• The total charge inside the Gauss surface should
• be zero from Gausss law and E0 on this
  surface.
• Therefore there must be a charge q
  distributed
• on the surface of the cavity.

-
-

-

-

-
-

• The similar argument can be used for the case
• where the conductor originally had a charge
  qC.
• In this case the total charge on the outer
  surface
• must be qqC after charge q is inserted in
  cavity.

Gauss surface
25

• Charges on Conductors

• Faradays ice pail experiment

charged conducting ball
conductor
(1) Faraday started with a neutral metal
ice pail (metal bucket) and an uncharged
electroscope. (2) He then suspended a
positively charged metal ball into the ice pail,
being careful not to touch the sides
of the pail. The leaves of the electroscope
diverged. Moreover, their degree of
divergence was independent of the metal ball's
exact location. Only when the metal
ball was completely withdrawn did the leaves
collapse back to their original position.
26

• Charges on Conductors

• Faradays ice pail experiment (contd)

charged conducting ball
conductor
(3) Faraday noticed that if the metal
ball was allowed to contact the inside surface
of the ice pail, the leaves of the
electroscope remained diverged (4)
Afterwards, when he completely removed the ball
from the inside of the ice pail, the
leaves remained diverged. However, the metal
ball was no longer charged. Since
the leaves of the electroscope that was attached
to the OUTSIDE of the pail did not
move when the ball touched the inside of the
pail, he concluded that the inner surface
had just enough charge to neutralize
the ball.
27

• Charges on Conductors

• Field at the surface of a conductor

• The electric field just outside a conductor has
• magnitude ? /?0 and is directed
  perpendicular to
• the surface.
• Draw a small pill box that extends
• into the conductor. Since there is
• no field inside, all the flux comes
• out through the top.
• EAq/?0 ?A/ ?0,
• E ? / ?0

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• Exercises

• Exercise 1

This is the same as the field due to a point
charge with charge 2Q
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• Exercises

• Exercise 1 (contd)

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• Exercises

• Exercise 2 A sphere and a shell of conductor

Q2-3Q1

• From Gausss law there can be no net charge
  inside the
• conductor, and the charge must reside on the
  outside
• surface of the sphere

Q2
Q1
R1

• There can be no net charge inside the conductor.
• Therefore the inner surface of the shell must
  carry a net
• charge of Q1 , and the outer surface must
  carry the
• charge Q1Q2 so that the net charge on the
  shell equals
• Q2 . These charges are uniformly distributed.

R2
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• Exercises

• Exercise 2 A sphere and a shell of conductor
  (contd)

Q2-3Q1
Q2
Q1
R1
R2
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• Exercises

• Exercise 3 Cylinder

An infinite line of charge passes directly
through the middle of a hallow charged conducting
infinite cylindrical shell of radius R. Lets
focus on a segment of the cylindrical shell of
length h. The line charge has a linear charge
density l, and the cylindrical shell has a net
surface charge density of stotal.
stotal
R
l
sinner
souter
h
33

• Exercises

• Exercise 3 Cylinder (contd)

The electric field inside the cylindrical shell
is zero. Therefore if we choose as a Gaussian
surface a cylinder, which lies inside the
cylindrical shell, the net charge enclosed is
zero. There is a surface charge density on the
inside wall of the cylinder to balance out the
charge along the line.
stotal
R
l
sinner
souter
h
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• Exercises

• Exercise 3 Cylinder (contd)

• The total charge on the enclosed portion (length
  h) of the line charge

• The charge on the inner surface of the
  conducting cylinder shell

stotal
R
l
sinner
souter
h
35

• Exercises

• Exercise 3 Cylinder (contd)

• The net charge density on the cylinder

• The outer charge density

stotal
R
l
sinner
souter
h
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• Exercises

• Exercise 3 Cylinder (contd)

• Draw a Gaussian surface surrounding the line
  charge of radius r (lt R)

stotal
R
l
sinner
souter
h
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• Exercises

• Exercise 3 Cylinder (contd)

• Draw a Gaussian surface surrounding the line
  charge of radius r (gtR)

• Net charge enclosed on the line

Net charge enclosed on the shell
stotal
R
l
sinner
souter
h