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Gausss Law

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Title: Gausss Law


1
Gausss Law Chapter 21 Summary Sheet 2
2
In this chapter youll learn
  • To represent electric fields using field-line
    diagrams
  • To explain Gausss law and how it relates to
    Coulombs law
  • To calculate the electric fields for symmetric
    charge distributions using Gausss law (EASY!)
  • To describe the behavior of charge on conductors
    in electrostatic equilibrium

3
21.1 Electric Field Lines
NB For electric field at P draw tail of vector at
point P
  • We can draw a vector at each point around a
    charged object -
  • direction of E is tangent to field line (in same
    direction as F)
  • E is larger where field lines are closer
    together
  • electric field lines extend away from positive
    charge (where they originate) and towards
    negative charge (where they terminate)

4
  • Electric field lines provide a convenient and
    insightful way to represent electric fields.
  • A field line is a curve whose direction at each
    point is the direction of the electric field at
    that point.
  • The spacing of field lines describes the
    magnitude of the field.
  • Where lines are closer, the field is stronger.

Vector and field-line diagrams of a point-charge
field
5
The field lines for two equal positive charges
The field lines for two charges equal in
magnitude but opposite in sign an electric
dipole
NB the electric field vector at a point is
tangent to the field line through the point
6
Field lines for simple charge distributions
There are field lines everywhere, so every
charge distribution has infinitely many field
lines.
  • In drawing field-line diagrams, we associate a
    certain finite number of field lines with a
    charge of a given magnitude.
  • In the diagrams shown, 8 lines are associated
    with a charge of magnitude q.
  • Note that field lines of static charge
    distributions always begin and end on charges, or
    extend to infinity.

7
Summary of electric field lines
8
Two conducting spheres what is the relative sign
and magnitude of the charges on the two spheres?
Large sphere 11 field lines leaving and 3
entering, net 8 leaving Small sphere 8
leaving Spheres have equal positive charge
Charge on small sphere creates an intense
electric field at nearby surface of large sphere,
where negative charge accumulates (3 entering
field lines).
9
Gausss Law A new look at Coulombs Law Flux
Flux of an Electric Field Gauss Law Gauss Law
and Coulombs Law
10
A new look at Coulombs Law
  • A new formulation of Coulombs Law was derived by
    Gauss (1777-1855).
  • It can be used to take advantage of symmetry.
  • For electrostatics it is equivalent to Coulombs
    Law. We choose which to use depending on the
    problem at hand.
  • Two central features are
  • a hypothetical closed surface a Gaussian
    surface usually one that mimics the symmetry of
    the problem, and
  • flux of a vector field through a surface

Gauss Law relates the electric fields at points
on a (closed) Gaussian surface and the net charge
enclosed by that surface
11
A surface or arbitrary shape enclosing an
electric dipole. As long as the surface encloses
both charges, the number of lines penetrating the
surface from inside is exactly equal to the
number of lines penetrating the surface from the
outside, no matter where the surface is drawn
12
A surface of arbitrary shape enclosing charges
2q and -q. Either the field lines that end on q
do not pass through the surface or they penetrate
it from the inside the same number of times as
from the outside. The net number that exit is
the same as that for a single charge of q, the
net charge enclosed by the surface.
The net number of lines out of any surface
enclosing the charges is proportional to the net
charge enclosed by the surface. Gauss Law
(qualitative)
13
Thats it! Gauss Law in words and pictures
14
21.2 Electric flux
  • Electric flux quantifies the notion number of
    field lines crossing a surface.
  • The electric flux ? through a flat surface in a
    uniform electric field depends on the field
    strength E, the surface area A, and the angle ?
    between the field and the normal to the surface.
  • Mathematically, the flux is given by
  • Here is a vector whose magnitude is the
    surface area A and whose orientation is normal to
    the surface.

15
Electric flux with curved surfaces and nonuniform
fields
  • When the surface is curved or the field is
    nonuniform, we calculate the flux by dividing the
    surface into small patches , so small that
    each patch is essentially flat and the field is
    essentially uniform over each.
  • We then sum the fluxes over
    each patch.
  • In the limit of infinitely manyinfinitesimally
    small patches,the sum becomes asurface integral

16
Flux of an electric field
Here we have an arbitrary (asymmetric) Gaussian
surface immersed in a non-uniform electric field.
The surface has been divided up into small
squares each of area ?A, small enough to be
considered flat. We represent each element of
area with a vector area ?A and magnitude ?A.
Each vector ?A is perpendicular to the Gaussian
surface and directed outwards
17
Electric field E may be assumed to be constant
over any given square Vectors ?A and E for each
square make an angle ? with each other Now we
could estimate that the flux of the electric
field for this Gaussian surface is ? ? E ? ?A
18
CHECKPOINT Gaussian cube of face area A is
immersed in a uniform electric field E that has
positive direction along z axis.
  • In terms of E and A, what is the flux through ..
  • ..the front face (in the xy plane)?
  • EA
  • 0
  • -EA
  • ..the rear face?
  • EA
  • 0
  • -EA
  • ..the top face?
  • EA
  • 0
  • -EA

Answers (a) EA (b) EA (c) 0 (d) 0
  • ..the whole cube?
  • EA
  • 0
  • -EA

19
  • The flux through side B of the cube in the
    figure is the same as the flux through side C.
    What is a correct expression for the flux through
    each of these sides?

End of Lecture 5
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