Transcript Document 782513

Chapter 24
Gauss’s Law
Gauss law, concept
EF lines come OUT of the surface  there are
some positive charges inside
EF lines come INTO of the surface  there
are some negative charges inside
Net flux outward  positive charges inside
Net flux inward  negative charges inside
zero Net flux  zero charge inside
less EF lines  less charges inside
More EF lines  more charges inside
examples: is there any charge inside?
Gauss Law:
in words:
“The amount of flux coming out of a closed surface is
proportional to the charge enclosed.”
The surface used in Gauss Law is closed, virtual, and with
arbitrary shape and size. It is called “gaussian surface”
Equation:
qin
E  
0
flux ~ charge enclosed
Flux: in words
The net flux through the surface is proportional to the net
number of lines leaving the surface.
This net number of lines is the number of lines leaving the
surface minus the number entering the surface.
Flux: Mathematical Definition
Units: N · m2 / C
—
 En dA
En is the component of the field
perpendicular to the surface
En
The surface integral means
the integral must be evaluated
over the surface in question.
the value of the flux will
depend both on the field
pattern and on the surface.
r
dA is a vector perpendicular
to the surface
Case of spherical symmetry
Example:
A positive point charge, q
We know that the EF is radial
We choose as a gaussian surface a
sphere of radius r centered at the
charge.
The magnitude of the electric field
everywhere on the surface of the
sphere is constant and normal:
E=En = keq / r2
 —
 En dA En —
 dA EnA
Section 24.2
Gauss’ Law
Gauss’ Law can be used as an alternative procedure for calculating electric
fields.
It is convenient for calculating the electric field of highly symmetric charge
distributions ONLY, where:
there is a gaussian surface such that
• the EF is perpendicular to the surface
• the EF is constant all over the surface
However, Gauss law is valid for ALL systems, symmetric or not.
Introduction
Gaussian Surface, Example 2
The charge is outside the closed
surface with an arbitrary shape.
Any field line entering the surface
leaves at another point.
Verifies the electric flux through a
closed surface that surrounds no
charge is zero.
Section 24.2
Applying Gauss’s Law
To use Gauss’s law, you want to choose a gaussian surface over which the
surface integral can be simplified and the electric field determined.
Take advantage of symmetry.
Remember, the gaussian surface is a surface you choose, it does not have to
coincide with a real surface.
Section 24.3
Conditions for a Gaussian Surface
Try to choose a surface that satisfies one or more of these conditions:
 The value of the electric field can be argued from symmetry to be constant
over the surface.
 The dot product of
EdA because and
can be expressed as a simple algebraic product
are parallel.
 The dot product is 0 because
and
are perpendicular.
 The field is zero over the portion of the surface.
If the charge distribution does not have sufficient symmetry such that a gaussian
surface that satisfies these conditions can be found, Gauss’ law is not useful for
determining the electric field for that charge distribution.
Section 24.3
Spherical Symmetry
Field Due to a Spherically Symmetric Charge Distribution
Select a sphere as the gaussian
surface.
For r >a
 E  EA  E4 r 2
Q
E
E 
0
Q
Q
 ke 2
2
4  0 r
r
same as the EF for a point charge!
Section 24.3
Spherically Symmetric, cont.
Select a sphere as the gaussian
surface, r < a.
qin < Q
Vin
r3
qin = Q
Q 3
Vtotal
a
Qr
Ek 3
a
Section 24.3
Spherically Symmetric Distribution, final
Inside the sphere, E varies linearly with r
 E → 0 as r → 0
The field outside the sphere is equivalent to
that of a point charge located at the center
of the sphere.
Section 24.3
Case of cylindrical symmetry
EF due to a line of charge with uniform linear density λ (uniform)
Select a cylindrical gaussian surface.
 The cylinder has a radius of r and a
length of ℓ.
is constant in magnitude and
perpendicular to the surface at every
point on the curved part of the surface.
E is parallel to the top and bottom (En=0)
  EAlateral  E2 rl
l


0 0
Q


E
 2k
2 0 r
r
Section 24.3
Plane Symmetry
Field due to an infinite Plane of Charge with uniform density σ
is perpendicular to the plane and
must have the same magnitude at all
points equidistant from the plane.
Choose a small cylinder whose axis is
perpendicular to the plane for the
gaussian surface.
is parallel to the curved surface
(En=0)
E is perpendicular to the top and
bottom surface
The flux through each end of the
cylinder is EA and so the total flux is
2EA.
Section 24.3
Plane Symmetry
Field due to an infinite Plane of Charge with uniform density σ
  2EA
A


0
0
Qin

E
2 0
The EF does not
depend on the distance
to the plane!
Section 24.3