当前课程知识点:College Physics II > Electrostatic Field > Gauss Law > The Gauss Law
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In this lecture
we are going to study the Gauss Law.
Our objectives are to understand the concept of the electric flux
be able to calculate electric flux for a given situation
and understand the Gauss’s law.
We know that the electric field lines begin on positive charges
and end on negative charges
The magnitude of the electric field intensity
is defined to be proportional to the number of field lines
in unit area perpendicular to the electric field
This figure shows a planar surface of area Aperp
that is perpendicular to a uniform electric field E
If we use symbol to represent
the total field lines penetrating Aperp
the definition of electric field lines
gives the electric field intensity
E=Φ/Aperp
and Φ=EAperp
HereΦ is also named as the electric flux
If a planar surface is not perpendicular to the field E
how could we represent the electric flux?
This figure shows a flat surface A in a uniform electric field
and Aperp is the projection of area A
in the direction perpendicular to the field
If the angle between A and Aperp is θ
Aperp=Acosθ
Clearlythe same number of field lines
crosses both A and Aperp
so the fluxes through both surfaces must be the same
The flux through surface A is therefore
Φ=EAperp=EAcosθ
To discuss the flux of a vector field
it is helpful to introduce an area vector A
The magnitude equals to the area
and the direction is along the normal direction n^
so A can be written as An^
and n^ is a unit vector perpendicular to the surface
You can easily see that
the angle between electr ic field
and the area vector also equals to θ
Eventually
we can define the electric flux of a uniform electric field
through a flat surface
as the scalar product of the electric field intensity
and the area vector
Φ =E·A
Note that there are two possible normal directions
at every point on that surface
Generally
for an open surface
we can choose either direction
as long as it’s consistent
over the entire surface
But for a closed surface
it’s a convention to use the outward normal direction
as the n direction
What is the expression of electric flux
of a non-flat surface in a non-uniform electric field?
In that case
we could divide the surface
into a large number of infinitesimal patches
which can be viewed as flat surfaces
At the same time
the electric field through every patch
can be seen as uniform
Now
we define the area vector dA
for every patch as the area of the patch
pointed in the normal direction
Then we could have the flux through that patch as
dΦ=E·dA
Clearly
the net flux is the sum of the infinitesimal flux elements
over the entire surface
With infinitesimal patches
the net flux through the whole surface
can be written as an integral of dΦ
then we could get the general expression of electric flux
as Φ=∫E·dA
In this expression
S could be any surface
As a special case
if the surface is a closed surface
the flux through that surface
can be written as Φ=∫E·dA
The circle on the integral symbol
simply means that the surface is closed
and we are integrating over the entire surface
In order to evaluate the above integral
we must first specify the surface
and then sum over the dot product of E and dA
Note that for a closed surface
the unit vector n^ is along the outward normal direction
As you can see from this figure
as the field lines leave the surface
the angle between E and dA is less than π/2
so the flux dΦ is positive
As the field lines enter the surface
the angle between E and dA is bigger than π/2
so the flux dΦ is negative
Remember the flux is a scalar
and the SI unit is N·m2/C
The characteristics of the electrostatic field lines
lead to an important mathematical equation
known as Gauss law
It states
the electric flux through a closed surface of any shape
equals to 1/ε0
times of the sum of charges enclosed within the surface
and the equation is written as
∫E·dA=Σqin/ε0
For example
for this charge distribution
the total flux through surface S
equals to (q1–q2–q5)/ε0
Now
take a look at the proof of the Gauss law
Let's first calculate the electric flux
through a spherical surface S around a positive point charge q
Suppose charge q is located at the center of the sphere
and the radius of the sphere is r
Recall that the electric field on the spherical surface
is E=(1/4πε0)q/r2r^
and r^ is a unit vector along the radial direction
The flux of this spherical surface
can be calculated by ∫E·dA
Replacing E by (1/4πε0)q/r2r^ gives ∫(1/4πε0)q/r2r^·dA
Since the radius r is a constant on the surface
we could move (1/4πε0)q/r2 out of the integral symbol
dA is in the same direction as r^
so the total flux
can be written as =(1/4πε0)q/r2∫dA
Clearly
∫dA means the total surface area of the spherical surface
and it equals to 4πr2
Eventually
the total flux through the closed spherical surface of radius r is q/ε0
This conclusion can be generalized to any closed surface enclosing charge q
For example
if another closed surface S' encloses charge q
it is clear that
the field lines crossing surface S must also cross surface S'
and this means that
the total number of field lines crossing these two surfaces is equal
so the total flux
through S' also equals to q/ε0
For the case
when a closed surface does not hold any charge
consider a typical electrostatic field
produced by charge q
As you can see from the figure
every line that enters the surface
must also leave that surface
If a field line enters the surface at dA1
and leaves at dA2
the flux at dA1 is d1=E1·dA1
and it is negative
The flux at dA2 is d2=E2·dA2
and it is positive
But these two fluxes should equal in magnitude
so
d2=–d1
Eventually
when no charges are included
the net flux through the closed surface is 0
Now let's take a look at the electric flux
through a random surface enclosing multiple charges
For charge system q1 q2 q3 and to qn
suppose surface S encloses some charges
The net electric field intensity anywhere on the surface
equals to the vector sum of Ei
The total flux is ∫E·dA
Replacing E by ΣEi gives ∫ΣEi·dA
We could change the order of integration and summation
and get Σ∫Ei·dA
Note that if qi is inside the surface
the integral of Ei·dA equals to qi/ε0
otherwise it is 0
So the total flux
through the closed surface equals to Σqenc/ε0
and Σqenc is the net charges inside the closed surface
Note that
Gauss law works for any closed surface in space
Generally
the closed surface is named as the Gaussian surface
It may coincide with the actual surface of a conductor
and
it may be an imaginary geometric surface
The only requirement for a Gaussian surface is that
it must be closed
Now we have proved the Gauss law as ∫E·dA=Σqenc/ε0
but we should have a clear understanding
of every term in the equation
Σqenc is the sum of charges inside the surface
Gauss law implies that
the total flux is determined by the sum of charges
inside the surface only
and it has nothing to do with the charges
outside the surface
or how the charges inside the surface distribute
However
the field E is the total electric field
on the Gaussian surface
and
it includes contributions
from charges both inside and outside the surface
We have obtained the Gauss law from the Coulomb law
but actually they are equivalent
so one could also get the Coulomb law based on the Gauss law
But
Gauss law gives a simple way
to determine the magnitude of electric field
for a given charge distribution
with sufficient symmetry
-Knowledge Map
-Introduction
--Introduction to Electromagnetism
-Coulomb Law
--Homework 1.1 Coloumb law
-Electric Field
--Calculation of Electric Field
--Homework Electric field calculation
-Gauss Law
--Homework 1.5 Gauss law
-Electric Potential
--Potential Energy due to Point Charges
--Electric Potential calculation
--Homework Electric Potential
-Electrostatic Induction
--Charge distribution in electrostatic equilibrium
--Homework Electrostatic Induction
-Capacitors
--Homework capacitor and capacitance
-Magnetic field and flux
--Magnetic force on moving charges
--Homework Magnetic force on charges
-Application of Lorentz force, The Hall Effect
--Application of Lorentz force
-Magnetic force on currents
--Force and torque on current loop
--Homework Magnetic field
-Biot-Savart Law
--Application of Biot-Savart Law
--Homework Biot-Savart Law
-Ampere's Law
--Ampere's Law and application
--Homework Ampere's law
-Faraday law of induction
--Lenz Law
--Homework Faraday law of induction
-Induction
--Energy Stored in Magnetic Field
--Homework Electromagnetic Induction
-Introduction
-Simple Harmonic Motions (SHM)
--Energy of Simple Harmonic Motion
-Energy of SHM, Physical pendulum, damped and forced oscillation
-Superposition of SHM
--Superposition of SHM of the same frequency
--Superposition of SHM with different frequencies
-Simple Harmonic Waves
--Characteristic quantities of waves
-Energy, and Interference of Waves
--Energy, Power and Intensity of a Wave
-Doppler Effect
--Homework Doppler Effect
-Youngs Double Slit Interference
--Young’s Double-slit Interference
--Homework Youngs Double Slit Interference
-Thin-film interference
--Homework Thin-film interference
-Newton's Ring, Michelson Interferometer
-Diffraction
--Fringe locations of diffraction
--Homework Diffraction
-Diffraction Gratings and X-ray Diffraction
--Homework Diffraction Gratings
-Polarization of Light
--Generation of Polarized Light
-Final Exam