The Gauss Law在线视频

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