Gauss' Law
Although the electric field of any given charge distribution can be
calculated using Coulomb's Law (as we did last week), the method involves
the evaluation of integrals, which, to put it mildly, is tedious.
Happy Happy Joy Joy there exists another method of calculating the
electric field, which relies on a theorem called Gauss' Law. This law is a
consequence of Coulomb's Law and there is no new physics but it does
contain some new math which provides an elegant shortcut for calculating
the electric field of a charge distribution PROVIDED that the distribution
has a certain amount of symmetry. Gauss' Law doesn't work in all cases
but when it does it is wondrous.
However, we need first to introduce a new concept: Electric Flux
Electric Flux and the number of field lines
Consider a mathematical
(that is, imagined surface
in the shape of a rectangle
of area A that is immersed
in a constant electric field
E. This electric field
makes an angle θ with the
surface and therefore has
a component tangential to
the surface and a
component normal, that is
perpendicular, to the
surface. The electric
flux Φ though the surface
is defined as the product of the area A by the magnitude of the normal
component of the electric field
Φ= EnA = E A cos θ
The quantity A cos θ can be seen as the projection of the area A onto a
plane perpendicular to the electric field, that is, A cos θ can be regarded
as that part of the area A the faces the electric field 'head on.'
Now the magnitude of the electric field must be proportional to the
magnitude of the charge, the number of field lines that we draw coming
out of a (positive) charge must be proportional to the charge. There is a
convention that the number of field lines coming out of a charge Q is Q/ εo
- so the number of field lines coming out of a coulomb of charge is 1/ εo =
1.13 x 1011. This is a useful number for the calculations we will be doing
although we would never draw that number.
Since the electric field is proportional to the charge and the number of
field lines is proportional to the charge, the electric flux is numerically
equal to the equal to the number of field lines intercepted by a unit area
facing the electric field. This means that E A cos θ must be numerically
equal to the number of field lines intercepted by the area A:
the electric flux
through an area is equal to the number of field lines
intercepted by the area.
[in the diagram the flux is 4]
More generally, consider a mathematical surface of arbitrary shape
immersed in an electric field of arbitrary strength. We divide the surface
up into small parts dS so
dΦ = E A cos θ dS
then we integrate both sides and get
According to this argument, this electric flux is again equal to the number
of field lines intercepted by the surface. Note that the lines going through
the surface in one direction give a positive contribution to the flux while
those going through the
surface in the opposite
direction give a negative
contribution.
In the picture to the right,
the flux is +2 - 3 = -1.
Now for a closed surface
without any charge inside
it but with charges
outside it. The field lines
enter the surface from one
direction making, say, a
positive contribution to
the flux but then leave on
the other side, making a
negative contribution. So
the net flux is zero.
Gauss' Law
We are now ready to prove Gauss Law which states
If the volume within an arbitrary closed surface holds a net charge Q, then
the electric flux through the surface is Q/ εo , that is
The small circle on the integral sign indicates that the integration is to
be done over a closed surface.
Here's the proof. The electric field in the equation is a sum of the
individual electric fields produced by some number of point charges. Some
of these charges are inside the closed surface and some are inside. If the
charge is outside, either the field lines miss the closed surface entirely
or they come in and go out. In either case they contribute nothing to the
flux.
For the charge inside the closed surface do contribute to the flux. If there
is a positive charge q inside the closed surface it will create q/ εo field
lines, all of which will pierce the surface and create a flux q/ εo . So the
net flux through the surface, or the net number of lines piercing the
surfaces, is equal to the net charge enclosed by the surface divided by εo
That's it for the proof.
Gauss' Law can be used to calculate the electric field provided the
distribution of charge has a high degree of symmetry. The surface
surrounding the charges is called the Gaussian surface.
Example 1
What is the electric charge of a point charge?
Let's pretend we do not know the electric field of a point charge. Now we
find a nice surface where the electric field is constant. (see figure below)
The electric field has to be spherically symmetric because the point is
symmetric. Therefore all points on the surface of a mathematical sphere
of radius r centered on q, the field is in the radial direction and has a
constant magnitude E
If the Gaussian surface coincides with the sphere of radius r, then the
angle is 90 and the cosine factor is 1 and E has a constant magnitude over
this surface
but the integral is simply the area of the sphere, 4πr
2
so E (4πr 2) = q/εo
E = q /(4π εor 2)
which is what we know it is.
A couple more examples
Remember the long straight wire with a charge density of λ
coulombs/meter. That one took us about half an hour. Now let's do it the
Gaussian way. First to find the surface over which E is either constant or
zero or the cosine is 0
The electric field emerges radially from the wire and the only surface
that satisfies the conditions that E is uniform is a cylinder (if not radial
field, there would be a distinction between the top at the bottom of the
wire). The E cos θ dS term vanishes at the caps because the field and the
area are perpendicular to each other.
Once again
and the integral is the surface of the outside of the cylinder - 2πxh
E (2πxh) = q/ εo
Now the only question is what is q - if the linear charge density is λ
coulombs/meter and the portion of the wire we are looking at is h long,
q= λ h
E (2πxh) = λ h / εo
E = λ /(2π εox )
the same result we sweated blood over last week.
The other one that nearly did us in was the electric field of a large (aka
infinite) sheet of charge with a charge density of σ coulombs/meter 2
The electric field must be perpendicular to the surface. Our Gaussian
surface is a cylinder with two caps of area A each. It sticks out a distance
z from the top and a distance z below (so that E will be the same at each
of the caps).
Along the sides of the cylinder the E is perpendicular to A, so the integral
vanishes so we have to do the integral only over the caps (or bases)
This is simply 2AE so
2AE = q/εo
Now what is q. There are σ coulombs/meter 2 and the area is A, so
q=σ A
2AE = σ A
E = σ/2
Again, what we sweated blood over last week.
Homework:
You have a sphere of radius R that has a total charge q which is uniformly
distributed over the VOLUME of this region [ meaning the charge density is
ρ = q/volume of sphere = q/(4 π R 3/3)
What is the electric field E at a radius r INSIDE the sphere (r < R)?