THE UNIVERSITY OF BRITISH COLUMBIA
Wed. 04 Jan. 2006 - finish by Wed. 11 Jan.
This first assignment was just for review, to make sure you hadn't
forgotten (or could quickly recall) what you learned in PHYS 301/354
(or earlier) about the E&M covered in the first 7 chapters
of our textbook: David Griffiths, "Introduction to Electrodynamics".
Almost all you need is on the inside front and back covers
of the textbook:
- MAXWELL'S EQUATIONS:
- Starting with Maxwell's equations in differential form,
derive Maxwell's equations in integral form.
ANSWER: Applying (4) to yields .
Plug in (1) to give
Similarly for with to give
We use (5) [also known as the CURL THEOREM]
on to give
and plug in (2) to get
Finally we apply (5) to to get
and plug in (3) to get
- Starting with Maxwell's generalization of Ampère's Law,
, derive the continuity equation, ,
which is the mathematical expression of charge conservation.
ANSWER: Just take the divergence of both sides of Eq. (3),
remembering that the divergence of a curl is always zero, and plug in
GAUSS' LAW for (see above). Immediately we have
which is the same
as the CONTINUITY EQUATION if we note that
where is the (unchanging) bound charge density.
- Starting with Maxwell's equations in free space
), show that and
each satisfy a wave equation. What is the speed of propagation
of the resulting wave in each case?
First take the time derivative of Eq. (2) to get
(where the order of differentiation
wrt time and space have been reversed). Then substitute
into Eq. (3) and take its curl to get
Expanding the double curl gives and GAUSS' LAW
says , so Eq. (7) becomes
At this point you were expected to impose the usual conditions
for propagating EM waves, namely no free charges or currents
). In a conductor, things get
a bit more complicated (see Section 9.4 of the textbook).
With , Eq. (8) has the same RHS
(right hand side) as the LHS of Eq. (6), yielding
- i.e. the WAVE EQUATION for ,
where the speed of propagation is
In free space, of course,
m/s [exactly, by definition].
is the index of refraction.
For we take the time derivative of Eq. (3),
this time setting at the outset. The result is
Then we take the curl of Eq. (2) to get
Again matching up RHS with LHS gives
- i.e. the same WAVE EQUATION for .
- CHARGED CONDUCTORS:
Two spherical cavities, of radii a and b, are hollowed out
from the interior of a solid neutral conducting sphere of radius R,
as shown in the figure. There are charges qa and qb
at the centres of the respective cavities.
- What is the electric field in the solid (shaded) conducting material?
- Find the surface charges
at the respective surfaces. ANSWER: Since the point charges are
centered on the spherical cavities, by GAUSS' LAW it takes
an equal and opposite uniform surface charge to cancel
inside the conductor. Thus
The net interior surface charge, -(qa+qb), leaves behind a
charge +(qa+qb) which will distribute itself uniformly over
the outer surface (to achieve maximum separation), giving an
exterior surface charge of
- What is the electric field outside the conductor
at a distance r>R from the centre of the large sphere?
- What are the electric fields inside cavities a and b?
where ra and rb are the distances from the centre of each cavity,
- What are the forces on qa and qb?
ANSWER: . They are centred in their cavities.
- If a third charge qc were brought near the conductor,
which (if any) would change:
- ? .
- ? .
- ? . An additional (nonuniform)
surface charge would be required to terminate electric field lines
coming from the third charge.
- The electric fields inside cavities a and b? .
- The electric field outside the conductor? . Doh!
- COAXIAL CAPACITOR:
A capacitor is constructed of two very long concentric cylindrical
conductors with their common axis horizontal, as shown in the diagram.
The space between them is exactly half filled with
a linear dielectric liquid with dielectric constant .
- Show that the electric field is radial
and is the same in the dielectric half
as in the vacuum half of the capacitor.
ANSWER: Symmetry with respect to z ensures that
can have only radial and aximuthal components Er and
and that neither one changes with z.
[I am using r rather than the textbook's s for the radial direction.]
Meanwhile each of the conductors is at constant potential,
forcing cylindrical symmetry upon
through its boundary conditions.
This symmetry is passed on to
so there can be no aximuthal component of :
and is radial everywhere. .
[Note that this does not mean a uniform distribution of
surface charge on the conductors! See next part.]
- Deduce the capacitance per unit length of this coaxial capacitor.
ANSWER: A capacitor filled with a dielectric has a larger surface charge
on the conductors for the same electric field in between
(i.e. the same voltage across) so its capacitance is increased
by a factor of
(the dielectric constant,
written "" in the textbook).
More charge will pile up on the conductors
where the dielectric is than elsewhere;
will be uniform within each region,
but a factor of higher where the
space is filled with dielectric.
The net capacitance is the average
of what one would get for the empty capacitor
and what one would get if it were completely filled:
- If the conductors carry free charges per unit length ,
find the polarization in the dielectric
at any point a distance r from the central axis, in terms of
ANSWER: We have
From Gauss' law on a cylinder of length and radius r
(with a<r<b) we have
- LINEAR CURRENTS:
Two very long parallel wires carry equal currents
in opposite directions, as illustrated in the figure.
Take the direction to be out of the page,
in the direction of the current in wire 1.
The field point P is located a distance r1 from wire 1
and a distance r2 from wire 2, as shown.
- Consider each wire separately and indicate the direction
of the vector potential in each case.
For thin wires we can replace the volume integral by a line integral:
where it is usually assumed that the current loop is closed.
In this case we pretend it is not, and take one wire at a time,
choosing the origin at point P for simplicity ().
is everywhere parallel to ,
at point P
- Show that the vector potential at the point P
is given by:
ANSWER: Choose the zero of arbitrarily at some radius R
from the wire of your choice. If we construct a rectangular loop
with sides of length parallel to the wire (along )
and the other sides radial from r<R out to R, the magnetic flux
through that loop,
around that loop is equal (since
by Stokes' theorem.
Meanwhile, since two sides are and one of the others is
at R where , the line integral of
is just . Setting these equal gives
for the wire of your choice.
The total is the sum of
- Show that the result in part (b) is consistent with
that obtained using Ampère's law.
ANSWER: Treat the wires separately. For the wire of your choice,
in cylindrical coordinates (again using r instead of s)
for each wire, the same as we get from AMPÈRE'S LAW.
(Note: since each wire's is different, the net field
is messy looking; that's why we just did it one at a time, relying
on the knowledge that we can superimpose them if needed.
Also note: the derivation of used the above,
supplied by AMPÈRE'S LAW, so consistency is hardly surprising!)
- LAPLACE'S EQUATION:
Consider an infinitely long metal pipe, of radius R,
which is placed at right angles to
an otherwise uniform electric field
Hint: Note that this situation has cylindrical symmetry (not spherical!),
with no z dependence, and hence simplifies to a 2-D plane polar problem.
- What is the "uniqueness theorem"
and why would you want to use it to solve for
the electric potential V?
ANSWER: The UNIQUENESS THEOREM says that if you have a
solution of Poisson's (or Laplace's) equation in some region
and it has the right values at all the boundaries of that region,
then it is the only solution in that region. It's obvious why
this is desirable: you can guess, and if it works, it's right!
Of course, educated guesses are a lot more efficient, and so we check
our list of all possible solutions for the specified
symmetry imposed by the boundary conditions. This quickly
eliminates all the really dumb gueses. Then we take advantage of the
linearity of the differential equation for V, which ensures that
any linear combination of the above list is allowed. Usually from there
we just have to find the constants that make it fit at the boundaries.
What's not to like?
- What are the boundary conditions on the electric potential V?
ANSWER: Let's measure r from the axis of the cylinder
and set in the direction.
The surface of the conductor at r = R must be an equipotential
(which we are free to set equal to zero),
independent of , but far from the cylinder
- Solve Laplace's equation for the potential V
outside the long metal pipe. You should obtain:
ANSWER: The most general form has all integer powers of
and , but the boundary condition at
tells us immediately that we need only use n=1 with
and correspondingly only r or 1/r (or both) for the r-dependence.
The most general reasonable solution is thus
Let's see if we can find values for a1 and b1 that satisfy the
boundary conditions. As we can ignore the effect of
the second term, leaving simply a1 = -E0. Good so far.
This can only be true if b1/R = E0 R or b1 = E0 R2.
Hey, we're done!
Jess H. Brewer