Consider a thin, hollow, fixed and overall uncharged metallic
sphere of radius r, with centre O. Now suppose a point of charge Q
is held at A. OA = d (d > r). Let N(x) be the charge : surface
area ratio at any point P on the surface of the sphere such that
the angle between the lines OA and OP is x. What possible function
N(x) minimises the total electrical potential energy of the
sphere?
Thanks,
Michael
disclaimer: this may be a load of rubbish
I am not particularly good at this type of thing, (so feel free to
point it out) but if you hold your charge outside the sphere, then
there will appear a section of opposite charge on the surface of
the sphere. Electrical energy is E: E = QV (I hope this is the
thing you mean) so lets say you have a +charge (unfeasible
assumptions coming up:)
then there will be a zone of negative charge (possibly an equal
amount?) on the surface of the sphere around the bit where the line
OA cuts it, and a zone of +charge diametrically opposite that part.
I'm not sure how you get the energy potential, but the test charge
will want to go towards the sphere, and vice-versa, so a force is
acting:
F=[1/4pe0][(Q1Q2/r2]
If the charges are equal, then it will be Q2
now E(or strictly work)=Fd
so E=[1/4pe0][(Q1Q2/r]
and then V=E/Q
so V is E but cancel one or other of the Qs (?)
er..
your use of the words charge:surface area ratio confuses me, but it
seems to be a similar idea to the patch of opposite charges. This
is total guesswork, but this ratio as you go through an angle x
(through the plane with P and the place where OA cuts the surface
on it, now called Z) may be described by -cos x, which is just an
idea, since the max -charge will appear at Z (ie x=0), and at x=pi,
will be the focus of +charge and at x=pi/2, you will be halfway
between the two focuses, so there will be no charge here. The bits
in between will have smaller amounts of charge because not all the
electrons will want to focus too near each other, so they will have
to space out, becoming more spaced out as you go back to the
x=pi/2
I don't see that you have mnuch choice of functions for N(x) if any
of this either A: helps (ha!) or B: makes you laugh then I have at
least tried to answer a fairly alien question. Was this in a
physics paper?
Hi! Thanks for the reply.
I think that perhaps I was a bit ambiguous when posing the
question. You said you don't understand what N(x) actually
represents. Well I'll try to explain.
Imagine you take any section on the surface of area A. Let the
total net charge on this section be Q. Let's describe Q/A as the
"average charge density" or "average charge:surface area ratio" in
this region. But what we actually want is the charge density at a
point. To do this, we take a sequence of areas converging on a
point and look at the limit of the average charge densities. (This
is analogous to the instantaneous speed of an object being the
limit of the average speed over continually shorter time
intervals.) So now we know what the charge density at any point
is.
N(x) simply denotes this charge density at any point at an angle x.
(Between OA and OP.) For any angle x there is a circle of points,
but it doesn't matter which point we take. The charge density will
be unaltered. By symmetry we can be sure of this.
So now we've got the definition of N(x). There are an infinite
number of possible functions N(x) and we want the one that
minimises the total potential energy. The only thing we know for
certain is that the total charge is zero. (Thus we could form an
integral equation for N(x).) I want to minimise the potential
energy because this will show the charge density at each point when
the electrons on the surface are in equilibrium. I have managed to
find the potential energy as double integral (neither of which I
can evaluate let alone minimise!)
When you say the ratio is -cos(x) then are you referring to N(x)?
I'm not too sure this would work as the minimum function should not
really be symmetric about the plane through the diameter
perpendicular to OA. But it's not a bad guess; it may be possible
to modify it (make it lopsided) and get what we require.
By the way, this was not in a physics paper. (The mathematics in
the sort of physics papers we do rarely goes beyond re-arranging F
= ma for m and a...) The reason I wanted to answer it was to find
out how large a dipole is induced on a sphere when a charge is
brought near it. This will always be done to minimise the
electrical potential energy. I admit it is an exceedingly hard
question. (It's my current favourite for your page on impossible
questions, although I doubt that this one would win the special
prize!) I'm wondering if there is any way of showing that the
answer I'm looking for cannot be written in closed form.
Many thanks,
Michael
Hi,
I'm still not sure what you mean by electrical potential energy,
and do any of the formulae for F, E, V etc that I gave serve any
purpose in this question?
Regards
Neil
Yes, they were what I was after.
The potential energy between two particles is
kQ1Q2/r like you said. The total potential
energy of any particle is the sum of the potential energy between
this particle and all the others in the system (both on and off the
sphere). Then the PE of the entire sphere is the sum of this over
all particles on the surface of the sphere! Not easy.
However, saying I wanted the potential energy minimised was just a
convenient way of saying that I wanted the function N(x) that would
leave the electrons in equilibrium on the sphere (given that the
sphere is fixed). Clearly the sphere will perceive an overall
attraction to the point charge, but if it is fixed then the
individual charge carriers are in equilibrium. (Because in a metal
they cannot leave the surface.) What I mean by equilibrium is that
they don't want to move along the surface of the sphere. It is
possibly best not to bring energy into it at all. This doesn't mean
that I have made any progress any other way... I just think it is
quite interesting that physicists use the idea of dipoles all the
time. (Moving a point charge close to a collection of charged
objects distorts the charge density.) Yet when it comes to actually
calculating the extent of the distortion, it seems almost
impossible.
Many thanks for your trouble,
Michael