Re: [Phys-l] drop a metal cylinder through a solenoid

[John Denker <jsd@av8n.com>]

In the context of a dielectric object in an applied field I wrote:

> > Numerically the charge is the same as you would get for
> > a capacitor, but the field-line pattern is dramatically different.

On 03/25/2012 07:02 PM, Carl Mungan wrote:

> I see the numerical part, as in my calculation above. But how is the 
> field-line pattern different?

I don't know whether the distinction is important or not, but here
goes:  

I feel safer calling it an anti-capacitor rather than a plain old
capacitor.  Here's how I think of the field lines:
  http://www.av8n.com/physics/img48/anti-capacitor.png

The diagram is approximately valid in the frame instantaneously 
comoving with the bob.  In this frame there is an electric field
(in addition to the magnetic field) ... in contrast to the lab 
frame, where there is only a magnetic field.  In the diagram I
cheated by representing the electric field as being sourced by 
charges, when in fact it is induced by rotating (boosting) the
lab-frame magnetic field.

By superposition, we have A + B = X as defined in the figure,
where A represents the applied field, B represents a plain
old capacitor, an X represents an anti-capacitor.
 *) For the plain old capacitor, the field is almost all inside
  the device, and almost zero outside.
 *) For the anti-capacitor, the field is almost all outside the
  device, and almost zero inside.

As for the field, it is linear, and superposition says that we
can understand the anti-capacitor by reference to the capacitor.
However, other things such as the energy are nonlinear.  In
particular, the energy of the capacitor is 1/2 C V^2 whereas
I reckon the energy of the anti-capacitor is C V^2 with no factor
of 1/2 out front ... as you can verify by freezing the charge in
place and flipping the anti-capacitor upside down ... which is
the standard easy way of finding the energy of a dipole in an
applied field.

Let's not worry about factors of two, since there is bigger game
afoot.  There is a factor of 10^30 that we are trying to understand.

For this reason, to my way of thinking, it is important to clearly
have the anti-capacitor picture firmly in mind.  In the steady
state, there is zero field inside the anti-capacitor, i.e. inside
the dielectric bob.

Now imagine that at time t=0 the external field changes by some 
infinitesimal amount.  This will try to set up an infinitesimal 
field inside the metal.  Immediately thereafter, the charges in
the metal will re-arrange themselves to cancel the field.  Some
small current will flow.  My point is that this current flows
across the _infinitesimal_ field.  The energy involved in this
re-arrangement is qualitatively different from the energy involved 
in charging a capacitor, where most of the charge (all except the 
first little bit) must be moved against a non-infinitesimal field.

I recommend thinking about the following analogy:  Get a wave
generator that puts out a low-frequency sine wave or symmetrical
triangle wave of amplitude Vmax.  Assume that has zero output 
impedance.  Run that through a resistor into a capacitor, such 
that the RC time constant is short compared to the period of the 
wave.  Calculate the energy dissipated in the resistor.  It will 
be tiny compared to the energy 1/2 C Vmax^2 transfered /through/ 
the resistor to and from the capacitor.

The system is almost reversible.  The voltage /drop/ across
the resistor will be small compared to Vmax, even though for 
most of the cycle the voltage relative to ground will be large
(i.e. comparable to Vmax). 

To say the same thing yet again: Suppose the two ends of the
resistor are at voltages Va and Vb.  The heating in the resistor 
goes like (Va-Vb)^2 ... not like Va^2 or Vb^2.  We apply this to
the dielectric (metallic) bob as follows:  You have to take into 
account the field due to the screening charges (part B of the 
diagram) and subtract that from the induced field (part A of the 
diagram) to find the actual "delta V" that affects the charges 
and currents inside the metal.

I suspect that's not entirely the explanation that was wanted,
but perhaps it's a start.  If there are remaining issues, please
re-ask or re-phrase the question.