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Re: I need help.



Regarding Ludwik's perplexity:

I do not know where the 0.14 mm comes from. In the first
description of your geometrical method (2/19/02) charges
+Q and -Q were located at the focal points.

True. But your silver dots are *not* the 'focal' *points*. They
are circular equipotential regions.

In the yesterday's
message they are shifted by 0.14 mm toward the center of
symmetry.

If you'd go back and reread my post of 19 Feb 2002 19:01:52 -0500
whose subject title is "Re: Calculating resistance" you would find
that I made a strong point that the equipotential circles were *not*
centered on the 'focal' points. In general, the centers of each of
the various equipotential circles are shifted off of its
corresponding (surrounded) 'focus' in the direction *away* from the
central point (i.e. the point that bisects the line segment
connecting the 'foci'). The amount of that outward shift is given by
R/(max(K^2,1/K^2) - 1) where K is the ratio of the distances to the
'foci' from the locus of the given equipotential circle, and R is the
distance between the 'foci'.

For the setup you described in your web page you had 2*a = 7.5 mm and
L (the distance between the centers of the equipotential circles
defined by the dots themselves) = 100.0 mm. For this setup
R = 99.718353 mm and each equipotential silver dot is shifted outward
from its 'focus' by 0.140823 mm since for your dots you have
K = 26.62911 for one dot and its reciprocal of K = 1/26.62911 =
0.03755288 for the other dot. Notice that L is given as
L = 99.718353 + 2*(0.140823) = 100.0000 mm

In practical terms this is not significant;

Of course.

I can not
locate equipotential points with an accurate less than 1 mm or
so.

Of course.

But I still want to know. Numerical illustrations are very
useful; they help to make sure messages are well understood.

That's why I quoted a ridiculously large number of sig figs in
my explanation above. (Caution, students, don't try this at
home.)

Following your first description I proceeded as follows.
After measuring DOP=50 V at the point (14,10) I calculated
K=7/3 =2.333.

For the 50 V equipotential circle the value of K is 2.27164 (assuming
that we take the right side circles to have K-values greater than 1
and left side circles to have K-values less than 1. For the 30 V
equipotential circle we have its K is given by K = 1/2.271638 =
0.4402109 . This also assumes that we place a potential of +80 V on
the equipotential defined by the right side dot (whose K = 26.62911)
and place a potential of 0 V on the left side dot (whose K =
1/26.62911 = 0.03755288). For this system the relationship between
the K-value and the potential V is given by

V = 40 + 12.18767*ln(K) volts .

(Note that when we plug in the K-value for the left dot the
potential goes to zero and when we plug in the K-value for the
right dot its potential goes to 80 V.)

This gives me the radius of the equipotential
circle = 5.250 cm.

Actually the radius is 54.448435 mm

Where is the center of this circle? It is at
x=19+1.837=21.837 cm (and y=10).

The center of this circle is 213.82797 mm (and y = 100 mm).

There were no 0.14 mm
shifting toward the center in this approach. What did I miss?
Ludwik Kowalski

It appears that you may have neglected the fact that the dots
themselves make equipotential circles of diameter 7.5 mm and
pretended instead that they were really zero size and sitting
exactly on top of the 'foci'. Of course this error has little actual
numerical consequence to the accuracy to which the measurements are
made. The reason the error made is small is simply because the
K-values for the dots themselves are quite large, i.e. 26.62911 (for
the right dot) and quite small, i.e. 0.03755288 (for the left dot).
If the right dot really was a single point its K-value would be
infinite and the K-value of the left dot would be 0 if it was a
single point. Since the shift of the center of an equipotential
circle from its surrounded 'focus' is R/(max(K^2,1/K^2) - 1) we see
that whenever the K-value is either large or small compared to 1 the
shift becomes negligible.

David Bowman
David_Bowman@georgetowncollege.edu