THIS IS NOT A RETALIATION FOR THE 2000 THREAD.
I was playing with numbers and would like to share them.
It is sort of coninuation of my previous message. How to
justify distributions of charges?
Here is a problem which should help. Four equally
charged beads can slide without friction on a horizontal
wire whose length is 20 cm. The charge on each bead
is 10 microC. Two beads are immobilized at the ends
of the wire, the other two are at a distance x from
the center of the wire. How does the potential energy
depend on x? Treat beads as point charges.
The answer --> PE has a minimum of 45.01 J at
x=3.333 cm, that is when the beads are equidistant.
Solve the same problem for 5 beads (8 microC each). This
time three beads are immobilized, one at the center and
one at each end. Two beads are free to slide, as before.
This time the minimum PE = 66.08 J occurs at x=5.28 cm.
The five beads are not equidistant at the bottom of the
"potential well"; each sliding bead is closer to the end
than to the center.
This problem is artificial but it does illustrate the
tendency of charges to "concentrate on sharp point".
For seven beads (three of them being fixed as above)
the potential energy becomes the function of two
parameters (due to symmetry) and I would expect the
minimum to occur when the consecutive distances
between beads become progressively smaller as one
moves away toward the end point beads.
What follows is my little program for 5 beads.
The beads are numbered 1,2,5,3,4 as one goes from
left to right along the wire. (The out-of-orderness
is due to the fact that it was a modification of the
simpler 4 bead problem.)
let L=0.2 ! wire length (meters)
let Q=8e-6 ! charge on each bead (C)
let PE14=9e9*Q*Q/L ! end to end
let PE15=9e9*Q*Q/(0.5*L) ! center to end
for x=0.052 to 0.053 step 0.00001
let PE23=9e9*Q*Q/(2*x)
let PE34=9e9*Q*Q/(0.5*L-x)
let PE52=9e9*Q*Q/x
let PE=PE14+PE23+2*(PE34+PE15+PE52)
print "x=";x;"meters. Pot. Energy=";PE;" joules"
next x
end