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Science Forum Index » Physics - Particle Forum » Neutron, electrostatic fields and strong force
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| Author |
Message |
| Steersman |
Posted: Tue Dec 12, 2006 12:32 pm |
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Guest
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Using basic electrostatic field theory, I did a piece-wise numerical
integration over all space of the interaction between the electrostatic
fields of two electrons, which agreed (as expected) with Coulomb's Law.
But I found that there is a region between the two electrons which is
attractive - this region has only about 1/5th of the force associated
with the rest of the field, so the net effect remains repulsive, and
agrees with Coulomb's Law. For example, midway between the two
electronc the electric field vectors are equal and opposite, cancelling
out both fields, giving a zero field strength, zero field energy
density, and hence attractive forces. In the space outside both
electrons the fields add, giving increased energy density and hence
repulsion.
This region of attraction interested me, so I tried truncating the
field at a certain radius "r" from the centre of the particle, and
found that:-
1. for separations greater than "2 r" there is no interaction.
2. as the separation falls from "2 r" to "r" the attractive force
increases to about "1.05 r" then falls to zero at "r"
3. below "r" the force increases rapidly.
The force/separation curve looks just like the "nuclear strong force"
curve. So if the electric field of a neutron was truncated at a radius
of roughly 0.35 femtometers, and the "unit charge" of the field was 13
times that of the electron, and of opposite sign, would it give the
strong force? Such a neutron would sit at the null point at a
separation of "r" from its neighbours, and they would tend to form
something like a face-centred-cubic matrix.
I put a detailed graphical explanation at...
http://uk.geocities.com/steersman@btinternet.com/neutron_electric_field.html
Anyone got any more ideas on this? |
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