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Topic: Formulating Coloumb's Law in terms of general inverse square law form:  (Read 1783 times)

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Twigg

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Formulating Coloumb's Law in terms of general inverse square law form:
« on: January 05, 2011, 03:40:22 PM »
I'm not sure if this is the right place for this question.
My textbook gives this form of Coloumb's law, which determines potential energy instead of force:

E=(1/(4*pi*E0))((q1*q2)/r) where E0 is the permittivity of free space.

I hardly know anything about electric fields, but I tried to reformulate the equation from an inverse square law equation:

SA=4*pi*r^2 where SA is the surface area of the spheres with radius r centered on either of the two particles
q/SA represents charge per unit area, so it follows that
q/(4*pi*r^2) also represents charge per unit area
but my question is, how do you introduce force from charge per unit area?

MrTeo

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Re: Formulating Coloumb's Law in terms of general inverse square law form:
« Reply #1 on: January 06, 2011, 09:41:41 AM »
Well, if we apply Gauss' law we get:

$request=http://www.forkosh.dreamhost.com/mimetex.cgi?{+\Phi_S\left(\vec{E}\right)=\frac{q_1}{\epsilon_0}+}&hash=7844cf18f09cdf9c065bd0ba59ccaaf1$

where ΦS is the flux of the electric field E through the surface S. In this case, as the direction of the field is always normal to the sphere's surface we can write ΦS=ES, so we get:

$request=http://www.forkosh.dreamhost.com/mimetex.cgi?{+E+\cdot+4\pi+r^2=\frac{q_1}{\epsilon_0}+}&hash=db66457f7e6c5ecaf9aa1b7001f6f40a$

As the electric field is F/q we have, at last (q2 is the charge we put in the electric field generated by the charge q1):

$request=http://www.forkosh.dreamhost.com/mimetex.cgi?{+F=q_2+\cdot+E=\frac{1}{4+\pi+\epsilon_0}+\cdot+\frac{q_1+q_2}{r^2}+}&hash=668bfc1f6b49f303cc5bb10afd4a25cf$

A more straightforward way to get the inverse square law would be to derive your expression (using r as the independent variable), as E is also dV/dr (where my V is your E, the electric potential):

$request=http://www.forkosh.dreamhost.com/mimetex.cgi?{+E=\frac{dV}{dr}=\frac{d}{dr}\left(\frac{1}{4+\pi+\epsilon_0}\cdot+\frac{q_1+q_2}{r}\right)=-\frac{1}{4+\pi+\epsilon_0}\cdot+\frac{q_1+q_2}{r^2}+}&hash=2254c1d7e15ab24088ddb36f239aaa4f$

(the "-" only means that the direction of the force vector is opposite to the increase in the electric potential)
Hope it's all clear now... what you did is also a common way to verify Gauss' law with an easy example...
The way of the superior man may be compared to what takes place in traveling, when to go to a distance we must first traverse the space that is near, and in ascending a height, when we must begin from the lower ground. (Confucius)

Twigg

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Re: Formulating Coloumb's Law in terms of general inverse square law form:
« Reply #2 on: January 06, 2011, 06:26:28 PM »
Yes that makes a lot more sense! Thanks!