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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Twigg on January 05, 2011, 03:40:22 PM

Title: Formulating Coloumb's Law in terms of general inverse square law form:
Post by: Twigg 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?
Title: Re: Formulating Coloumb's Law in terms of general inverse square law form:
Post by: MrTeo on January 06, 2011, 09:41:41 AM
Well, if we apply Gauss' law  (http://"http://en.wikipedia.org/wiki/Gauss's_law") we get:

(http://www.forkosh.dreamhost.com/mimetex.cgi?{ \Phi_S\left(\vec{E}\right)=\frac{q_1}{\epsilon_0} })

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:

(http://www.forkosh.dreamhost.com/mimetex.cgi?{ E \cdot 4\pi r^2=\frac{q_1}{\epsilon_0} })

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):

(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} })

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):

(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} })

(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...
Title: Re: Formulating Coloumb's Law in terms of general inverse square law form:
Post by: Twigg on January 06, 2011, 06:26:28 PM
Yes that makes a lot more sense! Thanks!