http://students.informatics.unimelb.edu.au/serve/cmcleod/stuff/elec.JPG(adsbygoogle = window.adsbygoogle || []).push({});

Eo = permittivity of free space

integral dV = (1/4(pi)Eo) integral (dQ)/a

Hopefully I have that formula correct.

so....

dQ = 2(pi)r(dr)(sigma)

Horizontal components cancel. So there will be a sin(theeta) term at the end of the integral.

sin(theeta) = z/a

and a = sqrt(z^2 + r^2)

and the integral will be between 0 and R

so the RHS becomes...

(1/4(pi)Eo) integral(0 -> R) (2(pi)r(dr)(sigma)/sqrt(z^2 + r^2)) * z/sqrt(z^2 + r^2)

Is this right? I have a feeling I have done something stupid.

Any help greatly appreciated.

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