http://students.informatics.unimelb.edu.au/serve/cmcleod/stuff/elec.JPG 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.