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I believe we are keeping ##\phi## constant, since ##R## is constrained to move along the angle ##\phi##. I think I know what you're getting at; I'm not calculating the differential correctly. I can't in general vary ##x## without varying ##y## since ##R## is constrained to move along ##\phi##...

Thanks for the reply - looks like I completely messed this one up! If I start from the start and now say that ##R## is the distance from the origin to the center of the disk, I see what you're saying about ##\frac{dy}{dx}=tan\phi##. Then, expanding ##dR##, I get (remembering to put in the...

Let ##R=\sqrt{x^{2} + y^{2}}##. Then \begin{align}v_{tangential}&=\frac{dR}{dt} \nonumber\\ &=\frac{dR}{dx}\frac{dy}{dt} + \frac{dR}{dy}\frac{dy}{dt} \nonumber\\ &=\frac{x}{R}\frac{dx}{dt} + \frac{y}{R}\frac{dy}{dt} \nonumber\\ &= cos\phi \frac{dx}{dt} + sin\phi \frac{dy}{dt}.\nonumber...

I see my mistake! Thanks for the correction guys :)

I'm resurrecting a zombie thread here, but I think the answer given for the dipole potential is incorrect. TSny provides the correct method for obtaining it, despite saying that OP's answer for the dipole potential is correct. To expand a little, if we find ##V_{dipole}(x, y, z)## we get...