i had a question. A thin wire is bent into semicircle, having a radius of curvature equal to 6.40cm. It's charged uniformly with charge density 33 x 10^-9 C/m. What is the potential at the center of curvature?

Question

anonymous · Answer

Doesn't Faraday' Principle have something to say about this?

anonymous · Answer

I'm pretty sure all those numbers are irrelevant.

anonymous · Answer

total charge is lambda*pi*r
since the point is equidistant from every charge element, the potential is just kq/r
k*(lambda*pi*r/r)= k*lambda*pi

anonymous · Answer

Or maybe it's Gauss' Law and not Faraday's .  .

anonymous · Answer

You need to find the electric field first, and then integrate it from infinity to the point in relation to the path. That gives you -V, where V is the potential.
I was thinking about using the coulomb law and integrate it to get E, I tried by gauss and couldn't do it.

anonymous · Answer

Yeah, now that I'm looking at the integrals, I don't think it works as nicely for a ring as it does for a sphere...

anonymous · Answer

I guess you are right

anonymous · Answer

the right answer to the question is 932v which comes out right by using the approach provided by algebraic!

anonymous · Answer

thanks for the medal.  I'll go cash it in for some internet prizes.

anonymous · Answer

@Algebraic! I don't think is that simple because the electric field created by the semicircle is diferent from the one crated by a particle.

anonymous · Answer

lol

anonymous · Answer

wrong way to do the problem, but w/e, if you want to do it that way, we can...

anonymous · Answer

or rather, you show me how you'd do it.  and we'll compare...

anonymous · Answer

Whats w/e?

anonymous · Answer

whatever

anonymous · Answer

you done?
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