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pattiecake
Feb29-04, 01:41 AM
This hmwk. problem has got me totally buggin...

A uniformly charged insulating rod of length 20cm is bent into the shape of a semicircle (so it looks like the letter "C"). The rod has a total charge of -8.50e-6C. Find the electric potential at a point P, in the center of the semicircle.

I know crazy things happen to electric potential when dealing with insulators...also, will this problem involve integrals?

If anyone has a clue I'd appreciate the guidance! Thanks in advance!

pattiecake
Mar2-04, 11:12 PM
Why will noone respond? =(

paul11273
Mar3-04, 12:20 PM
Yes, this requires and integral.

V=ke \int{dq/r}
V =ke(Q/r)

Where Q is the total charge of the rod and r is the radius this rod makes. Remember that circumference=2*pi*r, so use that to find r.

When you figure it out, post your answer.

I think no one has responded because you didn't post any sort of work that you did.
Post up what you have and we can take this further.

pattiecake
Mar7-04, 10:53 PM
thank u so much. i didn't mean to be an answer leech!

i used this formula before: V=Ke(Q/r): using Circumference=2piR to find the radius. The problem was after i got my answer: -2.39, i forgot we were dealing with a semicircle, and that i had to divide by 2. my final answer (which webassign confirmed) was -1.2MV. thanks so much for your help!