# Electric potential of insulating rod

1. Feb 29, 2004

### pattiecake

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!

2. Mar 2, 2004

### pattiecake

Why will noone respond? =(

3. Mar 3, 2004

### paul11273

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.

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.

4. Mar 7, 2004

### pattiecake

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!