# Homework Help: Electric Potential at center of semi-circular rod

1. Apr 9, 2006

### twiztidmxcn

About a month ago I asked what could be the part 1 of this question, and here's the part two.

Say you take a thin flexible rod with charge Q, bend it into a semicircle (half circle going from pi/2 to 3pi/2) with radius R.

Find an expression for the electric potential at the center of that semicircle.

I know that this has something to do with finding the electric field at the center point and then taking the negative integral to find electric potential.... but i'm not even sure how to start.

Normally, there would end up being cos/sin getting integrated due to there being a need for components when calculating electric field. However, when going to electric potential, how can the vector values be turned into scalars? Ahhhhhhhhhhh!

Any sort of help, starting help and a shove in the right direction would be awesome.

-twiztidmxcn

2. Apr 9, 2006

### Physics Monkey

Hi twiztidmxcn,

The important thing to remember is that the electric potential is always a scalar. You get the electric field by taking the negative gradient of the potential, and this gradient is where the vector nature of the electric field comes from. The fact that the potential is a scalar and hence mush easier to work with is part of the reason why it's such a useful concept. In your case, for example, one doesn't need to worry about components because the electric potential is a scalar and has no components.

To find the potential, you want to add up the contribution from each little piece of your semicircle. What do you need to know to do this? Well, you need to know the distance to the origin certainly. You also need to know how much charge each little infinitesimal piece of the semicircle carries. Do you need to know anything else? Once you have all the information, "add up" the contributions by integrating your result over the whole semicircle.

Hope this helps.

3. Apr 10, 2006

### Meir Achuz

Every point on the circle is R from the center, so the pot at the center is
Q/R (in Gaussian units).

4. Apr 11, 2006

### twiztidmxcn

thx to ya both, ended up getting it

dV = k*dq / r, integrate both sides, end up with V = Q/4*pi*epislon naught * r, which is what i needed