# Electric potential of a plastic rod

1. Nov 28, 2008

### noppawit

A plastic rod has been bent into a circle of radius R=8.20cm. It has a charge Q1 = +4.20pC uniformly distributed along one-quarter of its circumference and a charge Q2 = -6Q1 uniformly distributed along the rest of circumference. With V=0 at infinity, what is the electric potential a) at the center of the circle and b) at point P, which is on the central axis of the central axis of the circle at distance D = 6.71cm from the center?

In the problem, I tried to evaluate V = kq/r to solve this problem. I think that r and k, which is 8.99*10^9, is constant in this problem. I would like to know to integrate this equation and to make it relate with 2Pi*r

P.S.
Answer from textbook: a) = -2.3V and b) = -1.78V. I want to find out the solution.

2. Nov 28, 2008

### Staff: Mentor

You need to evaluate ∫k/r dq. Note that r and k are both constants, so the integral should be easy.

3. Nov 28, 2008

### noppawit

Now the problem is what is ramda, so I can change dq to ramda*ds. After that, I will be able to evaluate from 0 to 0.13 for 1/4 of circumference for q1 and evaluate from 0.13 to 0.515 for the rest of circumference for q2.

4. Nov 28, 2008

### Staff: Mentor

You can certainly solve the problem by finding the linear charge density λ, and then evaluating ∫(kλ)/r ds . λ is just the total charge divided by the length.

But that's the hard way. Since k, λ, and r are constant, you'd just end up evaluating ∫ds which should be trivial. The easy way is to recognize that ∫dq = Q, which is given.

But go ahead and do it either way.