Potential difference due to a continuous charge distribution

[Alec11]

This is my first time using this site so please excuse me if I missed any guidelines.

1. Homework Statement
A plastic rod having a uniformly distributed charge Q=-25.6pC has been bent into a circular arc of radius R=3.71cm and central angle ∅=120°. With V=0 at infinity, what is the electric potential at P, the center of curvature of the rod?

Homework Equations

V=∫dV=∫kdq/r

The Attempt at a Solution

V=∫dV=∫kdq/r=(kλ)∫dΘ=(kλ)(2π/3), which is wrong.

The issue that I'm having with this problem is that I don't understand why the integration is done the way that it is. The equation I'm given to solve it is V=∫kdq/r, which equals out to V=kq/r after the integration. I'm confused because this is just the same exact equation used to find V for point charges. Why are you not required to put the integral in terms of theta and integrate over the length of the arc? If the charge distribution were kept as a straight rod, then I would have to integrate over the entire distribution, but since its a circular arc that just goes out the window.

 

[TSny]

WELCOME TO PF!

V=∫dV=∫kdq/r=(kλ)∫dΘ=(kλ)(2π/3), which is wrong.

Why do you say this is wrong?

The equation I'm given to solve it is V=∫kdq/r, which equals out to V=kq/r after the integration. I'm confused because this is just the same exact equation used to find V for point charges. Why are you not required to put the integral in terms of theta and integrate over the length of the arc?

Consider V=∫kdq/r.
k is a constant. Is r also a constant for this integration?

 

[Alec11]

Because I tried doing that for this homework problem and it results in a wrong answer. I looked up how to do it and someone said that you just reduce the equation to V=kq/r, which gives the correct answer. This held true for every homework problem with circular arc shaped charge distributions. And yes, r is a constant for this integration.

 

[TSny]

Because I tried doing that for this homework problem and it results in a wrong answer.

What value did you use for λ?

And yes, r is a constant for this integration.

OK, so what does the integral ∫kdq/r reduce to if you "pull out" all the constants?

 

[Alec11]

What value did you use for λ?

I calculated λ=Q/(rθ) = (-25.6x10^(-12))/((2π/3)*(0.0371)) = (-3.29x10^(-10)) C/m

OK, so what does the integral ∫kdq/r reduce to if you "pull out" all the constants?

It becomes (k/r)∫dq

 

[TSny]

I calculated λ=Q/(rθ) = (-25.6x10^(-12))/((2π/3)*(0.0371)) = (-3.29x10^(-10)) C/m

Looks good. What value did you get for V when you used this value of λ?

It becomes (k/r)∫dq

Good. What does ∫dq evaluate to?

 

[Alec11]

Looks good. What value did you get for V when you used this value of λ?

I get V = -6.20V, which actually turns out to be right. Looking back at my previous attempts, it seems I accidentally went from dq => rθdθ, instead of dq => rdθ, which explains why I thought I was wrong originally. Thanks for helping me figure that out!

 

[TSny]

OK. Glad it got cleared up.