# Finding the Electric field at the Origin

## Homework Statement

I'm trying to find the electric field at the origin <0, 0, 0> due to a quarter of a ring with +Q charges on it. The quarter of the ring lies on the +y-axis. Due to symmetry, the x-components of the electric field will cancel out, leaving only the y-component of the electric field pointing in the negative y-direction. So my problem with this is finding the correct limit of integration to use for this problem. At first, I used the limits of integration from 3pi/4 to pi/4. When I used this on the x-component, it didn't give me an exact 0. It gave me like 0.2738 something like that. The answer for the x-components should have been exactly zero. So I don't know if this was the right limits of integration or is there other limits that I can use that I can't think of to make the x-components zero. If my limits of integration that I use doesn't make the x-components zero, then my answer for the electric field for the y-component would be wrong too. I know how to do everything else, but it's the limits that I am stuck on.

## Homework Equations

E = [(kΔq)/(πr^2)] * r hat
r hat = <-cosσ, -sinσ, 0>
Δq = (2(+Q)Δσ)/π

## The Attempt at a Solution

Integrating for the x-component: ∫[2k(+Q)(-cosσdσ)]/[πR^2] = 2k(+Q)/πR^2 ∫-cosσdσ

## Answers and Replies

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TSny
Homework Helper
Gold Member

## Homework Equations

E = [(kΔq)/(πr^2)] * r hat
r hat = <-cosσ, -sinσ, 0>
Δq = (2(+Q)Δσ)/π
OK, except I think you have a typo in the expression for E. Should there be ##\pi## in the denominator?

## The Attempt at a Solution

Integrating for the x-component: ∫[2k(+Q)(-cosσdσ)]/[πR^2] = 2k(+Q)/πR^2 ∫-cosσdσ
This looks correct. (Now there should be the factor of ##\pi## as you have it.)

I used the limits of integration from 3pi/4 to pi/4.
Why "from 3pi/4 to pi/4" rather than "from pi/4 to 3pi/4"?

Your integral should give exactly zero. Don't forget to put your calculator in radian mode if you are using your calculator to evaluate something like ##\sin(\pi/4)## (which you probably know without needing a calculator).