What is the electric field at the center of curvature of an arc?

[Thomas_]

Homework Statement

Determine the field at the center of curvature of an arc of arbitary angle \alpha

(\alpha is with the x-axis)

Homework Equations

E=\frac{kQ}{R^2}\widehat{r}

S=R\alpha

\lambda=\frac{Q}{S}

The Attempt at a Solution

I divide the arc into small pieces ds. ds=rd\alpha

\lambda = \frac{dQ}{ds}

This gives me: dE = \frac{k\lambda d \alpha}{R}\widehat{r}
I would have to integrate this / its components (X and Y)

the solution I'm given is: E_y=\frac{2k \lambda sin(\frac{\alpha}{2})}{R}

However, I don't understand the following: Why is there only a solution for Y? As far as I can tell the X components don't cancel since its an arc of arbitrary length, not a half or full circle. Also, shouldn't the Y component be in terms of cos? The angle alpha is with the X axis, I don't understand why E_y would be in terms of sin (since it become cos after integration).

Thank you!

 

[Doc Al]

Looks to me like they oriented the arc so that it's centered on the y-axis. (The statement that "\alpha is with the x-axis" does seem a bit odd.) Of course the easy way to solve the problem regardless of the orientation of the arc, is to align it with some axis, solve the integral as show, then transform it back to the given axis.

 

[Thomas_]

Well, they give me a picture with the arc starting from the positive x axis, going through an angle alpha which is counterclockwise from the positive x-axis. That's what is confusing me...

Also, shouldn't there be two components to the elctric field? X and y? Regardless of where the arc is oriented?

 

[Doc Al]

Well, they give me a picture with the arc starting from the positive x axis, going through an angle alpha which is counterclockwise from the positive x-axis. That's what is confusing me...

That does seem to contradict the answer.

Also, shouldn't there be two components to the elctric field? X and y? Regardless of where the arc is oriented?

By symmetry, you know the electric field must be along the line from the center of curvature to the midpoint of the arc. To find the magnitude, just choose an axis along that direction to do your integration.

Whether the field has x and y components depends on its orientation to the axis. If the arc were centered on the x-axis, it would only have an x-component. For whatever reason, in contradiction to the picture you were given, they chose to align the arc with the y-axis. But according to how you described the picture, it would have both an x and y component.