- #1

Thomas_

- 21

- 0

## Homework Statement

Determine the field at the center of curvature of an arc of arbitary angle [tex]\alpha[/tex]

([tex]\alpha[/tex] is with the x-axis)

## Homework Equations

[tex]E=\frac{kQ}{R^2}\widehat{r}[/tex]

[tex]S=R\alpha[/tex]

[tex]\lambda=\frac{Q}{S}[/tex]

## The Attempt at a Solution

I divide the arc into small pieces ds. [tex]ds=rd\alpha[/tex]

[tex]\lambda = \frac{dQ}{ds}[/tex]

This gives me: [tex]dE = \frac{k\lambda d \alpha}{R}\widehat{r}[/tex]

I would have to integrate this / its components (X and Y)

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

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!