Circular arc of charge, integration question

[dimensionless]

I have a circular arc of wire centered at the point (0,0). It has a radius of r, extends from \theta = -60 to \theta = 60 and also holds a charge q. For the differential electric field I have the following equation:

<br /> dE = \frac{\lambda ds}{4 \pi \epsilon_0 r^2}<br />
Where ds is the length of a differential element of the arc.

To the find the x component of the electric field I this equation:

<br /> dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} ds<br />

To integrate this I have to set ds = r d\theta so that the above equation reads:

<br /> dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} r d\theta<br />

Where does the r come from in the statement ds = r d\theta?

 

[nrqed]

I have a circular arc of wire centered at the point (0,0). It has a radius of r, extends from \theta = -60 to \theta = 60 and also holds a charge q. For the differential electric field I have the following equation:

<br /> dE = \frac{\lambda ds}{4 \pi \epsilon_0 r^2}<br />
Where ds is the length of a differential element of the arc.

To the find the x component of the electric field I this equation:

<br /> dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} ds<br />

To integrate this I have to set ds = r d\theta so that the above equation reads:

<br /> dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} r d\theta<br />

Where does the r come from in the statement ds = r d\theta?

That is just the relation between an arclength and the subtended angle in a circle. Recall that, as long as the angle is given in radians, we have s = r \theta, right? This is true for any arc fo a circle. If you tell me that an arc of a circle subtends {\pi \over 8}radians, for example, and that the circle has a radius of 20 cm, then the length of the arc is simply {5 \pi \over 2} cm.

For an infinitesimal subtended angle d \theta the relation is obviously that the infinitesima arc length is ds = r d \theta.