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:

$$dE = \frac{\lambda ds}{4 \pi \epsilon_0 r^2}$$
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:

$$dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} ds$$

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

$$dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} r d\theta$$

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:

$$dE = \frac{\lambda ds}{4 \pi \epsilon_0 r^2}$$
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:

$$dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} ds$$

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

$$dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} r d\theta$$

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 $ds = r d \theta$.