electric field

[ahuebel]

I haev a hw problem where I need to find the magnitude and direction of the electric field from a thin, uniformly charged 18cm rod bent into a semicircle. The test point is at the center of the semicircle.

It seems pretty basic but I am stuck on a point (assuming I am even on the right track). Because of symmetry I assume Ex = 0 so I just need to find Ey. the y component of dE is k*(dq/r^2) * sin (theta).
So the sum of the little dE's can be found by integrating across the semicircle. The integral would look something like 2k*dq/r^2 *[integral from 0 to pi/2]sin (theta). I know dq=lambda/dL where L is the length of the bent rod. How do I go from dL to d(theta)?

Thanks!

[lightgrav]

a differential can only be proportional to another differential;
if you make dL smaller, then dq will also be smaller!

Your error is : >I know dq=lambda/dL

if the total length is L (half a circle circumference...) and
the total charge is Q , then the "charge per unit length" lamda = Q/L.

[ahuebel]

yes you are right. That was a mistake on my part. dq=lambda*dL, but I still have the same issue of going from dL to d(theta) right?

[lightgrav]

The length of a segment of circle is L = theta * r .
you know the entire length (180 mm) and the entire angle (pi rad).

[ahuebel]

I must be missing something. This segment of a circle would seem to corrolate to arclength which may or may not be a distance from a point where theta = pi or theta = 0. It may be a small section -- like a sliver of a pie or something. This would mean this "dL" would be the arclength between two side of distance r (radius of the semicircle). If that is the case the segment would be close to 2rsin(theta') where theta' is the angle between two lines.

[FluxCapacitator]

Think about it this way. Each little bit of the semicircle is contributing to the overall electric field. Adding up all of those little bits will get you the final field. You can express those bits in many ways, from dq's to d\theta's. The math requires that you have an angle to integrate, thanks to the \sin\theta. So if you can find an expression for the charge per angle, then you can integrate that. You already found that, so integrating it will get you the proper answer.