Integrating an Electric Field Due to an Arc of Charge in Polar Coordinates

[quarkman]

How can I not do this??

I am trying to determine the electric field due to an arc of charge which spans the angles 0 to $$\pi$$ with a total charge q. Does anyone know why I cannot integrate this directly? The only way I get the right answer is to convert $$a_{\rho}$$ (unit vector in radial direction) to x,y coordinates and then integrate. I was just wondering if I could integrate with everything still expressed in polar form:

Begin with the following formula for the electric field:

$$\vec{E} = \int_{0}^{\pi} \frac{\rho_{L} d\ell (-a_{\rho})}{4 \pi \epsilon R^{2}}$$

(Where R is the radius of the arc and $$\rho_{L}$$ is the charge density of the arc.)

Then convert $$d\ell$$ to $$R d\phi$$ and integrate over the limits of the arc. This is all fine and dandy, (as $$\int^{\pi}_{0} d\phi$$ is obviously $$\pi$$ in this case) but when I convert from the radial unit vector to the rectangular ones $$a_{\rho} = \cos \phi a_{x} + \sin \phi a_{y}$$ I find myself very confused by the introduction of the angle $$\phi$$ again. I can do this and get the correct answer if I convert the radial unit vector to the rectangular ones first, but I want to know if it is possible to integrate without the introduction of rectangular coordinates (converting the polar electric field vector to rectangular coords after the integration is done). Thanks for any help. If I am not clear enough I can try to rephrase my problem as I enjoyed learning how to use the latex typesetting

 

[Galileo]

Your radial unit vector $\hat a_{\rho}$ is not a constant, but a function of $\phi$.
I`m not sure how you did it, but from your post it looks as if you integrated oer $\phi$ first, and THEN expressed the radial unit vector in rectangular coordinates.
This is wrong, because then you treated $\hat a_{\rho}$ as a constant,
but since it dependent on the angle you should change to rectangular coordinates first and then integrate (like you did when you found the correct answer).

The resultant electric field points in the negative y-direction (if q is positive).
Ofcourse, you can express that in polar coordinates.

 

[quarkman]

Your radial unit vector $\hat a_{\rho}$ is not a constant, but a function of $\phi$.
...
The resultant electric field points in the negative y-direction (if q is positive).
Ofcourse, you can express that in polar coordinates.

Why is the unit vector $$\hat a_{\rho}$$ not constant? Doesn't it always point in a radial direction? In this case I have specified it to point towards the origin using the minus sign.

I thought the purpose of polar coordinates was to simplyfy problems with circular symmetry. But here I have had to revert to rectangular ones to solve a problem with circular symmetry! Since the general form for the electric field due to a line of charge is:
$$\vec{E} = \int_{0}^{\pi} \frac{\rho_{L} d\ell (\hat a_{r})}{4 \pi \epsilon R^{2}}$$

where $$a_{r}$$ is the unit vector from the line of charge to the point we want to find the field. Using the definition for the unit vector we can write it in terms of the vector $$\vec{r}$$ and it's magnitude $$R$$. Like this:

$$\hat a_{r} = \frac{\vec{r}}{R}$$

My question/confusion is now how to express $$\vec{r}$$ in polar coordinates. Originally I thought this was:

$$\vec{r} = (-R a_{\rho}) + (\phi a_{\phi})$$

However this does not yield the correct answer. (You will recall that I want the field at the origin.)

 

[arildno]

As far as I can see, only the radial vector is a non-constant quantity in your original integral (which is to give the electric field at the origin)
Use therefore:
$$\int_{0}^{\pi}\vec{a}_{r}d\phi=(-\vec{a}_{\phi})|_{\phi=0}^{\phi=\pi}$$