Electric feild prob

1. Mar 6, 2005

Gale

two wires in the shape of a 9o degrees circular arc of radius R have charges
+(-) Q distributed uniformly on them. They are positioned opposite eachother in the second and 4th quadrants. Show that the electric feild at the origin is
$$\frac{4kQ}{\pi R^2}$$

I tried starting with the electric feild of a line
$$\frac{2k\lambda}{r}$$

$$\lambda=\frac{Q}{d\theta}$$

plugged that in, and then i wasn't sure what to do... integrate with respect to theta? but then i wasn't sure how to do that when theta was in the denom... help?

2. Mar 6, 2005

Hurkyl

Staff Emeritus
You're missing a term in your charge density. Check the units -- they should be charge per length.

Anyways, for the line approximation to be useful, you have to be close enough so that the wire is effectually an infinitely long, straight line. When you're way over at the origin, the wires look neither infinitely long, nor straight.

I would go all the way back to Coulomb's law.

3. Mar 6, 2005

Gale

:grumpy: you're right... back to the drawing board

4. Mar 6, 2005

Gale

ok so coulumbs law... i'm not sure how to get $$\pi$$ out of that...
can i say
$$E=\frac{kdQ}{r^2d\theta}$$

and then if so, can i integrate that... i'm sorry, i'm looking at it and it looks really obvious, but i can't get it to do anything...

5. Mar 6, 2005

Hurkyl

Staff Emeritus
Well, a lot of that expression is constant...

I don't think you have the expression right though... I still think you're missing a factor in the charge density, and you need to use the vector form.