Electric Field of 90° Arc of 2 Wires w/ Uniform Charge - Q/πR^2

[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 each other in the second and 4th quadrants. Show that the electric field at the origin is
$$\frac{4kQ}{\pi R^2}$$

I tried starting with the electric field 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?

 

[Hurkyl]

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.

 

[Gale]

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.

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

 

[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...

 

[Hurkyl]

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.