E field from half ring of charge.

1. Feb 19, 2007

seang

1. The problem statement, all variables and given/known data
How do I find the electric field from a semi-ring of charge at the center of the would-be circle? Does that make sense? Or another way: If the semi-circle of charge is centered at the origin, what is the electric field at the origin? I hope you know what I mean now.

the total charge of the ring is Q, the radius is R.

Oh yeah, the semi circle is in the left hand plane

2. Relevant equations
$$E = k\int_{}^{l}\frac{(r-r')\rho _l(r')}{|r-r'|^3} dl'$$
3. The attempt at a solution

$$E = k\int_{\pi/2}^{-\pi/2}\frac{(-r')\rho _l(r')}{|-r'|^3} d\theta$$

Is converting to cylindrical coordinates the right move? Is this even the equation I should be using? What should I use as the charge density?

Last edited: Feb 19, 2007
2. Feb 20, 2007

chaoseverlasting

Try applying Gauss's Theorem. Assume a gaussian surface enclosing the ring and $$\int E.ds=\frac{q}{\epsilon}$$

3. Feb 20, 2007

Meir Achuz

Gauss won't work here because there is not full circular symmetry
There is enough symmetry to see the direction of the E field.
Just put cos\theta into the numerator of your integral.
rho is really a linear density lambda=Q/2pi R.

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