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E field from half ring of charge.

  • Thread starter seang
  • Start date
  • #1
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Homework Statement


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



Homework Equations


[tex]
E = k\int_{}^{l}\frac{(r-r')\rho _l(r')}{|r-r'|^3} dl'

[/tex]

The Attempt at a Solution



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

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:

Answers and Replies

  • #2
Try applying Gauss's Theorem. Assume a gaussian surface enclosing the ring and [tex]\int E.ds=\frac{q}{\epsilon}[/tex]
 
  • #3
Meir Achuz
Science Advisor
Homework Helper
Gold Member
2,171
65
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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