E field from half ring of charge.

  • Thread starter seang
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  • #1
seang
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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
chaoseverlasting
1,039
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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
3,605
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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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