Nonuniform Semicircle of Charge

  • Thread starter GenDis
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  • #1
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Homework Statement



A non-uniformly charged semicircle of radius R=23.3 cm lies in the xy plane, centered at the origin, as shown (picture shows circle going from 0 to pi). The charge density varies as the angle (in radians) according to =−385, where has units of C/m. What is the x component of the electric field at the origin?

Homework Equations



dq = λ ds
dE = kdq/(r^2)


The Attempt at a Solution



I already found the total charge to be -4.43E-6 C. I know the charge is nonuniformly distributed, but given symmetry shouldn't the x component simply end up being 0?
 

Answers and Replies

  • #2
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No, it doesn't have to be zero. Imagine that most of the charge is located on one side, or if on one side it's more to the top and the other has more to the side; the x-component wouldn't be 0.
 
  • #3
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Ya that makes sense, I wasn't thinking. So here's what I got when I actually went after the problem.

dE = kq/(r^2), but since dq = λds we have

dE=kλds / r^2

integrating both sides from 0 to pi gave me

(k)(-3.85 E-9)*(pi^2) / R

Not the right answer, did I simply integrate incorrectly or was it not set up correctly? Do I need a cos(theta) in the integral?
 

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