Electric field at the semi-circle's center

  • Thread starter Colts
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Homework Statement



Charge Q is uniformly distributed around the perimeter of a semicircle of radius R. What is the electric field at the circle's center?

Homework Equations


E=kQ/R^2


The Attempt at a Solution


λ=Q/(∏R)

∫(kλ)/R^2 dθ from 0 to pi

and get the answer to be (kQ)/R^3

or is it

∫(kλ)θR/R^2 dθ from 0 to pi

and get (kλ∏^2)/2R

The second one I was trying to put the arc length in, but I'm not sure if you need to or not.
 

Answers and Replies

  • #2
TSny
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Remember that electric field is a vector quantity. Different elements of the semicircle produce electric field at the center in different directions. You need to take that into account. You have probably seen other examples where this is true, such as the field from a straight line of charge.
 
  • #3
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I forgot about that. Thanks. So Since it is a semi-circle. All the charges of the i or j direction (depending on how the circle is orientated) will cancel out. So would I multiply the first integral by cos(θ) to get the "vertical" component (once again depending how it is orientated, but I am thinking of it as a bowl right now). But what would the range of the integral? -pi/2 to pi/2?
 
  • #4
TSny
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I forgot about that. Thanks. So Since it is a semi-circle. All the charges of the i or j direction (depending on how the circle is orientated) will cancel out. So would I multiply the first integral by cos(θ) to get the "vertical" component (once again depending how it is orientated, but I am thinking of it as a bowl right now). But what would the range of the integral? -pi/2 to pi/2?
Yes, that sounds good.
 

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