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Continuous Distribution of Charges Problem

  • Thread starter BrainMan
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Homework Statement



Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in the figure (Figure 1) .

Find an expression for the electric field E⃗ at the center of the semicircle.
Hint: A small piece of arc length Δs spans a small angle Δθ=Δs /R, where R is the radius.
Express your answer in terms of the variables Q, L, unit vectors i^, j^, and appropriate constants.

upload_2017-2-24_20-18-25.png

Homework Equations




The Attempt at a Solution


upload_2017-2-24_20-21-28.png

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I drew a picture and established a coordinate system. Then I drew the vector of the electric fields at two point charges at point P. Then I saw that the y components of the field will cancel because of the symmetry. I then wrote what the field should be in terms of x. I'm not sure what to do now or how to get to a point where I could integrate.
 

Answers and Replies

  • #2
blue_leaf77
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I then wrote what the field should be in terms of x.
You mean ##x## as a coordinate? That's not necessary since you are asked to calculate the field on a fixed coordinate which is the center of the ring.
I'm not sure what to do now or how to get to a point where I could integrate.
After defining ##\lambda## to be the charge linear density ##Q/L = dQ/ds## you can integrate the last equation in your work.
 
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You mean ##x## as a coordinate? That's not necessary since you are asked to calculate the field on a fixed coordinate which is the center of the ring.

After defining ##\lambda## to be the charge linear density ##Q/L = dQ/ds## you can integrate the last equation in your work.
OK I substituted the linear charge density and integrated from 0 to L. Somethings still wrong though.
 
  • #4
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OK I substituted the linear charge density and integrated from 0 to L. Somethings still wrong though.
 
  • #5
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Here's what I did:

E = ∫dE = ∫(K*dQ*cosθ)/r^2 = ∫(K*λ*dS*cosθ)/r^2) = (Kλ)/r^2 ∫cos(S/r)ds = (Kλ/r) ∫ sin(S/r) ] 0 to L = (Kλ/r)[sin(L/r)-1]
 
  • #6
blue_leaf77
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∫(K*λ*dS*cosθ)/r^2) = (Kλ)/r^2 ∫cos(S/r)ds
Leave ##\theta## as it is and use ##ds = rd\theta## and integrate with the angle. So that it will be
$$
\int \frac{k\lambda}{r^2} \cos\theta \ r\ d\theta
$$
Upon doing the integration, you may just integrate the upper/lower half of the semicircle owing to the symmetry.
 
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Leave ##\theta## as it is and use ##ds = rd\theta## and integrate with the angle. So that it will be
$$
\int \frac{k\lambda}{r^2} \cos\theta \ r\ d\theta
$$
Upon doing the integration, you may just integrate the upper/lower half of the semicircle owing to the symmetry.
So I did (Kλ/r) ∫ cosθ dθ and got (Kλ/r) sinθ ] -pi/2 -> pi/2

So my final answer is 2Kλ/r = (2KQ) / (RL)

But my homework keeps saying "The correct answer does not depend on: KQ, RL."
 
  • #8
kuruman
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Q and L are the given quantities, other than the appropriate constants. R is not given. Can you express R in terms of L?
 
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  • #9
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Q and L are the given quantities, other than the appropriate constants. R is not given. Can you express R in terms of L?
So I changed my answer to 2KQπ/L^2 and it worked. Thanks!
 

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