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Calculate the x-component of the electric field

  • Thread starter xxaznitex
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  • #1
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yf_Figure_21_37.jpg


Positive charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positive x-axis at x=a+r, a distance r to the right of the end of Q.

Homework Statement


Calculate the x-component of the electric field produced by the charge distribution Q at points on the positive x-axis where x>a (i.e., r>0) in terms of some or all of the variables k, q, Q, a, and r, where k=\\frac{1}{4\\pi\\epsilon_0}.


The Attempt at a Solution


I'm having a problem visualizing the distance between the charge and the point. I understand that e=KQ/R^2. However, I have no idea how to get R.
I would do: (k*dQ)/(r+a-x)...
(kQ/a) [integral]1/(r+a-x)dx


But I don't think that looks right. Can anyone give me a few pointers?
 
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Answers and Replies

  • #2
Delphi51
Homework Helper
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Looks great! Except you forgot to square the distance (r+a-x) in the integral.
Also, something must be done about dq to express it in terms of dx so it will be possible to integrate. The charge on a segment dx would be dq = (q/a)dx because the charge per unit length is q/a.
Finally, the integral is awkward with the r and a in there. You can get it with math techniques of course (let R= r+a-x), but it might be better to think of the distance as R and integrate it from R = a+r to r. Of course dR = dx.
 
  • #3
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Thank you for your help. I needed someone to confirm the distance for me since I was completely lost on that part.
 

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