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Electric field of a rod

  1. Feb 8, 2012 #1
    I am not sure whether this is the correct section to post in but here it goes...

    I am really confused as to why the textbook has a "d" in denominator and why their integral is different from mine...I used the integral in 2nd column 2nd row but the textbook is different :S im not sure why difference in y (I used difference in L to represent the L's segment..) approaches Dy as the textbook states...Anyhow lets break the question into one main problem...I dont understand in the last few textbook lines why my integral is different from theirs even though I used the common integrals from appendix and also, how they got a d in denominator..

    I solved question on my own then looked at textbook's answer.

    http://img7.imageshack.us/img7/2696/integralsd.png [Broken]
    http://img834.imageshack.us/img834/5516/qsanspg1.jpg [Broken]
    http://img560.imageshack.us/img560/8483/qsanspg2.jpg [Broken]
    http://img195.imageshack.us/img195/525/qspg1.png [Broken]
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 8, 2012 #2


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    The textbook's derivation is using a different integral. Their integrand does not have a y in the numerator but instead has d. They use the first integral in the second column of the list you give and from there we see where the extra fact of d comes from. So what you need to do is look to see why your integrand differs from the textbook's. This seems to come from the beginning of the derivation as you can see that the d in the numerator comes from the expansion of \cos \theta_i in terms of y_i and d.
  4. Feb 8, 2012 #3
    Im so confused why is the integral the first one rather than the 2nd one? If I turned my difference in L to difference in y which becomes dy then d*dy is what numerator would be and 2nd integral looks closer match for it than the 1st integral.
  5. Feb 8, 2012 #4


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    Read through the textbook's explanation. It's really just the first step. They write out the electric field due to an infinitesimal length of charged rod and then convert that equation into Cartesian coordinates. Then they convert the charge element on the infinitesimal length into a charge density (assuming that the rod is uniformly charged). Finally, they just integrate along the length of the rod.

    But looking through your notes, it appears that you have an equivalent set of steps. Your integrand is the same, but you took the wrong integral. You should be taking the first integral.
  6. Feb 8, 2012 #5

    thats what im wondering, how do you know which integral to take because 1 and 2 are both very similar and 2 matches more closely to my equation so any ideas on why I should be taking integral 1?
  7. Feb 8, 2012 #6
    got it!! its because my numerator is dy and the constant next to it is d not "y" so I cant use the 2nd integral.
  8. Feb 8, 2012 #7
    Your integrand is the same, but you took the wrong integral. You should be taking the first integral.http://www.bosin.info/g.gif [Broken]
    Last edited by a moderator: May 5, 2017
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