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Electric field of a rod

  1. Feb 11, 2008 #1

    tony873004

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    Can someone tell me if I did this correctly, or if there's an easier way? Thanks!

    A uniformly charged rod of length [tex]\ell [/tex] with a charge density [tex]\lambda[/tex] lies along the x-axis, with its midpoint at the origin. Find the electric field at a point on the x-axis, with [tex]x > \ell /2[/tex]

    The electric field of a differential element is
    [tex]\overrightarrow {dE} = \frac{{kdQ}}{{r^2 }}[/tex]

    The charge of a differential element is
    [tex]dQ = \lambda \,dx[/tex]

    Let P be a point on the x-axis to the right of the rod.
    [tex]\overrightarrow E = \int\limits_{ - \ell /2}^{\ell /2} {\frac{{k\lambda \,dx}}{{r^2 }}} [/tex]

    The distance r from P to a differential element x is P-x
    Pull out the constants.
    [tex]\begin{array}{l}
    \overrightarrow E = k\lambda \int\limits_{ - \ell /2}^{\ell /2} {\frac{{\,1}}{{\left( {P - x} \right)^2 }}dx} \\
    \\
    \overrightarrow E = k\lambda \left[ {\frac{1}{{P - x}}} \right]_{ - \ell /2}^{\ell /2} = k\lambda \left( {\left( {\frac{1}{{P - \ell /2}}} \right) - \left( {\frac{1}{{P - \left( { - \ell /2} \right)}}} \right)} \right),\, - {\rm{\hat i}} \\
    \\
    \overrightarrow E = k\lambda \left( {\left( {\frac{1}{{P - \ell /2}}} \right) - \left( {\frac{1}{{P + \ell /2}}} \right)} \right) \\
    \end{array}[/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 11, 2008 #2
    IMO, its correct.
     
  4. Feb 11, 2008 #3

    tony873004

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    I just realized that the answer is in the back of the book. They give:
    [tex]\frac{{k\lambda \ell }}{{x_0^2 - \frac{1}{4}l^2 }}{\rm{\hat i}}[/tex]

    Where did I go wrong? I'm guessing that their x0 is what I was calling P. But the answers are not the same. I don't believe they are the same formula just written differently.

    Also, the direction given is i-hat. But since the problem never states what the charge of the rod is, how do I know that it is not negative i-hat?
     
  5. Feb 11, 2008 #4
    Since x > P, then r should be (x - P). I'm able to get the answer without the L in the numerator. Dont know where thats coming from.
     
  6. Feb 11, 2008 #5

    tony873004

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    x > L/2, meaning the point is to the right of the rod. I choose P as the point so x could be my integration variable. So, for example if the rod's length were 10, it would span from x= -5 to 5. And if P was 15, then the distance from P to each element in the rod would be P-x. For example P is 15-5=10 units from the right of the rod.

    I just verified using my calculator that my formula and the book's formula are the same. I just made up some numbers and plugged them in: k=12, lambda=13, L=14, P=15

    12*13*((1/(15-14/2))-(1/(15+14/2))) = 12.4090909090909
    12*13*14/(15^2-14^2/4) = 12.4090909090909

    The answers are the same. So its just a matter of algebra to arrive at their formula from my formula. But I don't know how. I'm wondering if I care. Is their formula any more correct than mine?
     
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