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Homework Help: Find an expression for the electric potential at P

  1. Jan 31, 2010 #1
    The thin, uniformly charged rod shown in the figure below has a linear charge density λ. Find an expression for the electric potential at P. (Use k_e for ke, lambda for λ, a, b, and L as necessary.)




    2. Relevant equations
    dV=k*dq/r
    lambda*dx=dq

    3. The attempt at a solution
    I integrated using the above equations and went from 0 to L. I thought this was the right way to do it but I keep getting the wrong answer.
     

    Attached Files:

  2. jcsd
  3. Jan 31, 2010 #2
    Hello Keithkent09.

    First: Show us your work and what you got for your integral.
    When you integrate your potential equation you are integrating over the
    location of charge correct? The integration does not happen from 0 to L. Your
    charge location is not from 0 to L do you see? What should it be?
     
  4. Jan 31, 2010 #3
    the integral of (k*dq)/r=(k*lambda+dx)/r
    r=sqrt(b^2+(a+x)^2) so...
    k*lambda*dx/(sqrt(b^2+(a+x)^2))...and was not sure what to do after this
     
  5. Feb 1, 2010 #4

    Redbelly98

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    The next step would be to do the integral. However ....

    ...here is a suggestion: let x=0 at the point directly below P. That will simplify the integral somewhat. Of course, this will change the limits on x: instead of 0 to L it will be ____ to ____ instead.
     
  6. Feb 1, 2010 #5
    Will the limits be from 0 to a+L?
    Also if x=0 under point P how does that make the integral any easier, does that mean that x is thrown out in the integral and all that remains is the dx?
     
    Last edited: Feb 1, 2010
  7. Feb 1, 2010 #6

    Redbelly98

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    If x=0 under point P, then the rod will lie between x=a and x=a+L, right?
    Well, set up the integral, using the approach you did before:
    What expression do you get for the integral in this case?
     
  8. Feb 1, 2010 #7
    is it just the integral of (k*lambda*dx)/(sqrt(a^2+b^2) from a to a+L?
     
  9. Feb 1, 2010 #8

    Redbelly98

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    Not quite, the (sqrt(a^2+b^2) part is wrong. There should be an x in there somewhere.

    Note, this term is equivalent to r, the distance from point P to some point along the wire. The point on the wire would be a distance x from the origin directly below P.
     
  10. Feb 1, 2010 #9
    Earlier I said that the denominator was sqrt(b^2+(a+x)^2) i do not understand why the placement of x=0 changes that up.
     
  11. Feb 1, 2010 #10

    Redbelly98

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    Never mind, let's go back to your expression which was correct and not confusing you:
    This needs to be integrated from x=0 to x=L.
    Try looking it up in a table of integrals.
    The following might be useful for doing this integral: try a change of variable, u=a+x.
     
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