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Derivation of the electric field from the potential

  1. Sep 26, 2004 #1
    :confused: I am studying for a test and i can't figure out for the life of me how my book derived the solution for this problem I know it has to be basic i just dont see it...

    An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2a... The dipole is along the x axis and is centered at the origin.
    calculate V and Ex if point P is located anywhere between the two charges.

    I understand the concept of this, and have calculated V, which is [ (2*(Ke)*q*x) / ((a^2) - (x^2)) ] and i know how to start the problem of Ex...

    Ex = - (dV/dx) = - [ (2*(Ke)*q*x) / ( (a^2) - (x^2) ) ....

    But I can't remember or figure out for the life of me how they got

    = - 2*(Ke)*q * [ { (a^2) + (x^2) } / { ( (a^2)-(x^2) )^2 } ]

    Can Anyone please help? :D
    Last edited: Sep 26, 2004
  2. jcsd
  3. Sep 26, 2004 #2


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    Science Advisor
    Homework Helper

    Your potential should look something like this
    [tex]q \left( \frac {1}{ {a-x}} - \frac {1}{ {a+x}}\right)[/tex]
    so just combine the fractions.
  4. Sep 26, 2004 #3
    yes that is correct :D.. but i already got past that point and found the answer for the electric potential, V ... which was the sum of that equation...

    and got (2KeqX) / ( a^2 - x^2) But that just gives me the potential... I needed to take it a step further and find the magnitude of the electric field from that... which is Ex ... which is = -dV/dx ... but i dont see how they derived the answer ...

    = - 2*(Ke)*q * [ { (a^2) + (x^2) } / { ( (a^2)-(x^2) )^2 } ]

  5. Sep 26, 2004 #4

    Doc Al

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    Staff: Mentor

    Is your problem that you don't know how to take the derivative?
  6. Sep 28, 2004 #5
    You have to your the quotient rule!!!!
  7. Sep 28, 2004 #6
    Sorry, I mean : You have to use the quotient rule!!!
  8. Sep 28, 2004 #7
    Quotient Rule

    Please see the attached file. You will see how the quotient rule is require to get that answer.

    Attached Files:

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