Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Electric Dipole derivation (Algebra)

  1. Aug 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that

    [tex] E = \frac {q}{4\pi\varepsilon(z-\frac{1}{2}d)^2} - \frac{q}{4\pi\varepsilon(z+\frac{1}{2}d)^2}
    [/tex]

    is

    [tex]E = \frac {q}{4\pi\varepsilon z^2} [(1-\frac{d}{2z})^{-2} - (1+\frac{d}{2z})^{-2}][/tex]



    2. Relevant equations



    3. The attempt at a solution

    After getting a common denominator and subtracting I got

    [tex] \frac {2qdz}{4\pi\varepsilon(z^4-\frac{1}{2}d^2z^2+\frac{1}{16}d^4)}[/tex]

    Stuck completely at this point. I've tried working backwards but that is even more confusing. Any help would be appreciated. Thanks.
     
  2. jcsd
  3. Aug 15, 2010 #2
    [tex]
    E = \frac {q}{4\pi\varepsilon(z-\frac{1}{2}d)^2} - \frac{q}{4\pi\varepsilon(z+\frac{1}{2}d)^2}
    [/tex]

    [tex]
    E = \frac {q}{4\pi\varepsilon}[(z-\frac{1}{2}d)^{-2}} - (z+\frac{1}{2}d)^{-2}}]
    [/tex]

    [tex]
    E = \frac {q}{4\pi\varepsilon}[(z(1-\frac{d}{2z}))^{-2}} - (z(1+\frac{d}{2z}))^{-2}}]
    [/tex]

    [tex]
    E = \frac {q}{4\pi\varepsilon}[z^{-2}(1-\frac{d}{2z})^{-2}} - z^{-2}(1+\frac{d}{2z})^{-2}}]
    [/tex]

    [tex]
    E = \frac {q}{4\pi\varepsilon z^2} [(1-\frac{d}{2z})^{-2} - (1+\frac{d}{2z})^{-2}]
    [/tex]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook