1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Dipole

  1. Jan 26, 2015 #1
    1. The problem statement, all variables and given/known data
    An electric dipole consists of two charges of equal magnitude ##q## and opposite sign, which are kept at a distance ##d## apart. The dipole moment is ##p= qd## .
    Let us next place two such dipoles, placed at distance ##r## apart, as shown in the accompanying figure.
    upload_2015-1-26_15-13-44.png
    a) Assuming that the potential energy for the charges while at infinity is zero, find the exact potential energy of the configuration in terms of ##d, r, q ## and fundamental constants.

    b)When ##d<<r## , approximate your previous result in terms of ##p, r ## and fundamental constants.

    2. Relevant equations
    ##V = \frac {1}{4\pi \epsilon_0} \frac{q}{r}##
    ##W = Vq##

    3. The attempt at a solution
    For the positve charge of A dipole, I calculated the potential energy
    ##= +q \cdot \frac{1}{4\pi \epsilon _0} (\frac{-q}{r} + \frac {+q}{\sqrt{r^2 +d^2}} + \frac{-q}{d}) ##, because, ##W = Vq##
    The potental energy for the other three charges are the same.
    So, the potential energy of the configuration is
    ## = 4q \cdot \frac{1}{4\pi \epsilon _0} (\frac{-q}{r} + \frac {+q}{\sqrt{r^2 +d^2}} + \frac{-q}{d})##
    ##= \frac{q^2}{\pi \epsilon _0} (- \frac{1}{r} + \frac {1}{\sqrt{r^2 +d^2}} - \frac{1}{d})##;

    Then , I can't find any way how to approximate the result when ##d << r##;
    In this case, I substituted ##r^2## for ##r^2 + d^2## ;
    So, two of the terms in bracket are cancelled.
    Then I plugged in ## q = \frac{p}{d}## ;
    But, still ##d## is there.
     
  2. jcsd
  3. Jan 26, 2015 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    This is not correct. The potential energy (q1q2/4πεr) is for a pair of charges separated by a given distance. The way you are implementing it, you are double counting each pair and end up with double the potential energy.

    The term proportional to 1/d in your parenthesis is the energy from the construction of the dipoles themselves. This is generally going to be much higher than the potential between the dipoles. Just substituting ##r^2## for ##r^2 + d^2## is going to make the potential between the dipoles disappear (as you noticed, the terms cancelled) and you will end up with an uninteresting result. Instead, you should make an expansion of the square root for small ##d##.
     
  4. Jan 26, 2015 #3
    I agree with you. So, the potential energy will be half of my calculation.
    So, I need to leave the term ##\frac {1}{d}## ?
    Then it becomes,
    ##\frac{1}{2} \cdot \frac{q^2}{\pi \epsilon _0}(- \frac{1}{r} + \frac{1}{\sqrt{d^2 + r^2}})##
    ##= \frac{q^2}{2 \pi \epsilon _0}(- \frac{1}{r} + \frac{1}{r} \cdot [1 + (\frac{d}{r})^2]^{- \frac{1}{2}})##
    ##= \frac{q^2}{2 \pi \epsilon _0} \frac{1}{r} (-1+ [1 - \frac{1}{2}(\frac{d}{r})^2] )##
    ##= \frac{q^2}{2 \pi \epsilon _0} \frac{1}{r} (- \frac{1}{2}(\frac{d}{r})^2))##
    ##= - \frac{p^2}{4 \pi \epsilon _0 r^3} ##
     
    Last edited: Jan 26, 2015
  5. Jan 26, 2015 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You have series expanded for r << d instead of r >> d. This
    should be
    $$
    \frac{q^2}{2 \pi \epsilon _0 r}\left(- 1 + \frac{1}{ \sqrt{1 + (\frac{d}{r})^2}}\right)
    $$
     
  6. Jan 26, 2015 #5
    It was a typo. Actually, typing in Latex is bothersome. Do you know about any easy-to-use tools for Latex typing?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Electric Dipole
  1. Electric dipole (Replies: 1)

  2. Electric dipole (Replies: 0)

  3. Electric dipole (Replies: 0)

  4. Electric dipole (Replies: 0)

  5. Electric Dipole (Replies: 8)

Loading...