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Electric dipole

  1. Feb 25, 2010 #1

    The electric dipole potential is

    [tex] V (\textbf{r}) = \frac{1}{4\pi\epsilon_o} \frac{\textbf{p}\bullet \hat{r} }{r^2}[/tex]

    I am trying to figure out the algebra in my book. How do you evaluate the scalar product of the dipole moment and the unit vector in the above equation?

    I get [tex]qd cos \theta[/tex] but i am not sure if that is right.

    Thank you
  2. jcsd
  3. Feb 25, 2010 #2
    You might expand the fraction with [tex]|\vec r|[/tex]!

    This yields

    [tex]\phi (\vec r) = \frac{1}{4\pi\varepsilon_o} \frac{\vec p \cdot |\vec r^{\,}|\hat{r} }{r^3}[/tex]​

    Maybe you cope with this!?
  4. Feb 25, 2010 #3
    I dont understand how you expand the fraction. The vector, r in brackets is the norm of r?

    I see that r-hat is the unit vector. Shouldnt r-hat = r / |r|
  5. Feb 25, 2010 #4
    you're right

    [tex]\hat r = \frac{\vec r}{|\vec r^{\,}|} \qquad \Leftrightarrow \qquad \hat r \cdot |\vec r^{\,}| = \vec r[/tex]​

    this applied to the primary equation yields

    [tex]\phi (\vec r) = \frac{1}{4\pi\varepsilon_o} \frac{\vec p \cdot |\vec r^{\,}|\hat{r} }{r^3} = \frac{1}{4\pi\varepsilon_o} \frac{\vec p \cdot \vec {r} }{r^3}[/tex]​

    it follows, that

    [tex]\vec p \cdot \vec r = p r \cos \vartheta[/tex]​

    do you agree?
  6. Feb 25, 2010 #5
    So does that mean that after evaluating the scalar product (where p = qd) the electric dipole potential would equal to

    [tex]V(r) = \frac{q d cos\theta}{4 \pi \epsilon_o r^2}[/tex]
  7. Feb 25, 2010 #6
    Absolutely right! I hope that is a sufficient answer to your question!?
  8. Feb 25, 2010 #7
    Yes I can now see the whole alegebra behind the equations in the book!

    Thanks very much Saunderson!
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