Electric dipole

  • #1
Hello

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
 

Answers and Replies

  • #2
62
1
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!?
 
  • #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|
 
  • #4
62
1
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?
 
  • #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]
 
  • #6
62
1
Absolutely right! I hope that is a sufficient answer to your question!?
 
  • #7
Yes I can now see the whole alegebra behind the equations in the book!

Thanks very much Saunderson!
 

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