# Electric dipole

Hello

The electric dipole potential is

$$V (\textbf{r}) = \frac{1}{4\pi\epsilon_o} \frac{\textbf{p}\bullet \hat{r} }{r^2}$$

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 $$qd cos \theta$$ but i am not sure if that is right.

Thank you

You might expand the fraction with $$|\vec r|$$!

This yields

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

Maybe you cope with this!?

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|

you're right

$$\hat r = \frac{\vec r}{|\vec r^{\,}|} \qquad \Leftrightarrow \qquad \hat r \cdot |\vec r^{\,}| = \vec r$$​

this applied to the primary equation yields

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

it follows, that

$$\vec p \cdot \vec r = p r \cos \vartheta$$​

do you agree?

So does that mean that after evaluating the scalar product (where p = qd) the electric dipole potential would equal to

$$V(r) = \frac{q d cos\theta}{4 \pi \epsilon_o r^2}$$

Absolutely right! I hope that is a sufficient answer to your question!?

Yes I can now see the whole alegebra behind the equations in the book!

Thanks very much Saunderson!