Evaluating Electric Dipole Potential: qd cos \theta?

[ronaldoshaky]

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

 

[saunderson]

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!?

 

[ronaldoshaky]

I don't 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|

 

[saunderson]

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?

 

[ronaldoshaky]

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}

 

[saunderson]

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

 

[ronaldoshaky]

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

Thanks very much Saunderson!