Electric Dipole - Calulating the electic field from Potential

[smoking-frog]

Homework Statement

Calculate the Electric field of a Dipole from its Potential.

\vec E=-\operatorname{grad}(\Phi_D)

Homework Equations

\Phi_D(\vec R)=\frac{Q}{4\pi\epsilon_0} \cdot \frac{\vec d \vec R}{R^3}

The Attempt at a Solution

Hi all!

I am trying to calculate the electric Field of a Dipole from its Potential.

Specifically, I am trying to understand the following equation:\operatorname{grad}(\Phi_D)=\frac{Q}{4\pi\epsilon_0R^3}\cdot((\vec d\cdot \vec R )\cdot \operatorname{grad}(\frac{1}{R^3})+\frac{1}{R^3} \cdot\operatorname{grad}(\vec d \cdot \vec R))

I tried re-writing the equation according to the Einstein Notation:
\operatorname{grad}(\Phi_D)=\frac{Q}{4\pi\epsilon_0}\cdot\frac{\partial}{\partial R_j}\frac{d_k R_l}{(R_i R_i)^{\frac{3}{2}}}

Is that correct? If so, how will I go about proving the above identity?

Thank you all for your help!

 

[clamtrox]

It is fairly straightforward to show, using the definition of gradient and partial derivatives, that gradient satisfies the Leibniz rule, \operatorname{grad}(fg) = f \operatorname{grad}(g) + g \operatorname{grad}(f)
where f and g are scalars

 

[smoking-frog]

But they are not scalar, are they?

\vec d \text{ and } \vec R are vectors...

The rule for vectors is, in my opinion, <br /> <br /> \operatorname{grad}(\vec a \cdot \vec b)=(\vec a \cdot \nabla) \cdot \vec b+(\vec b \cdot \nabla) \cdot \vec a+\vec a \times (\nabla \times\vec b)+\vec b \times (\nabla \times\vec a)

 

[clamtrox]

But they are not scalar, are they?

\vec d \text{ and } \vec R are vectors...

Correct. But I thought in \phi you had \vec{d} \cdot \vec{R}, which I believe is a scalar. Start from that and then note that \vec{d}*is constant, so differentiating \vec{d} \cdot \vec{R} is a bit simpler than that general formula suggests.

 

[smoking-frog]

Oh ok.

I actually tried that already, but the result was 0 for some reason...
But it's good to know I had the right idea and obviously just got lost in the math on the way.
I guess I'll simply try it again :)

Thanks for your help!