# Electric Dipole - Calulating the electic field from Potential

1. Feb 18, 2013

### smoking-frog

1. The problem statement, all variables and given/known data
Calculate the Electric field of a Dipole from its Potential.

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

2. Relevant equations
$$\Phi_D(\vec R)=\frac{Q}{4\pi\epsilon_0} \cdot \frac{\vec d \vec R}{R^3}$$

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!

2. Feb 18, 2013

### 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

3. Feb 18, 2013

### 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, $$\operatorname{grad}(\vec a \cdot \vec b)=(\vec a \cdot \nabla) \cdot \vec b+(\vec b \cdot \nabla) \cdot \vec a+\vec a \times (\nabla \times\vec b)+\vec b \times (\nabla \times\vec a)$$

Last edited: Feb 18, 2013
4. Feb 18, 2013

### clamtrox

5. Feb 18, 2013

### 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 :)