# Reversing electric field and potential relation

This is an ultra-basic issue. I'm trying to use the form of the eletric potential $$\phi = -\vec{E} \cdot \vec{r}$$

alongisde with the vector identity $$\nabla(\vec{a} \cdot \vec{b}) = \vec{a} \times(\vec{\nabla}\times \vec{b}) + \vec{b} \times(\vec{\nabla}\times \vec{a}) + (\vec{a} \cdot \vec{\nabla})\vec{b} +(\vec{b} \cdot \vec{\nabla})\vec{a}$$

to prove that a uniform and constant Eletric Field $$\vec{E}$$ holds $$\vec{E} = -\nabla \phi$$

Attempt:

Aplying the identity to the form of the potential gives us $$-\nabla \phi = -\nabla(-\vec{E} \cdot \vec{r}) = \nabla(\vec{E} \cdot \vec{r})$$ $$=\vec{E} \times(\vec{\nabla}\times \vec{r}) + \vec{r} \times(\vec{\nabla}\times \vec{E}) + (\vec{\nabla}\cdot \vec{E} )\vec{r} +(\vec{\nabla}\cdot \vec{r})\vec{E}$$

The fact that $$\vec{E}$$ is uniform tells us that it does not vary with the coordinates, so the curl and div of $$\vec{E}$$ is zero (two terms of the above expression are equal to zero).

$$\vec{\nabla}\times \vec{r}= 0$$ and because of this $$\vec{E} \times(\vec{\nabla}\times \vec{r}) = 0$$. The only term that is not zero is $$(\vec{\nabla}\cdot \vec{r})\vec{E}$$, and here comes the issue: $$(\vec{\nabla}\cdot \vec{r})= 3$$ isn't it?

I should get that $$-\nabla \phi = \vec{E}$$

I'm doing some very very basic errors, but at this time of the night, I'm not able to see it by myself. Hope you can see it.

• The electric field is not parallel to the position vector in general (think of a parallel plate capacitor and a position vector referred to an arbitrary point). Curl $\vec r$ is zero independently on the electric field. Apr 18, 2021 at 5:40
• Notice, however, that this kind of questions is not complying with the site policy Apr 18, 2021 at 5:44
• In that case, can you how I can handle that 3 factor. And about the policy: I'm really sorry, but I dont got it Apr 18, 2021 at 13:02
• There is no factor three. Check better the formula for the div of a scalar product. Apr 18, 2021 at 13:48
• The factor 3 comes from the div of position vector,right? I think the formula is right since it is the grad of a scalar product Apr 18, 2021 at 14:24

Rather than faffing around with vector calculus identities it is much easier to use index notation here. With $$\phi = -\mathbf{E}\cdot\mathbf{r} = -E_j r_j$$ we have $$(\nabla\phi)_i = \partial_i \phi = -\partial_i(E_j r_j) = -E_j\partial_i r_j = -E_j\delta_{ij} = -E_i = (-\mathbf{E})_i$$ so $$\mathbf{E} = -\nabla\phi$$.
• Ah, ok. Well the problem you have is that $(\mathbf{E}\cdot\nabla) \mathbf{r} \neq \mathbf{E}(\nabla\cdot\mathbf{r})$. The former is $E_j\partial_j r_i$ and the latter is $E_i \partial_j r_j$. Apr 18, 2021 at 14:52
• Oh, so I've made a mistake about the order of the things, thank you very much! I've found out that this identity can be put in a much more simple form: $\nabla(\vec{a} \cdot \vec{b}) = (\nabla a)\cdot \vec{b}+ (\nabla\vec{b})\cdot\vec{a}$ and from this form it's straight forward Apr 18, 2021 at 14:57