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 $$\vec{E} = -\nabla \phi$$


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.

  • $\begingroup$ 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. $\endgroup$ Apr 18, 2021 at 5:40
  • $\begingroup$ Notice, however, that this kind of questions is not complying with the site policy $\endgroup$ Apr 18, 2021 at 5:44
  • $\begingroup$ 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 $\endgroup$ Apr 18, 2021 at 13:02
  • $\begingroup$ There is no factor three. Check better the formula for the div of a scalar product. $\endgroup$ Apr 18, 2021 at 13:48
  • $\begingroup$ 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 $\endgroup$ Apr 18, 2021 at 14:24

1 Answer 1


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 $\mathbf{E} = -\nabla\phi$.

  • $\begingroup$ It's a great ideia bit I'm afraid that the use of the identity is mandatory in the case of the problem I'm trying to solve. $\endgroup$ Apr 18, 2021 at 13:00
  • 1
    $\begingroup$ 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 $E_i \partial_j r_j$. $\endgroup$
    – xzd209
    Apr 18, 2021 at 14:52
  • $\begingroup$ 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 $\endgroup$ Apr 18, 2021 at 14:57

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