Converting $vec{E} = - nabla phi$ into spherical coordinates

Question

Let's say I have an electric field $vec{E} = (0, 0, E_z)$, where $E_z$ in constant. Then the electric potential is $phi = - vec{E} cdot vec{r}$, where $vec{r} = (x, y, z)$.
Calculating $vec{E} = - nabla {phi}$ in cartesian coordinates is ok, we get the $vec{E}$ we started with.
But I have a problem with transforming this whole thing into spherical coordinates. Then the electric field is $vec{E} = (E_z cos{theta}, -E_zsin{theta}, E_z)$ and $nabla phi = (frac{partial phi}{partial r}, frac{1}{r} frac{partial phi}{partial theta}, frac{1}{r sin{theta}} frac{partial phi}{partial varphi})$.
I am not really sure whether $vec{r}$ should change to $(r sin{theta}cos{theta}, r sin{theta}sin{varphi}, rcos{theta})$ or $(r, rtheta, rsin{(theta)}varphi)$ or something else. Either way, I can't get the original $vec{E}$ from $vec{E} = - nabla {phi}$. The result is wrong.
What am I doing wrong? I assume it's the $vec{r}$ transformation but I am not sure.

Buraian · Answer

$phi = - vec{E} cdot vec{r}$,

This equation is only true for a constant electric field wrong. The correct relation between potential and field generally is :
$$ - nabla phi = vec{E}$$
The above is a differential equation which you can solve to obtain $phi$.

$$vec{E} =(E_zcosθ,−E_zsinθ,E_z) $$

The conversion for cartesian to polar basis is:
$$ hat{k} = hat{e_r} cos theta - hat{e_{theta} } sin theta$$
Hence,
$$ vec{E}  = E_z vec{k}  = E_z cdot ( hat{e_r} cos theta - hat{e_{theta} } sin theta)  $$
Now,
$$nabla phi = (frac{partial phi}{partial r}, frac{1}{r} frac{partial phi}{partial theta}, frac{1}{r sin{theta}} frac{partial phi}{partial varphi})$$
Hence,
$$-(frac{partial phi}{partial r}, frac{1}{r} frac{partial phi}{partial theta}, frac{1}{r sin{theta}} frac{partial phi}{partial varphi})=(E_z cos theta, 0 ,- E_z sin theta)$$
Now, all you have to do is find a potential which satisfies the above equation.
Refer my answer here

Converting $vec{E} = - nabla phi$ into spherical coordinates

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