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 $\vec{E} = (E_z \cos{\theta}, -E_z\sin{\theta}, E_z)$ 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 $(r \sin{\theta}\cos{\theta}, r \sin{\theta}\sin{\varphi}, r\cos{\theta})$ or $(r, r\theta, r\sin{(\theta)}\varphi)$ 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.

Answer 1

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

This equation is only true for a constant electric field. 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 $\phi$.

$$\vec{E} =(E_z\cosθ,−E_z\sinθ,E_z) $$

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,

$$-(\frac{\partial \phi}{\partial r}, \frac{1}{r} \frac{\partial \phi}{\partial \theta}, \frac{1}{r \sin{\theta}} \frac{\partial \phi}{\partial \varphi})=(E_z \cos \theta, - E_z \sin \theta,0)$$

Now, all you have to do is find a potential which satisfies the above equation.

Refer my answer here