Electric field and Legendre Polynomials

[Observer Two]

Homework Statement

I want to varify that the components of a homogenous electric field in spherical coordinates $\vec{E} = E_r \vec{e}_r + E_{\theta} \vec{e}_{\theta} + E_{\varphi} \vec{e}_{\varphi}$ are given via:

$E_r = - \sum\limits_{l=0}^\infty (l+1) [a_{l+1}r^l P_{l+1}(cos \theta) - b_l r^{-(l+2)} P_l cos(\theta)]$

$E_{\theta} = \sum\limits_{l=0}^\infty [a_{l+1}r^l + b_{l+1} r^{-(l+3)}]sin(\theta)P'_{l+1}(cos \theta)$

$E_{\varphi} = 0$

I have rotational symmetry about the z-axis (azimuthal symmetry).

Homework Equations

I know that the potential in charge-free space and with azimuthal symmetry can be given via the Legendre Polynomials:

$\Phi(r, \theta) = \sum\limits_{l=0}^\infty (a_l r^l + b_l r^{-(l+1)}) P_l(cos \theta)$

The Attempt at a Solution

Let's begin with $E_r$.

$\vec{E} \vec{e}_r = E_r$

And:

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

So basically what I have to do is apply the gradient (in spherical coordinates) and multiply with $\vec{e}_r$. In other words: Apply the $\vec{e}_r$ component of the gradient to the potential. Is this correct? If so: How exactly do I apply the gradient to a sum like (2)?

 

[strangerep]

In other words: Apply the $\vec{e}_r$ component of the gradient to the potential. Is this correct?

Not quite. First write out the representation of the gradient operator in spherical coordinates and apply it to your $\Phi$.

Does that get you any further?