Does $Curl(E) = 0 $ along an equipotential sphere require that the radial component of $E =$ constant on the sphere?

Question

I'm arriving at the conclusion that "$\nabla \times \vec{E} = 0$ on the surface of an equipotential sphere ($E_\theta = E_\phi = 0$) (as the field must be normal to an equipotential/conductor) implies $E_r = constant$ on said sphere.

$\left( \nabla \times \vec{E} \right)_r = \frac{1}{r \sin \theta} \left[ \frac{ \partial}{\partial \theta} \left( \sin \theta {E_\phi} \right) - \frac{\partial E_\theta}{\partial \phi}\right] = 0 $ since $E_\phi = E_\theta = 0$

$\left( \nabla \times \vec{E} \right)_\theta = \frac{1}{r} \left[ \frac{1}{\sin \theta}\left( \frac{\partial E_r}{\partial \phi}\right)- \frac{\partial}{\partial r} \left( r E_\phi \right) \right] = 0$ implies $\frac{\partial E_r}{\partial \phi} = 0$ since $E_\phi = 0$

$\left( \nabla \times \vec{E} \right)_\phi = \frac{1}{r} \left[ \frac{\partial}{\partial r} \left( r E_\theta \right) - \frac{\partial E_r}{\partial \theta}\right] = 0 $ implies $\frac{\partial E_r}{\partial \theta} = 0$ since $E_\theta = 0$

So we have that "If the field is normal to a surface of a conducting sphere (it has to be) and $\nabla \times \vec{E} = 0$ along the surface of the sphere (again it has to be) then

$ \frac{\partial E_r}{\partial \theta} = \frac{\partial E_r}{\partial \phi} = 0$ implies $E_r = constant$ along the sphere.

This seems mathematically sensible but I can think of examples where $E_r \neq constant$ along the surface of a conducting sphere. Namely when a point charge is put outside a conducting sphere. (Suppose we place the origin at the center of the conducting sphere in the picture below)

As you can see the field lines are not equally spaced. However the field is perpendicular to the surface.

What is my mistake in reasoning? What have I forgotten physically or mathematically?

EDIT: I made the mistake of assuming because the function value is $0$ at a point that its derivative is also $0$. I probably want to incorporate $\nabla \cdot \vec{E} = 0$ along with $ \nabla \times \vec{E} = 0$ implies $\frac{\partial E_r}{\partial \theta} = \frac{\partial E_r}{\partial \phi} = 0$

Answer 1

Inside any equipotential (conducting) object, the Electric Field is identically zero. This is because $\vec{E} = -\vec{\nabla} V$, which is zero because the potential is constant.

The problem seems to arise from an ambiguity regarding if the surface of the conductor obeys the equipotential equation or not. If it does, then the curl is really zero, and so is the electric field.

EDIT: I just realised that, by continuity, the potential must be constant on the surface. If this is correct then the first scenario is the correct one. Notice also that there is a discontinuity on the Electric Field at the surface:

$E_r = 0$, for points on the surface and inside the sphere ($r\le R$), and $E_r = f(\theta , \phi) \neq 0$ for points immediatelly out of the sphere ($r \gt R$).