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

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 $$\frac{\partial E_r}{\partial \theta} = \frac{\partial E_r}{\partial \phi} = 0$$

• A first look suggests that there is at least one issue with your calculation. The derivatives $\frac{\partial E_{\theta,\phi,r}}{\partial r}$ are not well-defined on a conductor surface as the electric field changes abruptly at the conductor surface. Your calculation implicitly assumes that these are well-defined and sets them to zero. This might or might not resolve your question but seems to be an issue nonetheless. – Dvij Mankad May 11 '19 at 4:42

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.
$$E_r = 0$$, 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 ($$r \gt R$$).