So, as noted, we use Poynting's theorem (and the static version has a short derivation) to get $$\frac{1}{\mu_0} \oint_{\partial V} (\vec{E} \times \vec{B}) \cdot d\vec{S} = -\int_V \vec{E} \cdot \vec{J} dV$$ The electric field is just the negative gradient of electric potential. We can use integration by parts in higher dimensions:

$$-\int_V \vec{E}\cdot\vec{J} dV = \int_V (\nabla \phi) \cdot \vec{J} dV = \oint_{\partial V} \phi \vec{J} \cdot d\vec{S} - \int_V \phi(\nabla \cdot \vec{J}) dV$$

The current density doesn't diverge. So it's just the first term, basically, the surface integral of voltage times current.

Of course your boundary condition says there's no current out of the surface, still, the previous statement holds in static situations even without that boundary condition.

So, as noted, we use Poynting's theorem to get: $$\frac{1}{\mu_0} \oint_{\partial V} (\vec{E} \times \vec{B}) \cdot d\vec{S} = -\int_V \vec{E} \cdot \vec{J} dV$$ [The static version of Poynting's theorem is just: divergence theorem, $\nabla \cdot (\vec{E} \times \vec{B}) = \vec{B} \cdot (\nabla \times \vec{E}) - \vec{E} \cdot (\nabla \times \vec{B})$, then $\nabla \times \vec{E} = 0$ and $\nabla \times\vec{B} = \mu_0\vec{J}$]

The electric field is just the negative gradient of electric potential. We can use integration by parts in higher dimensions:

$$-\int_V \vec{E}\cdot\vec{J} dV = \int_V (\nabla \phi) \cdot \vec{J} dV = \oint_{\partial V} \phi \vec{J} \cdot d\vec{S} - \int_V \phi(\nabla \cdot \vec{J}) dV$$

The current density doesn't diverge. So it's just the first term, basically, the surface integral of voltage times current.

Of course your boundary condition says there's no current out of the surface, still, the previous statement holds in static situations even without that boundary condition.

Because magnetic potential is static, it follows that the electric field is just the negative gradient of electric potential. We can use integration by parts in higher dimensions:

$$-\int_V \vec{E}\cdot\vec{J} dV = \int_V (\nabla \phi) \cdot \vec{J} dV = \oint_{\partial V} \phi \vec{J} \cdot d\vec{S} - \int_V \phi(\nabla \cdot \vec{J}) dV$$

Now the current density doesn't diverge, because the charge density is static. So it's just the first term, basically, the integral of voltage times current.

Of course your boundary condition says there's no current out of the surface, still, the previous statement holds in static situations even without that boundary condition.

So, as noted, we use Poynting's theorem (and the static version has a short derivation) to get $$\frac{1}{\mu_0} \oint_{\partial V} (\vec{E} \times \vec{B}) \cdot d\vec{S} = -\int_V \vec{E} \cdot \vec{J} dV$$ The electric field is just the negative gradient of electric potential. We can use integration by parts in higher dimensions:

$$-\int_V \vec{E}\cdot\vec{J} dV = \int_V (\nabla \phi) \cdot \vec{J} dV = \oint_{\partial V} \phi \vec{J} \cdot d\vec{S} - \int_V \phi(\nabla \cdot \vec{J}) dV$$

The current density doesn't diverge. So it's just the first term, basically, the surface integral of voltage times current.

Of course your boundary condition says there's no current out of the surface, still, the previous statement holds in static situations even without that boundary condition.