# On the electric field taken in the proof of the second uniqueness theorem

In the proof of uniqueness theorem, we consider $$\vec{E}_3 = \vec{E}_2 - \vec{E_1}$$ where $$\vec{E_2}$$ and $$\vec{E_1}$$ are electric fields which satisfy all the boundary condition required.

Now, it maybe noted that due to both $$\vec{E}_1$$ and $$\vec{E}_2$$ satisfying all the conditions that the difference $$\vec{E}_3$$ doesn't really satisfy any of the conditions.

Since $$\nabla \cdot \vec{E}_3 = 0$$ everywhere and $$\oint \vec{E_3} \cdot dA= 0$$ over every boundary surface.

This leads me to wonder, is the $$\vec{E}_3$$ we define meant to be a physical field or just an 'auxiliary' mathematical function used in proof? I ask this because from my understanding if an electric field is physical then it must equal the charge density by $$\epsilon_o$$ wherever there is charge.

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It's more or less just a mathematical construct. If you want to solve the problem of the electric field created by a charge density $$\rho$$, then you have $$\vec{\nabla} \cdot \vec{E} = \rho$$. If we assume that both $$\vec{E}_1$$ and $$\vec{E}_2$$ satisfy this equation, then their difference satisfies $$\vec{\nabla} \cdot (\vec{E}_2 - \vec{E}_1) = \vec{\nabla} \cdot \vec{E}_2 - \vec{\nabla} \cdot \vec{E}_1 = \rho - \rho = 0.$$ So it wouldn't be a solution to the "real problem" you're trying to solve.
You could, however, interpret it to be a solution to a different problem, one in which there are no charges present ($$\rho = 0$$) in the volume of interest. In other words, it's not entirely unphysical, it's just the solution to a different physical situation than the original one.