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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.

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  • $\begingroup$ Thank you for the help! $\endgroup$ May 26, 2021 at 20:48

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