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I have tried to follow "Symmetry, Uniqueness, and the Coulomb Law of Force" by Shaw (1965) in both asking and solving this question, but to no avail. Some of the mathematical arguments there are a bit too quick for me but, it suffices to say, the paper tries to make the "by symmetry" arguments of introductory electromagnetism rigorous.

My question is the following: Consider a 1D situation in which I have a charge distribution which obeys 𝜌(𝑧)=−𝜌(−𝑧). Then I conclude that there must exist another solution obtained from my original solution via this symmetry: 𝐸′(𝑧)=−𝐸(−𝑧). But by the uniqueness of solutions to electromagnetism problems we have 𝐸′(𝑧)=𝐸(𝑧) so that we have 𝐸(𝑧)=−𝐸(−𝑧). But this is absurd, since it implies that the electric field everywhere points toward the origin which makes no sense for a distribution obeying 𝜌(𝑧)=−𝜌(−𝑧) (dipole) so that it should point in one direction everywhere. Where have I erred in "using symmetry"?

Edit: On reconsidering, I now have the following. We first observe that the symmetry ##\rho(z) = -\rho(-z)## is equivalent to saying that the system must be invariant under a reflection ##\rho(z) \to \rho(-z)## followed by a "flipping" of charge ##\rho(-z) \to -\rho(-z)##. These transformations being symmetries of the system mean that we can obtain another solution to the problem by performing ##E(z) \to E'(z') = E'(-z) = -E(-z)## (this step is analogous to Shaw equation (2)) followed by ##E'(z') \to E''(z') = -E'(z')= -(-E(-z)) = E(-z)##. Then, by uniqueness, it must be that the solution ##E''(z) = E(z)##...but from the above this just says ##E''(z) = E(-(-z)) = E(z)## which is useless (in the last step I have tried to use an analogue to Shaw equation (1)?

My question is the following: Consider a 1D situation in which I have a charge distribution which obeys 𝜌(𝑧)=−𝜌(−𝑧). Then I conclude that there must exist another solution obtained from my original solution via this symmetry: 𝐸′(𝑧)=−𝐸(−𝑧). But by the uniqueness of solutions to electromagnetism problems we have 𝐸′(𝑧)=𝐸(𝑧) so that we have 𝐸(𝑧)=−𝐸(−𝑧). But this is absurd, since it implies that the electric field everywhere points toward the origin which makes no sense for a distribution obeying 𝜌(𝑧)=−𝜌(−𝑧) (dipole) so that it should point in one direction everywhere. Where have I erred in "using symmetry"?

Edit: On reconsidering, I now have the following. We first observe that the symmetry ##\rho(z) = -\rho(-z)## is equivalent to saying that the system must be invariant under a reflection ##\rho(z) \to \rho(-z)## followed by a "flipping" of charge ##\rho(-z) \to -\rho(-z)##. These transformations being symmetries of the system mean that we can obtain another solution to the problem by performing ##E(z) \to E'(z') = E'(-z) = -E(-z)## (this step is analogous to Shaw equation (2)) followed by ##E'(z') \to E''(z') = -E'(z')= -(-E(-z)) = E(-z)##. Then, by uniqueness, it must be that the solution ##E''(z) = E(z)##...but from the above this just says ##E''(z) = E(-(-z)) = E(z)## which is useless (in the last step I have tried to use an analogue to Shaw equation (1)?

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