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I'm reading Feynman's lectures about electromagnetism (specifically, vol. II, chapter 5, §2), and I came across a section in which he first proves there can't be a stable electrostatic equilibrium for a point charge, and then he says:

Our result has been obtained for a point charge. Does the same conclusion hold for a complicated arrangement of charges held together in fixed relative positions—with rods, for example? We consider the question for two equal charges fixed on a rod. Is it possible that this combination can be in equilibrium in some electrostatic field? The answer is again no. The total force on the rod cannot be restoring for displacements in every direction.

Call $\mathbf F$ the total force on the rod in any position─$\mathbf F$ is then a vector field. Following the argument used above, we conclude that at a position of stable equilibrium, the divergence of $\mathbf F$ must be a negative number. But the total force on the rod is the first charge times the field at its position, plus the second charge times the field at its position: $$ \mathbf F=q_1\mathbf E_1+q_2\mathbf E_2 \tag{5.1} $$ The divergence of $\mathbf F$ is given by $$ \nabla\cdot\mathbf F = q_1\nabla \cdot\mathbf E_1 + q_2\nabla \cdot\mathbf E_2. $$ If each of the two charges $q_1$ and $q_2$ is in free space, both $\nabla \cdot\mathbf E_1$ and $\nabla \cdot\mathbf E_2$ are zero, and $\nabla \cdot\mathbf F$ is zero—not negative, as would be required for equilibrium. You can see that an extension of the argument shows that no rigid combination of any number of charges can have a position of stable equilibrium in an electrostatic field in free space.

Can someone please explain me the last paragraph? How come the divergence of $\mathbf E_1$ and $\mathbf E_2$ has to be zero?

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Here $\mathbf E_1 = \mathbf E(\mathbf r_1)$ and $\mathbf E_2 = \mathbf E(\mathbf r_2)$ are the electric fields at the positions of the two charges, corresponding to the electric field genertated by some external configuration of charges - and ignoring the electric field produced by our test charges. As such, they consider the positions $\mathbf r_j$ to be in vacuum, and therefore they have a zero divergence as per Gauss's law.

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