If we place a conductor between the plates of a capacitor, the conductor reaches an electrostatic equilibrium with the surrounding electric field. At this equilibrium state, the charges within the conductor have redistributed such that the electric field inside the conductor is nullified. Now, what happens if we separate this conductor into two halves, each containing the redistributed charges corresponding to one side of the conductor (meaning we have one half that is positively charged and one half that is negatively charged). We did not take the conductors out of the field. Would there be an electric field between the separated halves of the conductor if we look at the complete system (Capacitor and its field and conductors and their interaction)? My book says there is no field because it would cancel out the external field of the capacitor but I always thought that electric field lines always end on charges, which would not make a superposition possible.
1 Answer
If I've understood you aright, the 'final' system consists of 4 flat conducting plates in 'stack' configuration but with gaps between them. Let's call them A, B, C and D. A and D are the original capacitor plates, with (I assume) equal and opposite charges (call them $Q$ and $-Q$ on their 'inner' surfaces (those facing B and C). B and C will have 'induced' charges, $-Q$ and $Q$ on their outer faces (those facing A and D). B and C will have no charge on their 'inner' surfaces (facing each other).
Symmetry shows that in the places where there is an electric field in this set-up it will be normal to the plates (as long as we're well away from the edges of the plates).
Now consider a Gaussian surface in the form of a box with four of its walls normal to plate B and partly inside plate B but sticking out into the gap between B and C. The other two walls are parallel to the plates. One of them (wall X, say) is inside B, the other (wall Y, say) is in the gap between B and C. No charge is enclosed in the box, because there is no charge on the inner surface of B. Therefore no electric flux enters or leaves the box. No flux crosses wall X because there is no electric field in the metal. No flux crosses the four walls normal to the plates either, therefore no flux crosses wall Y. There is therefore no electric field in the gap between B and C.
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$\begingroup$ Great explanation! If we were to put Wall X more towards the capacitor plate, so it includes the charges on B, then we'd calculate the flux between A and B since the fields there have nothing to do with the space between B and C? $\endgroup$ Commented Jun 23, 2023 at 23:20
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$\begingroup$ Yes: I'm interpreting "If we were to put Wall X more towards the capacitor plate, so it includes the charges on B" to mean putting wall X in the gap between A and B. Then indeed we could calculate the flux between A and B. But we'd need to have established (perhaps using my original Gaussian box) that no flux crosses wall Y of your extended box. $\endgroup$ Commented Jun 24, 2023 at 8:08
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$\begingroup$ What I didn't include in my answer is a nice use of a Gaussian surface to show that the induced charge on B's outer surface is equal and opposite to that on A's inner surface. Put the end-faces of the box inside the metal of A and B, with the side-walls normal to the plates and going from inside A, across the AB gap and into B. No flux through any of the walls, so no net charge inside. $\endgroup$ Commented Jun 24, 2023 at 8:14
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$\begingroup$ Wait, but if there is no flux between Inside A and Inside B, then there should be no electrical field, correct? Since Q = 0, because the charges are equal. But there absolutely should be a field from A to B. This seems counterintuitive at first. $\endgroup$ Commented Jun 24, 2023 at 8:43
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$\begingroup$ Are you talking about my second comment or my first? $\endgroup$ Commented Jun 24, 2023 at 8:54