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I was asked to determine the electric field in the spaces between the copper bar and the plates. The voltage across the plates was 30 V, so naturally I’d use $E=\Delta V/d$ to determine the field. However, the answer used 0.50 mm as $d$, rather than 1.00 mm. I do not understand why that is the case. I interpret 0.50 mm as being the $d$ corresponding to the total sum of empty space, but when you place conductors in electric fields, don’t they have no effect?

I learnt from my books that if you place a hollow metal container in a uniform electric field, it will seem as if a chunk of field had been removed at the container’s interior, but otherwise everywhere else the field remains the same—the field polarizes the walls of the container and causes the field to continue on the other side. See the diagram below.

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So I thought I would put 1.00 mm as $d$ as if the metal were of no consequence, but it was wrong. Why?

(Question from here, 3c(iv).)

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There is no electric field inside the copper conductor and so there is no potential difference between its upper and lower surfaces. If there were, then charge would move to eliminate the potential difference. Therefore the potential difference is dropped equally across the two gaps. i.e. if the top plate is at 30V, the top and bottom of the copper is at 15V and the bottom plate is at zero. So the electric field across each gap is 15V/0.25mm.

Yes, the copper bar has an effect. If you left the plates connected and withdrew the bar, then the electric field would be reduced to 30V/1mm.

Possibly you are confusing this with the other possibility of disconnecting the plates from the battery before inserting the bar. In which case, the charge on the plates stays the same and, by Gauss's law, the electric field in the gaps stays the same as it was before the bar was inserted - which means that the potential difference is increased.

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