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The question is: if I were to insert a brass plate between two charges, what will happen to the force between the charges? Would it increase, decrease or stay the same?

Does the brass plate increase the value of permittivity of the medium and therefore the force decreases?

The correct answer is that it will increase. But I do not understand how.

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The brass plate is a conductor, so the potential will be the same on both sides. The thickness of the brass plate therefore subtracts from the effective distance between the two charges, making the electric field strength higher in the remaining open space between the charges. This stronger field will cause more force to be experienced by each charge.

Another effect is that the shape of the field will change. Since the conductive plate has the same potential everywhere on its surface, each charge now sees a plane at half the voltage and half the distance, as apposed to the point charge before.

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  • $\begingroup$ But the electric field must be perpendicular to an ideal conductor. Brass is certainly not ideal, but it will certainly cause a change in shape of the field. I think this answer needs more work. $\endgroup$ – garyp Jul 8 '14 at 12:21
  • $\begingroup$ @garyp: Brass is a pretty good conductor (has low bulk resistivity). Since there is no current flowing, it is a "perfect" conductor for this purpose, meaning the potential will be equal everywhere on its surface. Yes, the field shape will also change. I'm adding to my answer accordingly. $\endgroup$ – Olin Lathrop Jul 8 '14 at 12:26
  • $\begingroup$ What's still not clear to me: since the shape of the field is different, the simple argument of equipotential within the conductor seems not enough. The field lines at the test particle are no longer radial straight lines. It's not clear to me if the field strength would have gotten larger or smaller. $\endgroup$ – garyp Jul 8 '14 at 14:03
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When you take a brass plate of considerable thickness and place it in between two charges, say positive and negative, induction takes place in the brass plate since it is a conductor: the electrons shift to the end near the positive charge while the cations stay near the negative charge. Now, induction occurs in order to make the field outside a certain region zero. Here the regions in question are the sets of points between the plate and the charges. Thus the magnitude of charge on each side of the plate tries to be as close as possible but opposite in sign to the charge that it is facing. Hence, the (here) attractive force increases, due to lesser distance (the brass plate is closer to a charged particle than the other charged particle since it is kept between them) while the repulsive force is comparatively weaker since the brass plate is of sizeable thickness and increases the distance between like charges (like charges being the charged particle and the similar charge on the opposite face of the plate). A similar question can be found in the problem book by SS Krotov.

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  • $\begingroup$ I think you mean polarization rather than induction. $\endgroup$ – BMS Jul 8 '14 at 17:57
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    $\begingroup$ I think both the terms fit, the plate is "polarized" while charges are "induced" in the plate. $\endgroup$ – Graviton Jul 9 '14 at 17:01

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