#### Question

A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor.

The battery will supply more charge.

The capacitance will increase.

The potential difference between the plates will increase.

Equal and opposite charges will appear on the two faces of the metal plate.

#### Solution

Equal and opposite charges will appear on the two faces of the metal plate.

The capacitance of the capacitor in which a dielectric slab of dielectric constant K, area A and thickness t is inserted between the plates of the capacitor of area A and separated by a distance d is given by `C = (∈_0A)/((d-t)+(t/K))`

Since it is given that the thickness of the sheet is negligible, the above formula reduces to C = `(∈_0A)/d`. In other words, there will not be any change in the electric field, potential or charge.

Only, equal and opposite charges will appear on the two faces of the metal plate because of induction due to the presence of the charges on the plates of the capacitor.