Question

Answer in brief:

A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?

Answer 1

Assume the parallel plate capacitor has capacitance C₀, area A plates, and a separation d. Suppose the introduced metal sheet has the same area A.

Case (1): Thickness t is finite. Free electrons in the sheet will travel to the positive plate of the capacitor. The metal sheet is subsequently drawn to the nearest capacitor plate and attached to it, giving it the same potential as that plate. When the gap between the capacitor plates is reduced to d - t, the capacitance increases.

Case (2): Thickness is negligible. The gap is divided into two thicknesses d₁ and d₂ of capacitances C₁ = ε₀A/d₁ and C₂ = ε₀A/d₂ in series by the thin metal sheet.

Their effective capacitance is

C = `("C"_1"C"_2)/("C"_1 + "C"_2) = (ε_0"A")/("d"_1 + "d"_2) = (ε_0"A")/"d" = "C"_0`

In other words, the capacitance remains constant.