Question

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

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

Case (1) : Finite thickness t. Free electrons in the sheet will migrate towards the positive plate of the capacitor. Then, the metal sheet is attracted towards whichever capacitor plate is closest and gets stuck to it, so that its potential is the same as that of that plate. The gap between the capacitor plates is reduced to d – t, so that the capacitance increases.

Case (2) : Negligible thickness. The thin metal sheet divides the gap into two of thicknesses d₁ and d₁ of capacitances C₁ = ε₀A/d₁ and C₂ = ε₀A/d₂ in series.
Their effective capacitance is
C = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{\varepsilon_{0} A}{d_{1}+d_{2}}=\frac{\varepsilon_{0} A}{d}\) = C₀
i.e., the capacitance remains unchanged.