A parallel-plate capacitor of capacitance q is made using two gold plates. Another parallel-plate capacitor of capacitance C2 is made using two aluminium plates with same plate separation, and all the four plates are of same area. If pg and pa are respectively the electrical resistivities of gold and aluminium, then which one of the following relations is correct?

The capacitance of a parallel-plate capacitor is determined solely by its physical dimensions and the properties of the dielectric material between the plates. According to the fundamental formula C = ε₀A/d, capacitance (C) depends on the permittivity of free space (ε₀) or the dielectric constant (κ), the overlapping surface area of the plates (A), and the separation distance between them (d). While the electrical resistivity (ρ) of the plate material (gold or aluminium) affects the resistance of the conductor itself, it does not enter the formula for capacitance. Since both capacitors have the same plate area and the same separation distance, and assuming the same dielectric (air or vacuum) is used, their capacitances are identical. Therefore, the capacitance of the gold-plate capacitor (q) is equal to the capacitance of the aluminium-plate capacitor (C2), making q = C2.

Sources

1. [1] Science , class X (NCERT 2025 ed.) > Chapter 11: Electricity > Activity 11.3 > p. 178