A parallel-plate capacitor of capacitance C₁ is made using two gold pl

A parallel-plate capacitor of capacitance C₁ 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 ρg and ρa are respectively resistivities of gold and aluminium, then which one of the following relations is correct?

”C₁
”C₂
”C₁ρa
”C₁
This question was previously asked in
UPSC CDS-1 – 2017
The capacitance of a parallel-plate capacitor is given by the formula C = ε * (A/d), where ε is the permittivity of the dielectric material between the plates, A is the area of one plate, and d is the distance between the plates.
– The problem states that both capacitors have the same plate separation (d) and the four plates are of the same area (meaning the plates for C₁ have area A, and the plates for C₂ also have area A).
– The plates are made of gold and aluminium. These materials are conductors and form the plates; they are not the dielectric material filling the space between the plates. The dielectric material between the plates is presumably the same for both capacitors (usually air or vacuum unless specified otherwise).
– Since A, d, and the dielectric material (and thus its permittivity ε) are the same for both capacitors, their capacitances must be equal.
– The resistivities of gold (ρg) and aluminium (ρa) are properties of the plate materials themselves and do not affect the capacitance value, which is determined by the geometry of the plates and the material *between* them.
The capacitance of a capacitor depends on its physical dimensions (area and separation of plates) and the dielectric properties of the material filling the space between the plates. The conductive material of the plates affects properties like resistance and conductivity, but not the capacitance itself in the standard parallel-plate model.