Two parallel-plate capacitors C1 and C2 are made using plates of gold and aluminium respectively, but all four plates have the same area and the same plate separation with air between them. If rho_g and rho_a are the electrical resistivities of gold and aluminium, what is the correct relation between C1 and C2?

Introduction / Context:
Capacitance of a parallel-plate capacitor depends primarily on the geometry of the plates and the dielectric material between them. It does not depend on which conducting material the plates themselves are made of, as long as the plates are good conductors. This question checks whether you can separate the role of conductor properties (like resistivity) from the factors that determine capacitance.

Given Data / Assumptions:
• Capacitor C1: made with two gold plates. • Capacitor C2: made with two aluminium plates. • All four plates have the same area A. • Plate separation d is the same for both capacitors. • The space between plates is air (same dielectric). • rho_g and rho_a are electrical resistivities of gold and aluminium.

Concept / Approach:
For a parallel-plate capacitor with air (or vacuum) as dielectric, capacitance is given by C = epsilon0 * A / d, where epsilon0 is permittivity of free space, A is area of each plate, and d is the plate separation. This formula does not involve the resistivity or conductivity of the plate material. As long as the plates are conductors, the electric field between them and the capacitance are determined solely by geometry and dielectric properties. Because both capacitors have the same plate area, same separation, and same dielectric, their capacitances must be equal, regardless of whether the plates are gold or aluminium.

Step-by-Step Solution:
Step 1: Write the capacitance formula for a parallel-plate capacitor: C = epsilon0 * A / d for air as dielectric. Step 2: For capacitor C1 (gold plates), C1 = epsilon0 * A / d. Step 3: For capacitor C2 (aluminium plates), C2 = epsilon0 * A / d. Step 4: Because A and d are identical and dielectric is air in both cases, C1 and C2 must be equal. Step 5: Therefore, C1 = C2, independent of rho_g and rho_a.

Verification / Alternative check:
Imagine gradually replacing gold with aluminium while keeping the same shape and size of plates and the same gap. The electric field in the dielectric region depends on how charges arrange on the surfaces, but both gold and aluminium are excellent conductors and will distribute charges on their surfaces to maintain equipotential conditions. As long as they behave as ideal conductors, the capacitance remains unchanged. Practical capacitors often use aluminium foils instead of more expensive metals because capacitance is determined by geometry and dielectric, not by plate resistivity.

Why Other Options Are Wrong:
Option a (C1 > C2) and option b (C2 > C1): These suggest that plate material changes capacitance, which is not true when geometry and dielectric are identical. Option c (C1 * rho_a = C2 * rho_g): This incorrectly mixes resistivity with capacitance; there is no such direct relation in basic capacitor theory. Option e (C1 / C2 = rho_g / rho_a): Again, this implies capacitance depends on resistivity ratio, which is not the case.

Common Pitfalls:
Students often confuse properties of conductors relevant to resistance (which involves resistivity and length) with properties relevant to capacitance. While resistivity affects how much a conductor heats up or how much voltage drop occurs along its length, capacitance in a parallel-plate configuration depends only on plate area, separation, and dielectric constant. Remember: geometry and dielectric determine capacitance; material resistivity affects resistance, not capacitance directly.

Final Answer:
Because both capacitors have the same geometry and dielectric, their capacitances are equal, so C1 = C2.