Multi-plate parallel-plate capacitor – dependence of capacitance on number of plates For a multi-plate capacitor with n identical plates (all of area A) arranged so that alternate plates are connected together to form two interleaved sets, the total capacitance is proportional to which quantity?

Introduction / Context:
Interleaved multi-plate capacitors are used to obtain large capacitance in compact volumes. Understanding how total capacitance scales with the number of plates helps in preliminary design and quick estimation for filters, timing circuits, and energy storage modules.

Given Data / Assumptions:

• There are n equally spaced, identical conducting plates of area A.
• Alternate plates are connected together to form two terminals.
• Dielectric is uniform; edge effects (fringing) are neglected.

Concept / Approach:

With alternate plates strapped, each adjacent pair of plates constitutes one elementary capacitor. The number of such gaps equals the number of inter-plate spaces between oppositely connected plates. For n plates, the count of effective capacitors in parallel is (n − 1). Capacitors in parallel add directly, so the total capacitance C_total is proportional to (n − 1) times the capacitance of one gap, and also proportional to the plate area A (with all other parameters fixed).

Step-by-Step Solution:

Verification / Alternative check:

Check small cases: for n = 2 (a single parallel-plate capacitor), (n − 1) = 1 → expected. For n = 3, there are two gaps in parallel → doubling the capacitance, consistent with the formula.

Why Other Options Are Wrong:

n A counts plates rather than gaps. (n + 1) A and A / n have no basis in the standard interleaved topology with alternate plates shorted.

Common Pitfalls:

Accidentally wiring adjacent plates together (which reduces the effective number of active gaps) or forgetting that only the spaces between oppositely connected plates store energy.

Final Answer:

(n - 1) A