Calculating the Capacitance — Question 8

Question 8

An alternating plate stack adds charge from all active gaps

A stack has N identical large metal plates, N at least 2, with area A and uniform vacuum gap d. Odd-numbered plates are connected to terminal A and even-numbered plates to terminal B. Ignore outer fields and fringing.

1. Count the active gaps and derive the terminal capacitance by summing their charge contributions.

2. For six plates find the factor relative to one gap and explain why the two faces of an interior plate must be included.

Question 8 - Solution

An alternating plate stack adds charge from all active gaps

(a) Method and calculation

$$N_{\mathrm{gaps}}=N-1,\quad Q_{\mathrm{gap}}=\frac{\epsilon_0A}{d}V.$$$$Q_A=(N-1)Q_{\mathrm{gap}}\Rightarrow C=(N-1)\epsilon_0A/d.$$

(b) Interpretation and further calculation

$$N=6:\quad C=5\epsilon_0A/d.$$$$Q_{\mathrm{interior\ plate}}=\pm2Q_{\mathrm{gap}},\quad Q_{\mathrm{end\ plate}}=\pm Q_{\mathrm{gap}}.$$

Physical check

The gap count is N minus one, not N; opposite faces of each interior plate border separate field regions.