Solution:

In a parallel plate capacitor made by stacking $n$ equally spaced plates connected alternatively:

Key Points:

1. The plates connected alternatively form a series of capacitors.
2. If there are $n$ plates, the number of gaps (capacitors in series) between them is $n - 1$.

For capacitors in series:

The total capacitance $C_{\text{total}}$ is given by:

$$\frac{1}{C_{\text{total}}} = \frac{1}{C} + \frac{1}{C} + \dots + \frac{1}{C} \, \, (\text{n - 1 times})$$ $$\frac{1}{C_{\text{total}}} = \frac{n - 1}{C}$$ $$C_{\text{total}} = \frac{C}{n - 1}$$

However, because the plates are connected alternatively, these effectively act as $n - 1$ capacitors in parallel.

For capacitors in parallel:

The equivalent capacitance is:

$$C_{\text{eq}} = (n - 1)C$$

Thus, the resultant capacitance is $(n - 1)C$.