1. Understand the arrangement

A stack of $n$ equally spaced plates is connected alternatively, i.e.:

• plates $1,3,5,\dots$ are connected together,
• plates $2,4,6,\dots$ are connected together.

So the whole system behaves like many capacitors connected between the same two terminals.

2. Capacitance between adjacent plates

The capacitance between any two adjacent plates is given as $C$.

Since there are $n$ plates, the number of gaps between adjacent plates is

$n-1$

Each gap forms one capacitor of capacitance $C$.

3. How are these capacitors connected?

Because alternate plates are joined together, every adjacent pair is connected across the same two nodes.

For example:

• between plates $1$ and $2$ there is a capacitor $C$,
• between plates $2$ and $3$ there is another capacitor $C$,
• between plates $3$ and $4$ there is another capacitor $C$,

and so on.

All these capacitors have their terminals effectively connected to the same two alternate groups of plates, so they are in parallel.

4. Equivalent capacitance

Capacitors in parallel add directly:

$C_{\text{eq}} = C + C + \cdots + C$

There are $n-1$ such capacitors, hence

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

5. Check options

• A: $(n+1)C$ ❌
• B: $(n-1)C$ ✅
• C: $nC$ ❌
• D: $C$ ❌

Therefore, the correct answer is

$\boxed{(n-1)C}$