A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is 'C' then the resultant capacitance is

Options

A. (a) `(n+1)C`
B. (b) `(n-1)C`
C. (c) `nC`
D. (d) `C`

Explanation

To solve the problem of finding the resultant capacitance of a parallel plate capacitor made by stacking \( n \) equally spaced plates connected alternatively, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Configuration**:
- We have \( n \) plates stacked on top of each other.
- The plates are connected alternatively, meaning that every two adjacent plates form a capacitor.

2. **Identifying the Number of Capacitors**:
- The number of pairs of adjacent plates (which form capacitors) can be calculated as \( n - 1 \). This is because for \( n \) plates, there are \( n - 1 \) gaps between them.

3. **Capacitance of Each Pair**:
- Each pair of adjacent plates has a capacitance of \( C \).

4. **Connecting Capacitors**:
- Since the plates are connected alternatively, the capacitors formed between the plates are in parallel.

5. **Calculating the Total Capacitance**:
- For capacitors in parallel, the total capacitance \( C_{\text{eq}} \) is the sum of the individual capacitances:
\[ C_{\text{eq}} = (n - 1) \times C \]

### Final Result
Thus, the resultant capacitance between any two adjacent plates in this configuration is:
\[ C_{\text{eq}} = (n - 1)C \]