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. C
B. `nC`
C. `(n-1)C`
D. `(n+1)C`

Answer

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 arranged in such a way that they are connected alternatively. This means that if we label the plates as P1, P2, P3, ..., Pn, then P1 is connected to P2, P3 is connected to P4, and so on.

2. **Identifying Capacitors**:
- The capacitance is defined between adjacent plates. Therefore, the capacitors formed will be:
- C1 between P1 and P2,
- C2 between P2 and P3,
- C3 between P3 and P4,
- ...,
- C(n-1) between P(n-1) and Pn.

3. **Counting the Capacitors**:
- Since there are n plates, the number of capacitors formed will be (n - 1). This is because each pair of adjacent plates forms one capacitor.

4. **Capacitance of Each Capacitor**:
- Each of these capacitors has a capacitance of C.

5. **Calculating the Total Capacitance**:
- Since all the capacitors are connected in parallel between the two points (let's call them A and B), the total capacitance (C_eq) can be calculated by simply adding the capacitances of the individual capacitors.
- Therefore, the equivalent capacitance is given by:
\[ C_{eq} = (n - 1) \cdot C \]

### Final Result:
The resultant capacitance between the two points A and B is:
\[ C_{eq} = (n - 1)C \]