A parallel plate capacitor has a parallel sheet of copper inserted between and parallel to the two plates, without touching the plates. The capacity of the capacitor after the introduction of the copper sheet is :

Options

A. minimum when the copper sheet touches one of the plates.
B. maximum when the copper sheet touches one of the plates.
C. invariant for all positions of the sheet between the plates.
D. greater than that before introducing the sheet.

Answer

To solve the problem of finding the capacity of a parallel plate capacitor after inserting a parallel sheet of copper between its plates, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a parallel plate capacitor with plates separated by a distance \(d\).
- A copper sheet is inserted between the plates, parallel to them, without touching either plate.

2. **Effect of the Copper Sheet**:
- The copper sheet acts as a conductor. When a voltage is applied across the capacitor, charges will induce on the surfaces of the copper sheet.
- The copper sheet effectively divides the capacitor into two capacitors in series.

3. **Capacitance Calculation**:
- The capacitance of a parallel plate capacitor is given by the formula:
\[ C = \frac{A \varepsilon_0}{d} \]
where \(A\) is the area of the plates, \(\varepsilon_0\) is the permittivity of free space, and \(d\) is the separation between the plates.

4. **New Configuration with Copper Sheet**:
- Let the thickness of the copper sheet be \(t\). The distance between the plates is now effectively divided into two parts: \(d_1 = \frac{d - t}{2}\) for the space above the copper sheet and \(d_2 = \frac{d - t}{2}\) for the space below it.
- The capacitance of each section can be calculated as:
\[ C_1 = \frac{A \varepsilon_0}{d_1} = \frac{A \varepsilon_0}{\frac{d - t}{2}} = \frac{2A \varepsilon_0}{d - t} \]
\[ C_2 = \frac{A \varepsilon_0}{d_2} = \frac{A \varepsilon_0}{\frac{d - t}{2}} = \frac{2A \varepsilon_0}{d - t} \]

5. **Total Capacitance in Series**:
- Since \(C_1\) and \(C_2\) are in series, the total capacitance \(C_{eq}\) is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
- Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{\frac{2A \varepsilon_0}{d - t}} + \frac{1}{\frac{2A \varepsilon_0}{d - t}} = \frac{(d - t)}{2A \varepsilon_0} + \frac{(d - t)}{2A \varepsilon_0} = \frac{(d - t)}{A \varepsilon_0} \]
- Therefore, we have:
\[ C_{eq} = \frac{A \varepsilon_0}{d - t} \]

6. **Final Result**:
- The capacitance of the capacitor after the introduction of the copper sheet is:
\[ C_{eq} = \frac{A \varepsilon_0}{d - t} \]