The area of parallel plates of an air capacitor is `0.2 m^(2)` and the distance between them is `0.01m` The potential difference between the plates, the potential difference between the plates is `3000 V`. When a `0.01 m` thick sheet of an insulating material is placed between the plates, the potential difference decrease to 1000 volt. Determine (i) capacitance of capacitance before placing the sheet (ii) charge on each plate (iii) dielectric constant of material (iv) capacitanc after placing the insulator (v) absoulate permittivity of the dielectric.

Answer

To solve the problem step by step, let's break down each part of the question systematically.

### Given Data:
- Area of the plates, \( A = 0.2 \, m^2 \)
- Distance between the plates, \( d = 0.01 \, m \)
- Initial potential difference, \( V_1 = 3000 \, V \)
- Potential difference after inserting the dielectric, \( V_2 = 1000 \, V \)
- Thickness of the dielectric sheet, \( d_{dielectric} = 0.01 \, m \)

### Part (i): Capacitance before placing the sheet

The formula for the capacitance \( C \) of a parallel plate capacitor is given by:

\[ C = \frac{\varepsilon_0 A}{d} \]

Where:
- \( \varepsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, F/m \).

Substituting the values:

\[ C = \frac{(8.85 \times 10^{-12} \, F/m) \times (0.2 \, m^2)}{0.01 \, m} \]

Calculating:

\[ C = \frac{(8.85 \times 10^{-12}) \times (0.2)}{0.01} = 1.77 \times 10^{-10} \, F = 17.7 \times 10^{-11} \, F \]

### Part (ii): Charge on each plate

The charge \( Q \) on each plate can be calculated using the formula:

\[ Q = C \times V \]

Substituting the values:

\[ Q = (1.77 \times 10^{-10} \, F) \times (3000 \, V) \]

Calculating:

\[ Q = 5.31 \times 10^{-7} \, C = 53.1 \times 10^{-8} \, C \]

### Part (iii): Dielectric constant of the material

The dielectric constant \( K \) can be found using the relationship between the initial and final potential differences:

\[ K = \frac{V_1}{V_2} \]

Substituting the values:

\[ K = \frac{3000 \, V}{1000 \, V} = 3 \]

### Part (iv): Capacitance after placing the insulator

The capacitance with the dielectric inserted is given by:

\[ C' = \frac{\varepsilon_0 K A}{d} \]

Substituting the values:

\[ C' = \frac{(8.85 \times 10^{-12} \, F/m) \times 3 \times (0.2 \, m^2)}{0.01 \, m} \]

Calculating:

\[ C' = \frac{(8.85 \times 10^{-12}) \times 3 \times (0.2)}{0.01} = 5.31 \times 10^{-10} \, F = 53.1 \times 10^{-11} \, F \]

### Part (v): Absolute permittivity of the dielectric

The absolute permittivity \( \varepsilon \) of the dielectric material is given by:

\[ \varepsilon = \varepsilon_0 K \]

Substituting the values:

\[ \varepsilon = (8.85 \times 10^{-12} \, F/m) \times 3 \]

Calculating:

\[ \varepsilon = 2.655 \times 10^{-11} \, F/m = 26.55 \times 10^{-12} \, F/m \]

### Summary of Results:
1. Capacitance before placing the sheet: \( C = 17.7 \times 10^{-11} \, F \)
2. Charge on each plate: \( Q = 53.1 \times 10^{-8} \, C \)
3. Dielectric constant: \( K = 3 \)
4. Capacitance after placing the insulator: \( C' = 53.1 \times 10^{-11} \, F \)
5. Absolute permittivity of the dielectric: \( \varepsilon = 26.55 \times 10^{-12} \, F/m \)