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.

Explanation

To solve the problem step by step, let's break down each part:

### Given Data:
- Area of the capacitor 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, \( V_2 = 1000 \, V \)

### (i) Capacitance before placing the sheet

The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where:
- \( \varepsilon_0 = 8.85 \times 10^{-12} \, F/m \) (permittivity of free space)

Substituting the values:
\[ C = \frac{(8.85 \times 10^{-12} \, F/m)(0.2 \, m^2)}{0.01 \, m} \]
\[ C = \frac{1.77 \times 10^{-12}}{0.01} \]
\[ C = 1.77 \times 10^{-10} \, F \]

### (ii) Charge on each plate

The charge \( Q \) on the plates can be calculated using the formula:
\[ Q = C \times V_1 \]
Substituting the values:
\[ Q = (1.77 \times 10^{-10} \, F)(3000 \, V) \]
\[ Q = 5.31 \times 10^{-7} \, C \]

### (iii) Dielectric constant of the material

The dielectric constant \( K \) can be determined using the relationship between the potential differences before and after inserting the dielectric:
\[ \frac{V_1}{V_2} = K \]
Substituting the values:
\[ K = \frac{3000 \, V}{1000 \, V} = 3 \]

### (iv) Capacitance after placing the insulator

The capacitance with the dielectric is given by:
\[ C' = K \times C \]
Substituting the values:
\[ C' = 3 \times (1.77 \times 10^{-10} \, F) \]
\[ C' = 5.31 \times 10^{-10} \, F \]

### (v) Absolute permittivity of the dielectric

The absolute permittivity \( \varepsilon \) of the dielectric material can be calculated using:
\[ \varepsilon = K \times \varepsilon_0 \]
Substituting the values:
\[ \varepsilon = 3 \times (8.85 \times 10^{-12} \, F/m) \]
\[ \varepsilon = 2.655 \times 10^{-11} \, F/m \]

### Summary of Results:
1. **Capacitance before placing the sheet**: \( C = 1.77 \times 10^{-10} \, F \)
2. **Charge on each plate**: \( Q = 5.31 \times 10^{-7} \, C \)
3. **Dielectric constant of material**: \( K = 3 \)
4. **Capacitance after placing the insulator**: \( C' = 5.31 \times 10^{-10} \, F \)
5. **Absolute permittivity of the dielectric**: \( \varepsilon = 2.655 \times 10^{-11} \, F/m \)