Question

A cube of side 1 cm contains 100 molecules each having an induced dipole moment of 0.2×10^-6C-m in an external electric field of 4NC^-1. The electric susceptibility of the materials is_____ C2N^-1m^-2

• A. 50
• B. 5
• C. 0.5
• D. 0.05

Answer 1

The Correct Option is B

Given:

• Side of the cube = 1 cm = \( 10^{-2} \, \text{m} \)
• Number of molecules = 100
• Induced dipole moment per molecule = \( 0.2 \times 10^{-6} \, \text{C} \cdot \text{m} \)
• Electric field = \( 4 \, \text{N/C} \)

Step 1: Calculate the polarization \( P \)

The total polarization is given by the formula:

\[ P = n \cdot \mu \]

where \( n \) is the number of molecules per unit volume, and \( \mu \) is the induced dipole moment. The number of molecules per unit volume \( n \) is given by: \[ n = \frac{\text{Number of molecules}}{\text{Volume of the cube}} = \frac{100}{(10^{-2})^3} = 10^6 \, \text{m}^{-3} \] Therefore, the polarization is: \[ P = 10^6 \times 0.2 \times 10^{-6} = 0.2 \, \text{C/m}^2 \]

Step 2: Calculate the electric susceptibility \( \chi_e \)

The electric susceptibility is related to the polarization and the electric field by the formula: \[ P = \epsilon_0 \chi_e E \] where: \[ \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \] and \( E = 4 \, \text{N/C} \). Rearranging to solve for \( \chi_e \): \[ \chi_e = \frac{P}{\epsilon_0 E} \] Substituting the values: \[ \chi_e = \frac{0.2}{8.85 \times 10^{-12} \times 4} = 5 \times 10^0 = 5 \]

Conclusion:

The electric susceptibility of the material is \({5} \, \text{C}^2 \, \text{N}^{-1} \, \text{m}^{-2} \), so the correct answer is (B) 5.

Answer 2

The electric susceptibility \( \chi_e \) is given by: \[ \chi_e = \frac{P}{\varepsilon_0 E} \] where \( P \) is the polarization (total dipole moment per unit volume) and \( E \) is the electric field. \[ P = n \times p \] where \( n \) is the number of molecules per unit volume, and \( p \) is the dipole moment of each molecule. Given:
\( n = \frac{100}{1^3} = 100 \) molecules per cm³ = \( 10^4 \) molecules per m³
\( p = 0.2 \times 10^{-16} \) C·m - \( E = 4 \) N·C⁻¹
\( \varepsilon_0 = 8.85 \times 10^{-12} \) C²N⁻¹m⁻²
Thus: \[ P = 100 \times (0.2 \times 10^{-16}) = 2 \times 10^{-14} \, \text{C/m}^2 \] \[ \chi_e = \frac{2 \times 10^{-14}}{8.85 \times 10^{-12} \times 4} = 5 \]

Thus, the electric susceptibility is 5 C²N⁻¹m⁻².