A dielectric slab of volume \(1\times10^{-4}\ \mathrm{m^{3}}\) is placed in a uniform electric field of \(4\times10^{4}\ \mathrm{V\,m^{-1}}\). The slab develops a net dipole moment of \(1.77\times10^{-10}\ \mathrm{C\,m}\). The electric susceptibility of the dielectric is (take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(5\)
Polarisation \( P \) is the dipole moment per unit volume of the dielectric. For a linear dielectric it is related to the field by \( P=\varepsilon_0\chi_eE \), where \( \chi_e \) is the electric susceptibility.
\( \begin{aligned} P &= \frac{\text{dipole moment}}{\text{volume}} \\ &= \frac{1.77\times10^{-10}}{1\times10^{-4}} \\ &= 1.77\times10^{-6}\ \mathrm{C\,m^{-2}} \end{aligned} \)
\( \begin{aligned} \chi_e &= \frac{P}{\varepsilon_0E} \\ &= \frac{1.77\times10^{-6}}{8.85\times10^{-12}\times4\times10^{4}} \\ &= 5 \end{aligned} \)
From the above, \( \chi_e=5 \). Hence, the electric susceptibility of the dielectric is 5. It is a pure number and has no unit.
NCERT (2026-27):
Dielectrics and Polarisation – “… The dipole moment per unit volume is called polarisation and is denoted by P. For linear isotropic dielectrics,”
\( P=\varepsilon_0\chi_eE \)\( \qquad (2.37) \)
“where χe is a constant characteristic of the dielectric and is known as the electric susceptibility of the dielectric medium.”