A linear isotropic dielectric has \(2\times10^{28}\) molecules per cubic metre. In an electric field, each molecule gets an induced dipole moment of \(8.85\times10^{-35}\ \mathrm{C\,m}\) along the field. The electric susceptibility of the dielectric is 4. The electric field inside 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) → \(5\times10^{4}\ \mathrm{V\,m^{-1}}\)
The induced dipole moments of all the molecules add up. So the dipole moment per unit volume, the polarisation, is the number of molecules per unit volume times the moment of one molecule. For a linear isotropic dielectric, \( P=\varepsilon_0\chi_eE \) then gives the field.
\( \begin{aligned} P &= n\,p \\ &= 2\times10^{28}\times8.85\times10^{-35} \\ &= 1.77\times10^{-6}\ \mathrm{C\,m^{-2}} \end{aligned} \)
\( \begin{aligned} E &= \frac{P}{\varepsilon_0\chi_e} \\ &= \frac{1.77\times10^{-6}}{8.85\times10^{-12}\times4} \\ &= 5\times10^{4}\ \mathrm{V\,m^{-1}} \end{aligned} \)
NCERT (2026-27):
Dielectrics and Polarisation – “The induced dipole moments of different molecules add up giving a net dipole moment of the dielectric in the presence of the external field.”
Dielectrics and Polarisation – “The dipole moment per unit volume is called polarisation and is denoted by P. For linear isotropic dielectrics,”
\( P=\varepsilon_0\chi_eE \)\( \qquad (2.37) \)