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Dielectrics are insulating materials commonly used in capacitors. When a dielectric material is placed in an electric field, the dipole moments of the individual molecules align with the field direction, resulting in a net polarization (P) of the material. This polarization creates an opposing electric field that partially cancels out the applied field.


The ability of a dielectric material to polarize in response to an electric field is characterized by its dielectric constant or relative permittivity. The dielectric constant is a measure of how much the electric field within a dielectric is reduced compared to the electric field in free space. Materials with higher dielectric constants can store more electric charge.

Electric Polarization

The electric polarization, P, of a dielectric material is defined as the electric dipole moment per unit volume of the material i.e., \begin{align} \vec{P} =\frac{\Sigma_{i} \vec{p}_i}{V} \end{align} where $V$ is the volume of the dielectric material, and $\vec{p}_i$ is the electric dipole moments of the ith molecules within the volume $V$.

The electric polarization is proportional to the applied electric field for a linear dielectric i.e., \begin{align} \vec{P} = \epsilon_0 \chi_e \vec{E} \end{align} where $\epsilon_0$ is the permittivity of free space and $\chi_e$ is the electric susceptibility of the material. The electric susceptibility is a dimensionless quantity that represents how much the polarization of the material changes in response to the electric field.


Problems from IIT JEE

Problem (JEE Mains 2021): If $q_f$ is the free charge on the capacitor plates and $q_b$ is the bound charge on the dielectric slab of dielcetric constant $k$ placed between the capacitor plates, then bound charge $q_b$ can be expressed as

  1. $q_b=q_f\left(1-\frac{1}{k}\right)$
  2. $q_b=q_f\left(1-\frac{1}{\sqrt{k}}\right)$
  3. $q_b=q_f\left(1+\frac{1}{k}\right)$
  4. $q_b=q_f\left(1+\frac{1}{\sqrt{k}}\right)$


  1. Topics
JEE Physics Solved Problems in Mechanics