Q. In a medium of dielectric constant $K$, the electric field is E$.$ If $\varepsilon_{0}$ is permittivity of the free space, the electric displacement vector is
A) $\frac{K\mathbf{E}}{\varepsilon_{0}}$
B) $\frac{\mathbf{E}}{{K\varepsilon}_{0\ }}$
C) $\frac{\varepsilon_{0}\mathbf{E}_{\ }}{K}$
D) ${K\varepsilon}_{0}\mathbf{E}$
Explanation
(d) The electric displacement field is a vector valued D that accounts for the effects of bound charges within materials. In general D is given by
$\mathbf{D} = \varepsilon_{0}\mathbf{E\ } + \mathbf{\ P}$
When E is electric field,$\varepsilon_{0}$ the vacuum permittivity and P the polarization density of the material.
In most ordinary terms
$\mathbf{D} = \varepsilon_{0}\mathbf{E}$
When dielectric is present $\varepsilon = K\varepsilon_{0}$
$\mathbf{\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D} = K\varepsilon_{0}\mathbf{E}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.