Q. In a medium of dielectric constant $K,$ the electric field is $\overrightarrow{E}.$ If $\varepsilon_{0}$ is permittivity of the free space, the electric displacement vector is
A) $\frac{K\overrightarrow{E}}{\varepsilon_{0}}$
B) $\frac{\overrightarrow{E}}{K\varepsilon_{0}}$
C) $\frac{\varepsilon_{0}\overrightarrow{E}}{K}$
D) $K\varepsilon_{0}\overrightarrow{E}$
Explanation
(d) Electric displacement vector $\overrightarrow{D} = \varepsilon\overrightarrow{E}$
But $\varepsilon = K\varepsilon_{0}$
Where, $K =$ dielectric constant and $\varepsilon_{0} =$ permittivity of free space
$\therefore\overrightarrow{D} = K\varepsilon_{0}\overrightarrow{E}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.