Question

In a medium of dielectric constant K, the electric field is $\overrightarrow{E}$ . If ${\epsilon }_0$ is permittivity of the free space, the electric displacement vector is

• A. $\frac{K\vec{E}}{{\epsilon }_0}$
• B. $\frac{\vec{E}}{K{\epsilon }_0}$
• C. $\frac{{\epsilon }_0\vec{E}}{K}$
• D. $K{\epsilon }_0\vec{E}$

Answer 1

Answer: (D)

The electric displacement field is a vector valued field D that accounts for the effects of bound charges within materials. In general D is given by $\overrightarrow{D}={\epsilon }_0\overrightarrow{E}+\overrightarrow{P}$ where $\overrightarrow{E}$ is electric field, ${\epsilon }_0$ the vacuum permittivity and P the polarization density of the material. In most ordinary terms $\overrightarrow{D}={\epsilon }_0\overrightarrow{E}$ when dielectric is present $\epsilon =K{\epsilon }_0$ $\therefore$ $\overrightarrow{D}=K{\epsilon }_0\overrightarrow{E}$