Question

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\vec{E}}{{{\varepsilon }_{0}}} $
• B. $ \frac{{\vec{E}}}{K{{\varepsilon }_{0}}} $
• C. $ \frac{{{\varepsilon }_{0}}\vec{E}}{K} $
• D. $ K{{\varepsilon }_{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}={{\varepsilon }_{0}}\overrightarrow{E}+\overrightarrow{P} $ where $ \overrightarrow{E} $ is electric field, $ {{\varepsilon }_{0}} $ the vacuum permittivity and P the polarization density of the material. In most ordinary terms $ \overrightarrow{D}={{\varepsilon }_{0}}\overrightarrow{E} $ when dielectric is present $ \varepsilon =K{{\varepsilon }_{0}} $ $ \therefore $ $ \overrightarrow{D}=K{{\varepsilon }_{0}}\overrightarrow{E} $