A) \[\frac{K\vec{E}}{{{\varepsilon }_{0}}}\]
B) \[\frac{{\vec{E}}}{K{{\varepsilon }_{0}}}\]
C) \[\frac{{{\varepsilon }_{0}}\vec{E}}{K}\]
D) \[K{{\varepsilon }_{0}}\vec{E}\]
Correct Answer: D
Solution :
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\[\varepsilon =K{{\varepsilon }_{0}}\] \[\therefore \] \[\overrightarrow{D}=K{{\varepsilon }_{0}}\overrightarrow{E}\]You need to login to perform this action.
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