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In a medium of dielectric constant \mathrm{k}, the electric field is E. If \varepsilon_0 is permittivity of the free space, the electric displacement vector is

Option: 1

\frac{K \mathbf{E}}{\varepsilon_0}


Option: 2

\frac{\mathbf{E}}{K \varepsilon_0}


Option: 3

\frac{\varepsilon_0 \mathbf{E}}{K}


Option: 4

\mathrm{K \varepsilon_0 \mathbf{E}}


Answers (1)

best_answer

The electric displacement field is a vector valued D that accounts for the effects of bound charges within materials. In general \mathbf{D} is given by

\mathrm{\mathbf{D}=\varepsilon_0 \mathbf{E}+\mathbf{P}}

When \mathrm{\mathbf{E}} is electric field, \mathrm{\varepsilon_0} the vacuum permittivity and \mathrm{\mathbf{P}} the polarization density of the material.   
In most ordinary terms
\mathrm{\mathbf{D}=\varepsilon_0 \mathbf{E}}
When dielectric is present \mathrm{ \varepsilon=K \varepsilon_0}
\mathrm{\therefore \quad \mathbf{D}=\mathrm{K} \varepsilon_0 \mathbf{E}}
 

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shivangi.bhatnagar

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