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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2006

  • question_answer
        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)  \[K{{\varepsilon }_{0}}\vec{E}\]

    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\[\varepsilon =K{{\varepsilon }_{0}}\] \[\therefore \]  \[\overrightarrow{D}=K{{\varepsilon }_{0}}\overrightarrow{E}\]


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