CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    SI unit of permittivity is:

    A)  \[{{C}^{2}}{{m}^{2}}{{N}^{2}}\]                                

    B)  \[{{C}^{2}}{{m}^{-2}}{{N}^{-1}}\]

    C)  \[{{C}^{2}}{{m}^{2}}{{N}^{-1}}\]                              

    D)  \[{{C}^{-1}}{{m}^{2}}{{N}^{-2}}\]

    Correct Answer: B

    Solution :

    From Coulombs law                 \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}\,{{q}_{2}}}{{{r}^{2}}}\] or            \[{{\varepsilon }_{0}}\frac{{{q}_{1}}\,{{q}_{2}}}{4\pi F{{r}^{2}}}\]                 \[\therefore \] Units of \[{{\varepsilon }_{0}}\] (permittivity)                 \[=\frac{{{C}^{2}}}{N-{{m}^{2}}}={{C}^{2}}{{N}^{-1}}{{m}^{-2}}\]


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