BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    Find the dimensions of electric permittivity,

    A) \[\text{ }\!\![\!\!\text{ }{{\text{A}}^{\text{2}}}{{\text{M}}^{\text{-1}}}{{\text{L}}^{\text{-3}}}{{\text{T}}^{\text{4}}}\text{ }\!\!]\!\!\text{ }\]   

    B)  \[\text{ }\!\![\!\!\text{ }{{\text{A}}^{\text{2}}}{{\text{M}}^{\text{-1}}}{{\text{L}}^{\text{-3}}}{{\text{T}}^{0}}\text{ }\!\!]\!\!\text{ }\]

    C)  \[\text{ }\!\![\!\!\text{ A}{{\text{M}}^{-3}}{{\text{L}}^{\text{-3}}}{{\text{T}}^{4}}\text{ }\!\!]\!\!\text{ }\]    

    D)         \[\text{ }\!\![\!\!\text{ }{{\text{A}}^{\text{2}}}{{\text{M}}^{0}}{{\text{L}}^{\text{-3}}}{{\text{T}}^{4}}\text{ }\!\!]\!\!\text{ }\]

    Correct Answer: A

    Solution :

                    From Coulombs law, The force of attraction or repulsion between two point charges q, q separated by distance r is \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{r}^{2}}}\] \[\Rightarrow \]               \[{{\varepsilon }_{0}}=\frac{1}{4\pi }\frac{{{q}^{2}}}{F{{r}^{2}}}\] where \[{{\varepsilon }_{0}}\]is electric permittivity. Dimensions of \[{{\varepsilon }_{0}}=\frac{{{[AT]}^{2}}}{[ML{{T}^{-1}}][{{L}^{2}}]}\]        \[{{\varepsilon }_{0}}=[{{A}^{2}}{{M}^{-1}}{{L}^{-3}}{{T}^{4}}]\]


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