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VIT Engineering VIT Engineering Solved Paper-2015

  • question_answer
               The capacitance of a parallel plate capacitor with air as medium is 3\[\mu F\], As a dielectric is introduced between the plates, the capacitance becomes 15\[\mu F\]. The permittivity of the  medium in \[{{C}^{2}}{{N}^{-1}}{{m}^{-2}}\] is

    A) \[8.15\times {{10}^{-11}}\]        

    B) \[0.44\times {{10}^{-10}}\]

    C) \[15.2\times {{10}^{12}}\]           

    D) \[1.6\times {{10}^{-14}}\]

    Correct Answer: B

    Solution :

    The capacitance of air capacitor is given by                     \[=1-\frac{3}{5}\]             ...(1) When a dielectric of permittivity \[C=\frac{10}{2\times {{10}^{4}}}=5\times {{10}^{-4}}\] and dielectric constant K is introduced between the plates, then                        \[\therefore \]       ....(ii) Dividing Eq. (ii) by Eq. (i) we get                      \[C=500\mu F\] \[{{C}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}=3\mu F\]            K = 5 So, the permittivity of the medium                       \[{{\varepsilon }_{r}}\]                        \[C=\frac{K{{\varepsilon }_{0}}A}{d}=15\mu F\]                        \[{{V}_{A}}:{{V}_{N}}={{10}^{15}}:1\]


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