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Punjab Medical Punjab - MET Solved Paper-2001

  • question_answer
    The dielectric strength of air at \[NTP\] is \[3\times {{10}^{6}}\,\,V/m\] then the maximum charge that can be given to a spherical conductor of radius \[3\,\,m\] is

    A) \[3\times {{10}^{-4}}\,\,C\]                        

    B) \[3\times {{10}^{-3}}\,\,C\]

    C) \[3\times {{10}^{-2}}\,\,C\]                        

    D) \[3\times {{10}^{-1}}\,\,C\]

    Correct Answer: B

    Solution :

    Using the relation for electric intensity                 \[F=\left( \frac{1}{4\pi {{\varepsilon }_{0}}} \right)\frac{q}{{{R}^{2}}}\]                                   ? (1) If\[q={{q}_{\max }}\]and\[F={{F}_{\max }}\] As \[{{E}_{\max }}=\]dielectric strength of air                 \[=3\times {{10}^{6}}\,\,V/m\]and\[R=3\,\,m\] So, from equation (1)                 \[3\times {{10}^{6}}=9\times {{10}^{9}}\times \frac{{{q}_{\max }}}{{{(3)}^{2}}}\] So,\[{{q}_{\max }}=3\times {{10}^{-3}}\,\,coulomb.\]


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