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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2000

  • question_answer
    The potential at a point, due to a positive charge of \[\text{100}\,\text{ }\!\!\mu\!\!\text{ C}\]at a distance of 9 m, is:

    A)  \[{{10}^{7}}\,\text{volt}\]                          

    B) \[{{10}^{6}}\,\text{volt}\]

    C)  \[{{10}^{5}}\,\text{volt}\]                          

    D)  \[{{10}^{4}}\,\text{volt}\]

    Correct Answer: C

    Solution :

     Charge, \[Q=10\mu C=100\times {{10}^{-6}}C={{10}^{-4}}C\] Distance, d = 9 m              (given) The relation for the potential is given by                                 \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{Q}{d}=9\times {{10}^{9}}\times \frac{{{10}^{-4}}}{9}\]                                 \[={{10}^{5}}\,\text{volt}\]


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