RAJASTHAN ­ PET Rajasthan PET Solved Paper-2003

  • question_answer
    An electron is accelerated from a potential difference of\[{{10}^{4}}V\]at distance 5 cm. The force on electron will be

    A)  \[3.2\,\times \,{{10}^{-14}}N\]

    B)  \[4N\]

    C)  \[3.2\,\times \,{{10}^{-12}}N\]

    D)  None of these

    Correct Answer: A

    Solution :

     \[E=\frac{V}{r}\]and\[F=qE\] \[F=\frac{q}{r}.V\] \[F=\frac{e}{r}.V=\frac{1.6\times {{10}^{-19}}\times {{10}^{4}}}{5\times {{10}^{2}}}\] \[=3.2\times {{10}^{-14}}N\]


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