JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    The radius of nucleus of silver (atomic number = 47) is \[3.4\times {{10}^{-14}}m\]. The electric potential on the surface of nucleus is \[(e=1.6\times {{10}^{-19}}C)\]                                                       [Pb. PET 2003]

    A)                    \[1.99\times {{10}^{6}}\,volt\]

    B)                                      \[2.9\times {{10}^{6}}\,volt\]

    C)                    \[4.99\times {{10}^{6}}\,volt\]

    D)                                      \[0.99\times {{10}^{6}}\,volt\]

    Correct Answer: A

    Solution :

      \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{(Ze)}{r}=9\times {{10}^{9}}\times \frac{47\times 1.6\times {{10}^{-19}}}{3.4\times {{10}^{-14}}}=1.99\times {{10}^{6}}V\]


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