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

  • question_answer
    Two plates are \[2\,cm\] apart, a potential difference of\[10\ volt\] is applied between them, the electric field between the plates is                                                                                                    [MP PET 1994; DPMT 2002]

    A)                    \[20\ N/C\]

    B)                                      \[500\,N/C\]

    C)                    \[5\,N/C\]

    D)                                      \[250\ N/C\]

    Correct Answer: B

    Solution :

                 \[E=\frac{V}{d}=\frac{10}{2\times {{10}^{-2}}}=500\,N/C\]


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