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

  • question_answer
    The work done in carrying a charge of \[5\mu \,C\] from a point A to a point B in an electric field is 10mJ.  The potential difference \[({{V}_{B}}-{{V}_{A}})\] is then [Haryana CEE 1996]

    A)            + 2kV

    B)            ? 2 kV

    C)                    + 200 V

    D)                                      ? 200 V

    Correct Answer: A

    Solution :

                 Work done \[W=Q({{V}_{B}}-{{V}_{A}})\Rightarrow \,({{V}_{B}}-{{V}_{A}})=\frac{W}{Q}\] \[=\frac{10\times {{10}^{-3}}}{5\times {{10}^{-6}}}J/C=2\,kV\]                         


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