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JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • question_answer
    On moving a charge of 20 C by 2 cm, 2 J of work is done, then the potential difference between the points is                [AIEEE 2002]

    A) \[0.1\] V

    B) 8V

    C) 2 V

    D) 3V

    Correct Answer: A

    Solution :

    [a] Potential difference between two points in an electric field is
                                        \[{{V}_{A}}-{{V}_{B}}=\frac{W}{{{q}_{0}}}\]
    where, W is work done by moving charge \[{{q}_{0}}\] from point A to B.
    So. \[{{V}_{A}}-{{V}_{B}}=\frac{2}{20}\](here, \[W=2\,J,{{q}_{0}}=20C\])
                \[=0.1\,V\]


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