2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Electrostatics & Capacitance

JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Self Evaluation Test - Electrostatic Potential and Capacitance

  • question_answer
    In moving from A to B along an electric field line, the work done by the electric field on an electron is \[6.4\times {{10}^{-19}}J.\] If \[{{\phi }_{1}}\] and \[{{\phi }_{2}}\]are equipotential surfaces, then the potential difference \[{{V}_{C}}-{{V}_{A}}\] is  

    A) -4V      

    B) 4V  

    C) zero      

    D) 6.4V

    Correct Answer: B

    Solution :

    [b] \[{{W}_{el.}}=q({{V}_{i}}-{{V}_{f}})\]  or \[6.4\times {{10}^{-19}}=-1.6\times {{10}^{-19}}\left( {{V}_{A}}-{{V}_{B}} \right)\] or \[{{V}_{A}}-{{V}_{B}}=-4V\] or \[{{V}_{A}}-{{V}_{C}}=-4V\text{     }\left( \because {{V}_{B}}={{V}_{C}} \right)\] or \[{{V}_{C}}-{{V}_{A}}=4V\]


You need to login to perform this action.
You will be redirected in 3 sec spinner