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

  • question_answer
    A capacitor of capacity \[{{C}_{1}}\] is charged up to V volt and then connected to an uncharged capacitor of capacity \[{{C}_{2}}.\] Then final potential difference across each will be

    A) \[\frac{{{C}_{2}}V}{{{C}_{1}}+{{C}_{2}}}\]

    B) \[\left( 1+\frac{{{C}_{2}}}{{{C}_{1}}} \right)V\]

    C) \[\frac{{{C}_{1}}V}{{{C}_{1}}+{{C}_{2}}}\,\]

    D) \[\,\left( 1-\frac{{{C}_{2}}}{{{C}_{1}}} \right)V\]

    Correct Answer: C

    Solution :

    [c] Common potential \[V=\frac{{{C}_{1}}V+{{C}_{2}}\times 0}{{{C}_{1}}+{{C}_{2}}}\] \[=\frac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}}V\]


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