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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    The radius of two metallic spheres \[A\] and \[B\] are \[{{r}_{1}}\] and \[{{r}_{2}}\] respectively \[({{r}_{1}}>{{r}_{2}})\]. They are connected by a thin wire and the system is given a certain charge. The charge will be greater

    A)                    On the surface of the sphere \[B\]

    B)                    On the surface of the sphere \[A\]

    C)                    Equal on both

    D)                    Zero on both

    Correct Answer: B

    Solution :

                       After connection of wire, potential becomes equal \ \[\frac{{{Q}_{1}}}{{{r}_{1}}}=\frac{{{Q}_{2}}}{{{r}_{2}}}\] Þ \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\] when r1 > r2, then \[{{Q}_{1}}>{{Q}_{2}}\]


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