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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Charge and Coulombs Law

  • question_answer
    Force of attraction between two point charges \[Q\] and ? Q separated by \[d\,metre\] is \[{{F}_{e}}\]. When these charges are placed on two identical spheres of radius \[R=0.3\,d\] whose centres are \[d\,metre\]apart, the force of attraction between them is  [AIIMS 1995]           

    A)                    Greater than \[{{F}_{e}}\]                                          

    B)                    Equal to \[{{F}_{e}}\]           

    C)                    Less than \[{{F}_{e}}\]     

    D)                    Less than \[{{F}_{e}}\]

    Correct Answer: A

    Solution :

                 Separation between the spheres is not too large as compared to their radius so due to induction effect redistribution of charge takes place. Hence effective charge separation decreases so force increases.           


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