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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    Two spheres \[A\]and \[B\] of radius \[4cm\] and \[6cm\] are given charges of \[80\mu c\] and \[40\mu c\]respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is                                                                     [MP PET 1991]

    A)                    \[20\mu C\]from \[A\]to \[B\]                                 

    B)            \[16\mu C\] from \[A\]to \[B\]

    C)                    \[32\mu C\]from \[B\] to \[A\]                                

    D)            \[32\mu C\] from \[A\]to \[B\]

    Correct Answer: D

    Solution :

                 Total charge \[Q=80+40=120\,\mu \,C\]. By using the formula \[{{Q}_{1}}'=Q\,\left[ \frac{{{r}_{1}}}{{{r}_{1}}+{{r}_{2}}} \right]\]. New charge on sphere A is \[Q_{A}^{'}=Q\,\left[ \frac{{{r}_{A}}}{{{r}_{A}}+{{r}_{B}}} \right]\]\[=120\,\left[ \frac{4}{4+6} \right]\,=48\,\mu \,C\]. Initially it was \[80\mu C\] i.e., \[32\,\mu \,C\] charge flows from A to B.


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