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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Mock Test - Capacitance

  • question_answer
    In the circuit shown in figure,\[{{C}_{1}}=6\,\mu F,\]\[{{C}_{2}}=3\,\mu F\], and battery B = 20 V. The switch \[{{S}_{1}}\] is first closed. It is then opened, and \[{{S}_{2}}\] is closed. What is the final charge on\[{{C}_{2}}\]?

    A) \[120\mu C\]     

    B) \[80\mu C\]

    C) \[40\mu C\]       

    D) \[20\mu C\]

    Correct Answer: C

    Solution :

    [c] After closing \[{{S}_{1}}\] charge on \[{{C}_{1}}\] is\[q=6\times 20=120\mu C\]. Now, \[{{S}_{1}}\] is opened, on closing\[{{S}_{2}}\], charge q will be distributed between \[{{C}_{1}}\]and \[{{C}_{2}}\]according to their copacitances. So charge on \[{{C}_{2}}\]is \[{{q}_{2}}=\frac{{{C}_{2}}q}{{{C}_{1}}+{{C}_{2}}}=\frac{3\times 120}{3+6}=40\mu C\]


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