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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Critical Thinking

  • question_answer
    Condenser \[A\] has a capacity of \[15\mu F\] when it is filled with a medium of dielectric constant 15. Another condenser \[B\] has a capacity of \[1\mu F\]with air between the plates. Both are charged separately by a battery of\[100\ V\]. After charging, both are connected in parallel without the battery and the dielectric medium being removed. The common potential now is                                                               [MNR 1994]

    A)                    \[400V\]

    B)                                      \[800\ V\]                              

    C)                    \[1200\ V\]                    

    D)            \[1600\ V\]

    Correct Answer: B

    Solution :

               Charge on capacitor A is given by \[{{Q}_{1}}=15\times {{10}^{-6}}\times 100=15\times {{10}^{-4}}C\] Charge on capacitor B is given by \[{{Q}_{2}}=1\times {{10}^{-6}}\times 100={{10}^{-4}}C\] Capacity of capacitor A after removing dielectric \[=\frac{15\times {{10}^{-6}}}{15}=1\mu F\] Now when both capacitors are connected in parallel their equivalent capacitance will be Ceq \[=1+1=2\mu F\] So common potential \[=\frac{(15\times {{10}^{-4}})+(1\times {{10}^{-4}})}{2\times {{10}^{-6}}}=800V.\]


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