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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1m. If the electric field between the plates is \[100\,\,N\,\,{{C}^{-1}},\]the magnitude of charge on each plate is  \[\left( \text{Take}\,{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}\frac{{{C}^{2}}}{N{{m}^{2}}} \right)\] [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A)      \[8.85\times {{10}^{-10}}C\]                            

    B) \[7.85\times {{10}^{-10}}C\]

    C)             \[9.85\times {{10}^{-10}}C\]                            

    D) \[6.85\times {{10}^{-10}}C\]

    Correct Answer: A

    Solution :

    The electric field between two plates is \[E=\frac{\sigma }{{{\varepsilon }_{0}}}=\frac{q}{A{{\varepsilon }_{0}}}\Rightarrow q=EA{{\varepsilon }_{0}}\] \[=(100)(1)(8.85\times {{10}^{-12}})\] \[q=8.85\times {{10}^{-10}}C\]


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