2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains \[4\mu C\] charge, its radius will be:  [Take:\[\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}\]]                                  [JEE Online 08-04-2017]

    A)  32 mm                

    B)  16 mm

    C)  28 mm                

    D)  20 mm

    Correct Answer: B

    Solution :

    Energy of sphere\[=\frac{{{Q}^{2}}}{2C}\]                 \[4.5=\frac{16\times {{10}^{-12}}}{2C}\]                 \[C=\frac{16\times {{10}^{-12}}}{9}=4\pi {{\varepsilon }_{0}}R\]                 \[R=\frac{16\times {{10}^{-12}}}{9}\times \frac{1}{4\pi {{\varepsilon }_{0}}}\]                 \[=9\times {{10}^{9}}\times \frac{16}{9}\times {{10}^{-12}}\]                 \[=16\times {{10}^{-3}}=16mm\]


You need to login to perform this action.
You will be redirected in 3 sec spinner