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EAMCET Medical EAMCET Medical Solved Paper-2006

  • question_answer
    The electrical potential on the surface of a sphere of radius r due to a charge \[3\times {{10}^{-6}}C\]is 500 V. The intensity of electric field on the surface of the sphere is  \[\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} \right]\]\[(in\text{ }N{{C}^{-1}})\]:

    A)  \[<10\]                               

    B)  >20

    C)  Between 10 and 20       

    D)  <5

    Correct Answer: A

    Solution :

                     \[{{V}_{s}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{R}\] \[\therefore \]  \[500=\frac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}}{R}\] \[\Rightarrow \]               \[R=\frac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}}{500}\] \[\Rightarrow \]               \[R=54\,\,m\] \[\therefore \]  Electric field on the surface \[{{E}_{s}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{{{R}^{2}}}\] \[=\frac{{{V}_{S}}}{R}=\frac{500}{54}=9.25<10\,N{{C}^{-1}}\]


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