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JEE Main & Advanced AIEEE Solved Paper-2009

  • question_answer
    Let\[p(r)=\frac{Q}{\pi {{R}^{4}}}r\]be the charge density distribution for a solid sphere of radius R and total charge Q. For a point 'p' inside the sphere at distance\[{{r}_{1}}\]from the centre of sphere, the magnitude of electric field is     AIEEE  Solved  Paper-2009

    A) 0                                             

    B) \[\frac{Q}{4\pi {{\varepsilon }_{0}}r_{1}^{2}}\]  

    C)        \[\frac{Qr_{1}^{2}}{4\pi {{\varepsilon }_{0}}{{R}^{4}}}\]

    D)        \[\frac{Qr_{1}^{2}}{3\pi {{\varepsilon }_{0}}{{R}^{4}}}\]

    Correct Answer: C

    Solution :

    \[E=\frac{Q}{4\pi {{\varepsilon }_{0}}{{R}^{4}}}\int\limits_{r=0}^{{{r}_{1}}}{\frac{4\pi {{r}^{3}}dr}{\pi {{R}^{4}}}}\] \[=\frac{Q}{4\pi {{\varepsilon }_{0}}}\frac{r_{1}^{2}}{{{R}^{4}}}\].


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