JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • question_answer
    A solid ball of radius R has a charge density \[\rho \] given by \[\rho ={{\rho }_{o}}\left( 1-\frac{r}{R} \right)\] for \[0\le r\le R\]. The electric field outside the ball is: [JEE Online 15-04-2018 (II)]

    A) \[\frac{{{\rho }_{0}}{{R}^{3}}}{{{\in }_{0}}{{r}^{2}}}\]

    B) \[\frac{4{{\rho }_{0}}{{R}^{3}}}{3{{\in }_{0}}{{r}^{2}}}\]

    C) \[\frac{3{{\rho }_{0}}{{R}^{3}}}{4{{\in }_{0}}{{r}^{2}}}\]

    D) \[\frac{{{\rho }_{0}}{{R}^{3}}}{12{{\in }_{0}}{{r}^{2}}}\]

    Correct Answer: D

    Solution :

    [d] \[\rho ={{\rho }_{0}}\left( 1-\frac{r}{R} \right)\]
    \[dq=\rho dv\]
    \[{{q}_{in}}=\int_{{}}^{{}}{dq}\]
    \[=\rho dv\]
    \[{{\rho }_{0}}\left( 1-\frac{r}{R} \right)4\pi {{r}^{2}}dr\]
    \[\because dv=4\pi {{r}^{2}}dr\]
    \[=4\pi {{\rho }_{0}}\int_{0}^{R}{\left( 1-\frac{r}{R} \right){{r}^{2}}dr}\]
    \[=4\pi {{\rho }_{0}}\int_{0}^{R}{{{r}^{2}}dr-\frac{{{r}^{2}}}{R}dr}\]
    \[=4\pi {{\rho }_{o}}\left[ \left[ \frac{{{r}^{3}}}{3} \right]_{o}^{R}-\left[ \frac{{{r}^{4}}}{4R} \right]_{o}^{R} \right]\]
    \[=4\pi {{\rho }_{o}}\left[ \frac{{{R}^{3}}}{3}-\frac{{{R}^{4}}}{4R} \right]\]
    \[=4\pi {{\rho }_{o}}\left[ \frac{{{R}^{3}}}{3}-\frac{{{R}^{3}}}{4} \right]\]
    \[=4\pi {{\rho }_{o}}\left[ \frac{{{R}^{3}}}{12} \right]\]
    \[q=\frac{\pi {{\rho }_{o}}{{R}^{3}}}{3}\]
    \[E.4\pi {{r}^{2}}=\left( \frac{\pi {{\rho }_{o}}{{R}^{3}}}{3{{\in }_{o}}} \right)\]
    \[\Rightarrow E=\frac{{{\rho }_{o}}{{R}^{3}}}{12{{\in }_{o}}{{r}^{2}}}\]


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