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

  • question_answer
    Let there be a spherically symmetric charge distribution with charge density varying as\[\rho (r)={{\rho }_{0}}\left( \frac{5}{4}-\frac{r}{R} \right)\]upto\[r=R,\]and\[\rho (r)=0\]for\[r>R,\]where r is the distance from the origin. The electric field at a distance \[r(r<R)\]from the origin is given by -                    [AIEEE 2010]

    A) \[\frac{{{\rho }_{0}}r}{3{{\varepsilon }_{0}}}\left( \frac{5}{4}-\frac{r}{R} \right)\]

    B) \[\frac{4\pi {{\rho }_{0}}r}{3{{\varepsilon }_{0}}}\left( \frac{5}{3}-\frac{r}{R} \right)\]

    C) \[\frac{{{\rho }_{0}}r}{4{{\varepsilon }_{0}}}\left( \frac{5}{3}-\frac{r}{R} \right)\]

    D) \[\frac{4{{\rho }_{0}}r}{3{{\varepsilon }_{0}}}\left( \frac{5}{4}-\frac{r}{R} \right)\]

    Correct Answer: C

    Solution :

    [c] \[r<R\]
    \[\oint{E.dS}=\frac{\int{{{\rho }_{v}}dv}}{{{\varepsilon }_{0}}}\]
    \[E.4\pi {{r}^{2}}=\int\limits_{0}^{r}{\frac{{{\rho }_{0}}}{{{\varepsilon }_{0}}}}\left( \frac{5}{4}-\frac{r}{R} \right)4\pi {{r}^{2}}dr\]
    \[E.4\pi {{r}^{2}}=\frac{{{\rho }_{0}}4\pi }{{{\varepsilon }_{0}}}\left[ \int\limits_{0}^{r}{\frac{5}{4}{{r}^{2}}dr-\int\limits_{0}^{r}{\frac{{{r}^{3}}}{R}dr}} \right]\]
    \[E.4\pi {{r}^{2}}=\frac{4\pi {{\rho }_{0}}}{{{\varepsilon }_{0}}}\left[ \frac{5}{4}\frac{{{r}^{3}}}{3}-\frac{{{r}^{4}}}{4R} \right]\]
    \[E=\frac{{{\rho }_{0}}}{{{\varepsilon }_{0}}}\left[ \frac{5}{4}\frac{r}{3}-\frac{{{r}^{2}}}{4R} \right]\]
    \[\]


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