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