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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