A) \[\frac{{{\rho }_{0}}r}{4{{\varepsilon }_{0}}}\left( \frac{5}{3}-\frac{r}{R} \right)\]
B) \[\frac{4\pi {{\rho }_{0}}r}{3{{\varepsilon }_{0}}}\left( \frac{5}{3}-\frac{r}{R} \right)\]
C) \[\frac{{{\rho }_{0}}r}{4{{\varepsilon }_{0}}}\left( \frac{5}{4}-\frac{r}{R} \right)\]
D) \[\frac{{{\rho }_{0}}r}{3{{\varepsilon }_{0}}}\left( \frac{5}{4}-\frac{r}{R} \right)\]
Correct Answer: C
Solution :
Let us consider a spherical shell of radius \[x\]and thickness\[dx\]. |
Charge on this shell, |
\[dq=\rho .4\pi {{x}^{2}}dx={{\rho }_{0}}\left( \frac{5}{4}-\frac{x}{R} \right).4\pi {{x}^{2}}dx\] |
\[\therefore \] Total charge in the spherical region from centre to \[r\left( r<R \right)\]is |
\[q=\int{dq=4\pi {{\rho }_{0}}\int\limits_{0}^{r}{\left( \frac{5}{4}-\frac{x}{R} \right){{x}^{2}}dx}}\]\[=4\pi {{\rho }_{0}}\left[ \frac{5}{4}.\frac{{{r}^{3}}}{3}-\frac{1}{R}.\frac{{{r}^{4}}}{4} \right]=\pi {{\rho }_{0}}{{r}^{3}}\left( \frac{5}{3}-\frac{r}{R} \right)\] |
\[\therefore \] Electric field at \[r,E=\frac{1}{4\pi {{\in }_{0}}}.\frac{q}{{{r}^{2}}}\]\[=\frac{1}{4\pi {{\in }_{0}}}.\frac{\pi {{\rho }_{0}}{{r}^{3}}}{{{r}^{2}}}\left( \frac{5}{3}-\frac{r}{R} \right)=\frac{{{\rho }_{0}}r}{4{{\in }_{0}}}\left( \frac{5}{3}-\frac{r}{R} \right)\] |
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