A) \[\frac{{{\rho }_{o}}}{4{{\varepsilon }_{0}}}\left( \frac{r}{3}-\frac{{{r}^{2}}}{4R} \right)\]
B) \[\frac{{{\rho }_{o}}}{{{\varepsilon }_{o}}}\left( \frac{r}{3}-\frac{{{r}^{2}}}{4R} \right)\]
C) \[\frac{{{\rho }_{o}}}{3{{\varepsilon }_{o}}}\left( \frac{r}{3}-\frac{{{r}^{2}}}{4R} \right)\]
D) \[\frac{{{\rho }_{o}}}{12{{\varepsilon }_{o}}}\left( \frac{r}{3}-\frac{{{r}^{2}}}{4R} \right)\]
Correct Answer: B
Solution :
Let us consider a spherical shell of radius x and thickness dx. Charge on this shell \[dq=\rho .4\pi {{x}^{2}}dx={{\rho }_{0}}\left( 1-\frac{x}{R} \right).4\pi {{x}^{2}}dx\] \[\therefore \]Total charge in the spherical region from centre to r (r < R) is \[q=\int_{{}}^{{}}{dq=4\pi {{\rho }_{0}}}\int\limits_{0}^{r}{\left( 1-\frac{x}{R} \right){{x}^{2}}dx}\] \[=4\pi {{\rho }_{0}}\left[ \frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4R} \right]_{0}^{r}\]\[=4\pi {{\rho }_{0}}\left[ \frac{{{r}^{3}}}{3}-\frac{{{r}^{4}}}{4R} \right]\] \[=4\pi {{\rho }_{0}}{{r}^{3}}\left[ \frac{1}{3}-\frac{r}{4R} \right]\] \[\therefore \]Electric field at \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{{{r}^{2}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{4\pi {{\rho }_{0}}{{r}^{3}}}{{{r}^{2}}}\left[ \frac{1}{3}-\frac{r}{4R} \right]\] \[=\frac{{{\rho }_{0}}}{{{\varepsilon }_{0}}}\left[ \frac{r}{3}-\frac{{{r}^{2}}}{4R} \right]\]You need to login to perform this action.
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