A) \[\frac{Q}{4\pi {{\in }_{0}}r_{1}^{2} }\]
B) \[\frac{Qr_{1}^{2}}{4\pi {{\in }_{0}}{{R}^{4}} }\]
C) \[\frac{Qr_{1}^{2}}{3\pi {{\in }_{0}}{{R}^{4}} }\]
D) 0
Correct Answer: B
Solution :
Let us consider a spherical shell of thickness \[dx\]and radius\[x\]. The volume of this spherical shell\[4\pi {{x}^{2}}dx\]. The charge enclosed within shell |
\[=\left[ \frac{Q.x}{\pi {{R}^{4}}} \right]\left[ 4\pi {{x}^{2}}dx \right]=\frac{4Q}{{{R}^{4}}}{{x}^{3}}dx\] |
The charge enclosed in a sphere of radius \[{{r}_{1}}\] is |
\[=\frac{4Q}{{{R}^{4}}}\int\limits_{0}^{{{r}_{1}}}{{{x}^{3}}dx=\frac{4Q}{{{R}^{4}}}\left[ \frac{{{x}^{4}}}{4} \right]_{0}^{{{r}_{1}}}=\frac{Q}{^{{{R}^{4}}}}}{{r}_{1}}^{4}\] |
\[\therefore \]The electric field at point p inside the sphere at a distance\[{{r}_{1}}\], from the centre of the sphere is |
\[E=\frac{1}{4\pi {{\in }_{0}}}\frac{\left[ \frac{Q}{{{R}^{4}}}{{r}_{1}}^{4} \right]}{{{r}_{1}}^{2}}\] |
\[=\frac{1}{4\pi {{\in }_{0}}}\frac{Q}{{{R}^{4}}}{{r}_{1}}^{2}\] |
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