A) \[\frac{Q}{4\pi \,{{\in }_{0}}\,{{r}_{1}}^{2}}\]
B) \[\frac{Q{{r}_{1}}^{2}}{4\pi \,{{\in }_{0}}\,{{R}^{4}}}\]
C) \[\frac{Q{{r}_{1}}^{2}}{3\pi \,{{\in }_{0}}\,{{R}^{4}}}\]
D) Zero
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 total charge enclosed within shell \[=\left[ \frac{Qx}{\pi {{R}^{4}}} \right]\,[4\pi {{x}^{2}}dx]=\frac{4Q}{{{R}^{4}}}{{x}^{3}}dx\] The total 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 \[E=\frac{1}{4\pi {{\in }_{0}}}\frac{\left[ \frac{Q}{{{R}^{4}}}{{r}_{1}}^{4} \right]}{{{r}_{1}}^{2}}=\frac{1}{4\pi {{\in }_{0}}}\frac{Q}{{{R}^{4}}}{{r}_{1}}^{2}\]You need to login to perform this action.
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