A) \[{{r}^{1/2}}\]
B) \[{{r}^{-1}}\]
C) \[r\]
D) \[{{r}^{-2}}\]
Correct Answer: C
Solution :
Key Idea: Sphere is uniformly charged, hence charge density is constant throughout the sphere. When point P lies inside the sphere at a distance from the centre 0, then from Gauss theorem \[\oint{E.ds}=\frac{Q'}{{{\varepsilon }_{o}}}\] \[\therefore \] \[E.4\pi {{r}^{2}}=\frac{Q'}{{{\varepsilon }_{o}}}\] \[\Rightarrow \] \[E=\frac{1}{4\pi {{r}^{2}}}\frac{Q'}{{{\varepsilon }_{o}}}\] ?? (i) Change density\[=\rho =\frac{Q}{\frac{4}{3}\pi {{R}^{3}}}=\frac{Q'}{\frac{4}{3}\pi {{r}^{3}}}\] \[\Rightarrow \] \[Q'=Q{{\left( \frac{r}{R} \right)}^{3}}\] ...(2) Putting this value in Eq. (1), we have \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{Qr}{{{R}^{3}}}\] \[(for\,\,r<R)\] \[\Rightarrow \] \[E\propto r\]You need to login to perform this action.
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