question_answer

A charge Q is distributed uniformly in a sphere (solid). Then, the electric field at any point r where\[r<R\](r is the radius of sphere) varies as:

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                 \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{Qr}{{{R}^{3}}}\]                       \[(for\,\,r<R)\] \[\Rightarrow \]               \[E\propto r\]