question_answer

A system consists of a uniform charged sphere of radius R and a surrounding medium filled by a charge with the volume density \[\rho =\frac{\alpha }{r},\] where \[\alpha \]a positive constant is and r is the distance from the center of the charge. The charge of the sphere for which the electric field intensity E outside the sphere is independent of r is-

A) \[\pi {{R}^{2}}\alpha \]           

B) \[4\pi {{R}^{2}}\alpha \]

C) \[2\pi {{R}^{2}}\alpha \]          

D)        \[3\pi {{R}^{2}}\alpha /4\]

Correct Answer: C

Solution :

[c] \[\oint{{{E}_{p}}dS=\frac{Q+Q'}{{{\varepsilon }_{0}}}}\]  Where Q' is the charge outside the sphere \[Q{{'}_{1}}\int\limits_{R}^{r}{dV}=\,\int\limits_{R}^{r}{\frac{\alpha }{r}\times 4\pi {{r}^{2}}dr}\] \[=4\pi \alpha {{\left( \frac{{{r}^{2}}}{2} \right)}^{r}}=4\pi \alpha \left( \frac{{{r}^{2}}}{2}-\frac{{{R}^{2}}}{2} \right)=2\pi \alpha \left( {{r}^{2}}-{{R}^{2}} \right)\] \[{{E}_{p}}4\pi {{r}^{2}}=\frac{Q+2\pi \alpha ({{r}^{2}}-{{R}^{2}})}{{{\varepsilon }_{0}}}\] \[{{E}_{p}}=\frac{Q}{4\pi {{r}^{2}}{{\varepsilon }_{0}}}+\frac{\alpha }{2{{\varepsilon }_{0}}}-\frac{\alpha {{R}^{2}}}{2{{r}^{2}}\varepsilon {{}_{0}}}\]  E is independent of r if \[\frac{Q}{4\pi {{r}^{2}}{{\varepsilon }_{0}}}-\frac{\alpha {{R}^{2}}}{2{{r}^{2}}{{\varepsilon }_{0}}}=0\] \[Q=2\pi \,{{R}^{2}}\alpha \]