Electrostatics Question 736
Question: 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-
Options:
A) $ \pi R^{2}\alpha $
B) $ 4\pi R^{2}\alpha $
C) $ 2\pi R^{2}\alpha $
D) $ 3\pi R^{2}\alpha /4 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \int _{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 {{( \frac{r^{2}}{2} )}^{r}}=4\pi \alpha ( \frac{r^{2}}{2}-\frac{R^{2}}{2} )=2\pi \alpha ( r^{2}-R^{2} ) $
$ 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}}{2r^{2}\varepsilon {{} _{0}}} $
E is independent of r if
$ \frac{Q}{4\pi r^{2}{\varepsilon _{0}}}-\frac{\alpha R^{2}}{2r^{2}{\varepsilon _{0}}}=0 $
$ Q=2\pi R^{2}\alpha $
