question_answer

In spherical distribution, the charge density varies A as \[\rho (r)=\frac{A}{2}\] for \[a<r<b\] (as shown) where A is a constant A point charge Q lies at the centre of the sphere at \[r=0\]. The electric field in the region \[a<r<b\] has a constant magnitude for            

A)  \[A=0\]                                  

B)  \[A=Q\]

C)  \[A=\frac{Q}{2\pi {{a}^{2}}}\]       

D) \[A=\frac{Q}{4\pi {{a}^{2}}}\]

Correct Answer: C

Solution :

If we consider a thin shell of radius r of thickness dr such that a < r < b, then the charge on this shell is: \[dq=\rho 4\pi {{r}^{2}}dr\] \[=4\pi A\,r\,dr\] The total charge in the region of radius a to r is: \[q=\int{dq=4\pi A\int\limits_{a}^{r}{rdr=2\pi A({{r}^{2}}-{{a}^{2}})}}\] If E is the field at distance r in the region a < r < h then by Gauss law \[E(4\pi {{r}^{2}})=\frac{q+Q}{{{\varepsilon }_{0}}}=\frac{2\pi A({{r}^{2}}-{{a}^{2}})+Q}{{{\varepsilon }_{0}}}\] \[\Rightarrow E=\frac{A}{2{{\varepsilon }_{0}}}+\frac{Q-2\pi A{{a}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] Since, the field has a constant magnitude, in the region\[a<r<b\]. We have \[\frac{Q-2\pi A{{a}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[\therefore A=\frac{Q}{2\pi {{a}^{2}}}\]