$(a)\;\large\frac{q}{2 \pi a^2}\qquad(b)\;\large\frac{q}{4 \pi a^2}\qquad(c)\;\large\frac{q}{ \pi a^2}\qquad(d)\;\large\frac{2q}{ \pi a^2}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer : (a) $\;\large\frac{q}{2 \pi a^2}$

Explanation :

By symmetry electric field in the region between two shells is radial . Consider a shell of radius r . Flux through this shell is :

$\phi = \oint \overrightarrow{E}\;. d \overrightarrow{S}$

$\phi=E 4 \pi r^2=\large\frac{q_{inc}}{\in_{0}}$

Net charge between r=a to r=r would be

$Q=\int_{a}^{r}\;\rho (4 \pi r^2)\;dr=\int_{a}^{r}\;\large\frac{C}{r}\;4 \pi r^2\;dr$

$=\large\frac{4 \pi C }{2} \;[r^2]_{a}^{r}$

$Q=2 \pi C (r^2-a^2)$

$E\; 4 \pi r^2=\large\frac{Q+q}{\in_{0}}$

$E=\large\frac{k\;(2 \pi C (r^2-a^2)+q)}{r^2}$

$E=2 \pi \;k\;C+\large\frac{k}{r^2}\;(-2 \pi C\;a^2+q)$

Since E is constant . It should be independent of r

$q-2 \pi C a^2=0$

$C=\large\frac{q}{2 \pi a^2}\;.$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...