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The region between two concentric shells of radii a and b $\;(b > a)\;$ contains volume charge density $\;\rho(r)=\large\frac{C}{r}\;$ , where C is a constant and r is the radial distance , as shown in figure . A point charge q is placed at the origin , r=0 . Find the value of C for which the electric field in the region between the spheres is constant.

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

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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
$q-2 \pi C a^2=0$
$C=\large\frac{q}{2 \pi a^2}\;.$
answered Feb 18, 2014 by yamini.v

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