# The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density  where A is a constant and r is the distance from the centre.  At the centre of the spheres is a point charge Q.  The value of A such that the electric field in the region between the spheres will be constant, is

Qenc - charge enclosed by closed surface.

$\int \vec{E} \cdot \vec{ds} = \frac{Q+q}{\epsilon_o} \Rightarrow E \times 4 \pi r^2 = \frac{Q+q}{\epsilon_o} \:\:\:\:\:\: -(i)$

$q = \int_{a}^{r}\frac{A}{x} 4 \pi x^2 dx = 4 \pi A \int_{a}^{r}xdx$

$= 4 \pi A \left [ \frac{x^2}{2} \right ]_{a}^{r} = 2 \pi A(r^2-a^2)$

Now putting the value of q in equation (i)

$E \times 4 \pi r^2 = \frac{1}{\epsilon_o}\left [ Q + 2 \pi A(r^2-a^2) \right ]$

$E = \frac{1}{4 \pi \epsilon_o}\left [\frac{Q}{r^2} + 2 \pi A-\frac{2 \pi A a^2}{r^2} \right ]$

E will be constant if it is independent of r

$\therefore \frac{Q}{r^2}= \frac{2 \pi A a^2}{r^2} or A = \frac{Q}{2 \pi a^2}$

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