The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has volume charge density $$\rho = \frac{A}{r}$$, 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:

First, let us identify the geometry. We have two concentric spheres with radii $$a$$ and $$b$$ satisfying $$a < r < b$$. The shell lying between these two radii is filled with a volume charge whose density varies as $$\rho = \dfrac{A}{r}$$, while the very centre (at $$r = 0$$) carries a point charge $$Q$$.

We wish to make the electric field $$E(r)$$ anywhere in the region $$a < r < b$$ the same for all values of $$r$$, that is, independent of $$r$$. To obtain the required value of $$A$$ we use Gauss’s law.

Gauss’s law in integral form states

$$\displaystyle \oint_S \vec E \cdot d\vec S \;=\;\frac{1}{\varepsilon_0}\,Q_{\text{enc}}.$$

For spherical symmetry, we choose a Gaussian surface that is itself a sphere of radius $$r$$ (with $$a < r < b$$). On this surface the magnitude $$E(r)$$ is the same everywhere and the flux becomes

$$E(r)\,(4\pi r^2)\;=\;\frac{1}{\varepsilon_0}\,Q_{\text{enc}}(r).$$

So

$$E(r)=\frac{Q_{\text{enc}}(r)}{4\pi\varepsilon_0\,r^2}.$$

Now we calculate the total charge enclosed inside radius $$r$$, namely

$$Q_{\text{enc}}(r)=Q_{\text{point}}+Q_{\text{shell up to }r}.$$

The point charge contributes simply $$Q_{\text{point}} = Q.$$

For the shell portion (from $$a$$ out to $$r$$) we integrate the given volume charge density. The differential charge in a thin spherical shell of radius $$r'$$ and thickness $$dr'$$ is

$$dQ = \rho(r')\,d\tau = \frac{A}{r'}\;\bigl(4\pi r'^2\,dr'\bigr)=4\pi A\,r'\,dr'.$$

Integrating from $$r' = a$$ to $$r' = r$$ gives

$$Q_{\text{shell up to }r}=4\pi A\int_{a}^{r}r'\,dr'$$

$$\hspace{3.55cm}=4\pi A\left[\frac{r'^2}{2}\right]_{a}^{r}$$

$$\hspace{3.55cm}=2\pi A\bigl(r^2 - a^2\bigr).$$

Hence

$$Q_{\text{enc}}(r)=Q+2\pi A\bigl(r^2-a^2\bigr).$$

Substituting this into the earlier result for $$E(r)$$ we obtain

$$E(r)=\frac{Q+2\pi A\bigl(r^2-a^2\bigr)}{4\pi\varepsilon_0\,r^2}.$$

We now enforce the requirement that $$E(r)$$ must not depend on $$r$$. For that to happen, the numerator must be directly proportional to $$r^2$$ so that the $$r^2$$ in the denominator cancels out and leaves a constant.

Let us write the numerator in the desired proportional form:

$$Q+2\pi A\bigl(r^2-a^2\bigr)=k\,r^2,$$

where $$k$$ is some constant (its exact value is immaterial—only the absence of any other $$r$$-dependence matters).

Comparing coefficients of like powers of $$r$$ on both sides, we see:

 • For the $$r^2$$ terms: $$2\pi A=k.$$
 • For the terms independent of $$r$$: $$Q-2\pi A a^2=0.$$

The second equality immediately gives us the needed value of $$A$$:

$$Q-2\pi A a^2=0\;\;\Longrightarrow\;\;A=\frac{Q}{2\pi a^2}.$$

This is a unique choice; any other value would leave residual $$r$$-dependence in the field.

Hence, the correct answer is Option C.