In the figure, the inner (shaded) region A represents a sphere of radius $r_A=1$, within which the electrostatic charge density varies with the radial distance r from the center as ${\rho }_A=kr$, where k is positive. In the spherical shell B of outer radius $r_B$, the electrostatic charge density varies as ${\rho }_B=\frac{2k}{r}$. Assume that dimensions are taken care of. All physical quantities are in their SI units.

Which of the following statement(s) is (are) correct?

Correct Option: If $r_B=\frac{3}{2}$, then the electric potential just outside B is $\frac{k}{{\epsilon }_0}$.

Step-by-Step Explanation:

1. Charge in the Inner Sphere A:
The inner shaded region A is a sphere of radius $r_A=1$ with volume charge density ${\rho }_A=kr$. We calculate the total charge $Q_A$ inside sphere A by integrating over thin spherical shells of radius $r$ and thickness $dr$:

$Q_A={\int }_0^{r_A}{\rho }_A\left(4\pi r^{2}dr\right)={\int }_0^1\left(kr\right)\left(4\pi r^{2}dr\right)=4\pi k{\int }_0^1{r}^{3}dr$

Evaluating the integral gives:

$Q_A=4\pi k{\left[\frac{r^4}{4}\right]}_0^1=\pi k$

2. Charge in the Spherical Shell B:
The outer shell B extends from inner radius $r_A=1$ to outer radius $r_B$ with volume charge density ${\rho }_B=\frac{2k}{r}$. The total charge $Q_B$ in region B is given by:

$Q_B={\int }_{r_A}^{r_B}{\rho }_B\left(4\pi r^{2}dr\right)={\int }_1^{r_B}\left(\frac{2k}{r}\right)\left(4\pi r^{2}dr\right)=8\pi k{\int }_1^{r_B}rdr$

Evaluating the integral gives:

$Q_B=8\pi k{\left[\frac{r^2}{2}\right]}_1^{r_B}=4\pi k(r_B^2-1)$

3. Total Enclosed Charge of the Configuration:
The total enclosed charge $Q_{\text{total}}$ within outer radius $r_B$ is:

$Q_{\text{total}}=Q_A+Q_B=\pi k+4\pi k(r_B^2-1)=4\pi k{r}_B^2-3\pi k$

4. Electric Potential Just Outside B:
For a spherically symmetric charge distribution, the electric potential just outside the surface at $r=r_B$ is given by:

$V\left(r_B\right)=\frac{1}{4\pi {\epsilon }_0}\frac{Q_{\text{total}}}{r_B}=\frac{1}{4\pi {\epsilon }_0{r}_B}(4\pi k{r}_B^2-3\pi k)$

Substituting $r_B=\frac{3}{2}$:

$Q_{\text{total}}=4\pi k{\left(\frac{3}{2}\right)}^2-3\pi k=4\pi k\left(\frac{9}{4}\right)-3\pi k=9\pi k-3\pi k=6\pi k$

Now, calculating the potential $V$ at $r_B=\frac{3}{2}$:

$V=\frac{1}{4\pi {\epsilon }_0}\frac{6\pi k}{3/2}=\frac{1}{4\pi {\epsilon }_0}\left(4\pi k\right)=\frac{k}{{\epsilon }_0}$

Thus, if $r_B=\frac{3}{2}$, the electric potential just outside B is indeed $\frac{k}{{\epsilon }_0}$.