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 ρ A = k r , where k is positive. In the spherical shell B of outer radius r B , the electrostatic charge density…

Question

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 $ρ_{A} = k r$ , where $k$ is positive. In the spherical shell $B$ of outer radius $r_{B}$ , the electrostatic charge density varies as $ρ_{B} = \frac{2 k}{r}$ . Assume that dimensions are taken care of. All physical quantities are in their SI units.

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

Aicharya · Accepted Answer

Since both the densities are positive, the total charge can not be zero. Hence, option A is incorrect.
For total charge,
$Q_{Total} = \int_{0}^{r_{A}} k r (4 π r^{2}) d r + \int_{r_{A}}^{r_{B}} \frac{2 k}{r} (4 π r^{2}) d r$
$= \frac{4 π k}{4} r_{A}^{4} + \frac{8 π k}{2} (r_{B}^{2} - r_{A}^{2})$
$= π k + 4 π k (r_{B}^{2} - r_{A}^{2})$
If $r_{B} = \frac{3}{2}$
$Q_{Total} = π k + 4 π k (\frac{9}{4} - 1)$
$= π k + 4 π k (\frac{5}{4}) = 6 π k$
The potential just outside $B$ will be,
$V = \frac{1}{4 π ε_{0}} \frac{Q_{total}}{r_{B}} = \frac{1}{4 π ε_{0}} \frac{6 π k}{r_{B}} = \frac{3 k}{2} \frac{2}{3 ε_{0}} = \frac{k}{ε_{0}}$
Hence, option B is correct.
If $r_{B} = 2$
$Q_{Total} = π k + 4 π k (4 - 1)$ $= 13 π k$
Hence, option C is incorrect
If $r_{B} = \frac{5}{2}$
$Q_{Total} = π k + 4 π k (\frac{25}{4} - 1)$
$= π k + π k (21)$
$= 22 π k$
Therefore, the electric field just outside B will be,
$E = \frac{1}{4 π ε_{0}} \frac{Q_{Total}}{{r_{B}}^{2}} \frac{1}{4 π ε_{0}} \frac{22 π k}{25} \times 4$ $= \frac{22 k}{25 ε_{0}}$
Hence, option D is incorrect.

In the figure, the inner (shaded) region A represents a sphere of radius r A = 1 , within which the…

Solution