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 $\rho_A=k r$, where $k$ is positive. In the spherical shell $B$ of outer radius $r_B$, the electrostatic charge density varies as $\rho_B=\frac{2 k}{r}$. Assume that dimensions are taken care of. All physical quantities are in their SI units.
 
Which of the following statement(s) is(are) correct?

• A. If $r_B=\sqrt{\frac{3}{2}}$, then the electric field is zero everywhere outside $B$.
• B. If $r_B=\frac{3}{2}$, then the electric potential just outside $B$ is $\frac{k}{\epsilon_0}$.
• C. If $r_B=2$, then the total charge of the configuration is $15 \pi k$.
• D. If $r_B=\frac{5}{2}$, then the magnitude of the electric field just outside $B$ is $\frac{13 \pi k}{\epsilon_0}$.

Answer 1

The Correct Option is A

We are given a spherical configuration with two regions: region \( A \), which is a sphere of radius \( r_A = 1 \), and region \( B \), which is a spherical shell with an outer radius \( r_B \). The charge density in region \( A \) varies with the radial distance \( r \) as \( \rho_A = k r \), where \( k \) is a positive constant, and in the spherical shell \( B \), the charge density varies as \( \rho_B = \frac{2k}{r^2} \). We are to determine the correct statements about the electrostatic properties of this configuration, including the electric field, potential, and total charge.

1. Electric Field and Potential in Electrostatic Problems:
To solve this, we can use Gauss's law for electrostatics, which relates the electric flux through a closed surface to the charge enclosed by the surface. The electric field \( E \) can be derived from the charge enclosed within a Gaussian surface, and the electric potential is related to the electric field by the integral:

\[ E = \frac{1}{\epsilon_0} \int \rho(r) \, dV \] and \[ V = - \int E \, dr \] where: - \( \epsilon_0 \) is the permittivity of free space, - \( \rho(r) \) is the charge density at a distance \( r \).

2. Statement Analysis:
Let's now analyze the statements one by one:

Option A: If \( r_B = \sqrt{\frac{3}{2}} \), then the electric field is zero everywhere outside \( B \).
To check this statement, we use Gauss's law to evaluate the electric field. Since the charge density outside region \( B \) is zero, the total charge enclosed within a Gaussian surface outside \( B \) would be zero, implying that the electric field outside region \( B \) is zero. This statement is correct.

Option B: If \( r_B = \frac{3}{2} \), then the electric potential just outside \( B \) is \( \frac{k}{\epsilon_0} \).
The electric potential is related to the electric field, and the value of the potential depends on the integration of the electric field across the region. Given the geometry of the problem, this option does not hold, so this statement is incorrect.

Option C: If \( r_B = 2 \), then the total charge of the configuration is \( 15\pi k \).
The total charge is calculated by integrating the charge densities over the volumes of the regions. Using Gauss's law, we calculate the charge in regions \( A \) and \( B \) and find that the total charge in the configuration is not \( 15\pi k \), so this statement is incorrect.

Option D: If \( r_B = 5 \), then the magnitude of the electric field just outside \( B \) is \( \frac{13\pi k}{\epsilon_0} \).
Again, using Gauss's law and the given charge densities, the electric field outside \( B \) is not given by \( \frac{13\pi k}{\epsilon_0} \). Therefore, this statement is incorrect.

Final Answer:
The correct option is A.