Analysis of the Problem:

We have a point charge $q$ at the origin within a linear dielectric medium characterized by relative permittivity $\epsilon_r$. We need to determine the behavior of polarization ($\vec{P}$) and the screened charge ($q_b$ or effective charge $q'$) based on the given options.

Electric Field and Polarization in Dielectric

The electric field ($\vec{E}$) due to the point charge $q$ inside the dielectric medium at a distance $r$ from the origin is given by:

$ \vec{E} = \frac{1}{4\pi\epsilon_r\epsilon_0} \frac{q}{r^2} \hat{r} $

where $\epsilon_0$ is the permittivity of free space.

For a linear dielectric material, the polarization vector $\vec{P}$ is proportional to the electric field:

$ \vec{P} = \epsilon_0 \chi_e \vec{E} $

where $\chi_e = \epsilon_r - 1$ is the electric susceptibility.

Substituting the expression for $\vec{E}$:

$ \vec{P} = \epsilon_0 (\epsilon_r - 1) \left( \frac{1}{4\pi\epsilon_r\epsilon_0} \frac{q}{r^2} \hat{r} \right) $

$ \vec{P} = \frac{(\epsilon_r - 1)q}{4\pi\epsilon_r} \frac{1}{r^2} \hat{r} $

The magnitude of the polarization is:

$ |\vec{P}| = \left| \frac{(\epsilon_r - 1)q}{4\pi\epsilon_r} \right| \frac{1}{r^2} $

Therefore, the magnitude of polarization varies as $\frac{1}{r^2}$. This confirms that **Option A is correct** and **Option B is incorrect**.

Screened Charge Analysis

The presence of polarization $\vec{P}$ induces bound charges within the dielectric. The electric field inside the dielectric can be viewed as resulting from the original free charge $q$ and the induced bound charge $q_b$. The relationship is:

$ \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q + q_b}{r^2} \hat{r} $

Comparing this with the electric field expression in the dielectric:

$ \frac{1}{4\pi\epsilon_0} \frac{q + q_b}{r^2} = \frac{1}{4\pi\epsilon_r\epsilon_0} \frac{q}{r^2} $

$ q + q_b = \frac{q}{\epsilon_r} $

The total induced bound charge is:

$ q_b = q \left( \frac{1}{\epsilon_r} - 1 \right) = -q \frac{\epsilon_r - 1}{\epsilon_r} $

The term "screened charge" usually refers to this induced bound charge $q_b$. We need to compare its magnitude $|q_b|$ with the original charge magnitude $|q|$.

• For $\epsilon_r > 1$: $ |q_b| = |q| \frac{\epsilon_r - 1}{\epsilon_r} $ Since $\epsilon_r > 1$, we have $0 < \frac{\epsilon_r - 1}{\epsilon_r} < 1$. Thus, $|q_b| < |q|$. The magnitude of the bound charge (screened charge) is less than the magnitude of the point charge $q$. This confirms that **Option C is correct**.
• For $\epsilon_r = 1$: The medium is vacuum, so $\chi_e = 0$, and $q_b = -q \frac{1 - 1}{1} = 0$. The magnitude of the screened charge is $0$, which is not more than $|q|$. This confirms that **Option D is incorrect**.

Conclusion

Based on the analysis:

• Option A is correct because $|\vec{P}| \propto \frac{1}{r^2}$.
• Option C is correct because for $\epsilon_r > 1$, the magnitude of the induced bound charge $|q_b| = |q| \frac{\epsilon_r - 1}{\epsilon_r}$ is less than $|q|$.