Question

A dielectric sphere of radius \( R \) has constant polarization \( \mathbf{P} = P_0 \hat{z} \) so that the field inside the sphere is \( \mathbf{E} = - \frac{P_0}{3 \epsilon_0} \hat{z} \). Then, which of the following is (are) correct?

• A. The bound surface charge density is \( P_0 \cos\theta \).
• B. The electric field at a distance \( r \) on the z-axis varies as \( \frac{1}{r^2} \) for \( r \gg R \).
• C. The electric potential at a distance \( 2R \) on the z-axis is \( \frac{P_0 R}{12 \epsilon_0} \).
• D. The electric field outside is equivalent to that of a dipole at the origin.

Hint:

For a polarized dielectric sphere, the electric field outside behaves like a dipole field, decaying as \( \frac{1}{r^2} \) at large distances, and the surface charge density is proportional to \( \cos\theta \).

Answer 1

The Correct Option is A, C, D

Step 1: Understanding the polarization of the dielectric sphere.
For a dielectric sphere with constant polarization, the bound surface charge density is given by \( \sigma_b = \mathbf{P} \cdot \hat{n} \), where \( \hat{n} \) is the unit normal to the surface. Since the polarization is along the z-axis, the bound surface charge density is \( P_0 \cos\theta \). Outside the sphere, the electric field behaves like a dipole field, which decays as \( \frac{1}{r^2} \) at large distances. The electric potential at \( 2R \) on the z-axis can be derived using the appropriate formula, yielding \( \frac{P_0 R}{12 \epsilon_0} \).
Step 2: Conclusion.
Thus, the correct answers are options (A), (C), and (D).