Question

A spherical dielectric shell with inner radius \(a\) and outer radius \(b\), has polarization \(\vec{P} = \dfrac{k}{r^2}\hat{r}\), where \(k\) is a constant and \(\hat{r}\) is the unit vector along the radial direction. Which of the following statements is/are correct?

• A. The surface density of bound charges on the inner and outer surfaces are \(-k/a^2\) and \(+k/b^2\), respectively. The volume density of bound charges inside the dielectric is zero.
• B. The surface density of bound charges is zero on both inner and outer surfaces. The volume density of bound charges inside the dielectric is \(+k\).
• C. The surface density of bound charges on the inner and outer surfaces are \(-k/a^2\) and \(k/b^2\), respectively. The volume density of bound charges inside the dielectric is zero.
• D. The surface density of bound charges is zero on both inner and outer surfaces. The volume density of bound charges inside the dielectric is \(\dfrac{3k}{4\pi(b^3 - a^3)}\).

Hint:

For radial polarization \(\vec{P}(r)\), check signs of surface charge densities using direction of \(\hat{n}\) on inner and outer surfaces.

Answer 1

The Correct Option is C

Step 1: Bound charge densities.
The surface bound charge density is given by \[ \sigma_b = \vec{P} \cdot \hat{n} \] At \(r = a\), the normal \(\hat{n}\) is inward, so \(\sigma_b(a) = -\dfrac{k}{a^2}\). At \(r = b\), the normal \(\hat{n}\) is outward, so \(\sigma_b(b) = +\dfrac{k}{b^2}\).

Step 2: Volume bound charge density.
The volume bound charge density is \[ \rho_b = -\nabla \cdot \vec{P} \] Since \(\vec{P} = \dfrac{k}{r^2}\hat{r}\), \[ \nabla \cdot \vec{P} = \dfrac{1}{r^2} \dfrac{d}{dr}(r^2 P_r) = \dfrac{1}{r^2} \dfrac{d}{dr}(r^2 \times \dfrac{k}{r^2}) = 0 \] Thus, \(\rho_b = 0\).

Step 3: Conclusion.
Hence, surface charge densities are \(-k/a^2\) and \(+k/b^2\), and there is no volume bound charge inside.