ELECTROSTATICS O14

Solution

We are asked to find the surface density of polarization charges ($\sigma_p$) on the outer surface of the dielectric layer ($r = r_2$).

1. Electric Displacement ($D$):
The displacement field depends only on the free charge $q$. At a radius $r$ ($r_1 < r < r_2$):

$$D = \frac{q}{4\pi r^2}$$

2. Electric Field ($E$):
Inside the dielectric with constant $\epsilon_r$:

$$E = \frac{D}{\epsilon_0 \epsilon_r} = \frac{q}{4\pi \epsilon_0 \epsilon_r r^2}$$

3. Polarization ($P$):
The polarization vector magnitude is given by:

$$P = (\epsilon_r – 1)\epsilon_0 E = (\epsilon_r – 1)\epsilon_0 \frac{q}{4\pi \epsilon_0 \epsilon_r r^2} = \frac{(\epsilon_r – 1)q}{4\pi \epsilon_r r^2}$$

4. Surface Polarization Charge Density ($\sigma_p$):
On the outer surface, the normal vector points outward, parallel to $P$. Thus, $\sigma_p = \vec{P} \cdot \hat{n} = P(r_2)$.

$$\sigma_p = \frac{(\epsilon_r – 1)q}{4\pi \epsilon_r r_2^2}$$

Note: The total bound charge on the outer surface is positive, shielding the negative bound charge on the inner surface.

Answer: (d) $\frac{(\epsilon_r – 1)q}{4\pi \epsilon_r r_2^2}$