The bound charge density is given by

$\begin{align*}\sigma_b = \mathbf{P} \cdot \hat{\mathbf{n}}\end{align*}$
where $\mathbf{P}$ is the polarization vector in the dielectric and $\hat{\mathbf{n}}$ is a normal vector that points from the dielectric to the conductor. The polarization vector in the dielectric is
$\begin{align*}\mathbf{P} = \epsilon_0 \chi_e \mathbf{E} = \epsilon_0 \left(K - 1 \right) \mathbf{E} = \epsilon_0 \left(K - 1 ...$
The electric displacement produced by the charge density $\sigma$ is $\sigma$ , and it is in the direction opposite $\hat{\mathbf{n}}$ . Thus,
$\begin{align*}\sigma_b = \mathbf{P} \cdot \hat{\mathbf{n}} = \boxed{\frac{1 - K}{K} \sigma}\end{align*}$
Therefore, answer (E) is correct.

Solution to 1986 Problem 54