Question #42654

Two point charges, each of magnitude $+Q$, are fixed at $x = 0$ and $x = L$ in a vacuum. A solid isotropic dielectric slab of thickness $t = L/4$ is placed such that its left face is fixed at $x = 0$. The initial relative permittivity of the slab is $\epsilon_r = 4$. A small test charge $+q$ is placed at a point $x = x_0$ ($t < x_0 < L$) such that it is in electrostatic equilibrium. (Assume the effective distance in vacuum for a thickness $t$ of dielectric $\epsilon_r$ is $t\sqrt{\epsilon_r}$).\The temperature of the slab is then increased by $\Delta T$. The slab has a coefficient of linear thermal expansion $\alpha$. The relative permittivity $\epsilon_r$ of the slab depends on its density $\rho$ according to the relation $\epsilon_r - 1 = K \rho$, where $K$ is a positive constant. Assuming the slab expands freely, find the shift in the equilibrium position of the test charge, $\Delta x = x_0' - x_0$. (Assume $\alpha \Delta T \ll 1$).