Question

The gap between the plates of a parallel-plate capacitor is filled with isotropic dielectric whose permittivity $\epsilon$ varies linearly from $\epsilon_1$ to $\epsilon_2$ ($\epsilon_2 > \epsilon_1$) in the direction perpendicular to the plates. The area of each plate equals $S$, the separation between the plates is equal to $d$. Find

a) the capacitance of the capacitor;

b) the space density of the bound charges as a function of $\epsilon$ if the charge of the capacitor is $q$ and the field $\mathbf{E}$ in it is directed toward

Answer 1

a) $C = \frac{S(\epsilon_2 - \epsilon_1)}{d \ln\left(\frac{\epsilon_2}{\epsilon_1}\right)}$, b) $\rho_b(\epsilon) = -\frac{q \epsilon_0 (\epsilon_2 - \epsilon_1)}{S d \epsilon^2}$

Answer 2

The problem involves a parallel-plate capacitor with a non-uniform dielectric. The permittivity varies linearly between the plates.

a) Capacitance of the capacitor: Let the plates be parallel to the xy-plane, located at z=0 and z=d. Let the area of each plate be S. The permittivity $\epsilon$ varies linearly from $\epsilon_1$ at one plate (say z=0) to $\epsilon_2$ at the other plate (z=d). The linear variation of permittivity with position z is given by: $\epsilon(z) = \epsilon_1 + \frac{\epsilon_2 - \epsilon_1}{d} z$. We can consider the capacitor as being composed of infinitesimally thin layers of thickness dz, stacked in series along the z-direction. Each layer at position z has an area S and permittivity $\epsilon(z)$. The capacitance of such a thin layer is $dC = \frac{\epsilon(z) S}{dz}$. Since these layers are in series, the reciprocal of the total capacitance C is the sum of the reciprocals of the capacitances of the layers: $\frac{1}{C} = \int_0^d \frac{1}{dC} = \int_0^d \frac{dz}{\epsilon(z) S}$. Substitute the expression for $\epsilon(z)$: $\frac{1}{C} = \int_0^d \frac{dz}{(\epsilon_1 + \frac{\epsilon_2 - \epsilon_1}{d} z) S} = \frac{1}{S} \int_0^d \frac{dz}{\epsilon_1 + \frac{\epsilon_2 - \epsilon_1}{d} z}$. Let $u = \epsilon_1 + \frac{\epsilon_2 - \epsilon_1}{d} z$. Then $du = \frac{\epsilon_2 - \epsilon_1}{d} dz$, so $dz = \frac{d}{\epsilon_2 - \epsilon_1} du$. When $z=0$, $u = \epsilon_1$. When $z=d$, $u = \epsilon_1 + \frac{\epsilon_2 - \epsilon_1}{d} d = \epsilon_1 + \epsilon_2 - \epsilon_1 = \epsilon_2$. The integral becomes: $\frac{1}{C} = \frac{1}{S} \int_{\epsilon_1}^{\epsilon_2} \frac{1}{u} \frac{d}{\epsilon_2 - \epsilon_1} du = \frac{d}{S(\epsilon_2 - \epsilon_1)} \int_{\epsilon_1}^{\epsilon_2} \frac{1}{u} du$. $\frac{1}{C} = \frac{d}{S(\epsilon_2 - \epsilon_1)} [\ln|u|]_{\epsilon_1}^{\epsilon_2} = \frac{d}{S(\epsilon_2 - \epsilon_1)} (\ln \epsilon_2 - \ln \epsilon_1) = \frac{d}{S(\epsilon_2 - \epsilon_1)} \ln\left(\frac{\epsilon_2}{\epsilon_1}\right)$. The capacitance is $C = \frac{S(\epsilon_2 - \epsilon_1)}{d \ln\left(\frac{\epsilon_2}{\epsilon_1}\right)}$. This formula assumes $\epsilon_1$ and $\epsilon_2$ are absolute permittivities. If they were relative permittivities, the absolute permittivity would be $\epsilon(z) = \epsilon_0 \epsilon_r(z)$, and the capacitance would be $C = \frac{\epsilon_0 S(\epsilon_2 - \epsilon_1)}{d \ln\left(\frac{\epsilon_2}{\epsilon_1}\right)}$. Based on the phrasing "permittivity $\epsilon$", we assume $\epsilon_1, \epsilon_2$ are absolute permittivities.

b) Space density of bound charges: The space density of bound charges $\rho_b$ is related to the polarization $\mathbf{P}$ by $\rho_b = -\nabla \cdot \mathbf{P}$. In a linear, isotropic dielectric, the polarization is related to the electric field $\mathbf{E}$ by $\mathbf{P} = \epsilon_0 \chi_e \mathbf{E} = \epsilon_0 (\epsilon_r - 1) \mathbf{E} = (\epsilon - \epsilon_0) \mathbf{E}$, where $\chi_e$ is the electric susceptibility and $\epsilon_r = \epsilon/\epsilon_0$ is the relative permittivity. The electric field $\mathbf{E}$ inside the parallel-plate capacitor is directed perpendicular to the plates. Let the field be in the +z direction (from the plate with charge +q to the plate with charge -q). The problem states the field is directed "toward", which is ambiguous. Let's assume the standard convention where the field points from positive to negative charge, so if the charge is q, the field is directed from the positive plate to the negative plate. Let the positive plate be at z=0 and the negative plate at z=d. Then $\mathbf{E} = E(z) \hat{\mathbf{z}}$. The displacement field $\mathbf{D}$ is uniform within the dielectric and is related to the free charge density $\sigma_f = q/S$ on the plates by $D = \sigma_f = q/S$. The relationship between $\mathbf{D}$ and $\mathbf{E}$ is $\mathbf{D} = \epsilon \mathbf{E}$. So, $E(z) = D/\epsilon(z) = \frac{q/S}{\epsilon(z)}$. $\mathbf{E}(z) = \frac{q}{S \epsilon(z)} \hat{\mathbf{z}}$. The polarization is $\mathbf{P}(z) = (\epsilon(z) - \epsilon_0) \mathbf{E}(z) = (\epsilon(z) - \epsilon_0) \frac{q}{S \epsilon(z)} \hat{\mathbf{z}} = \frac{q}{S} \left(1 - \frac{\epsilon_0}{\epsilon(z)}\right) \hat{\mathbf{z}}$. The space density of bound charges is $\rho_b = -\nabla \cdot \mathbf{P}$. Since $\mathbf{P}$ is in the z-direction and depends only on z, the divergence is $\frac{\partial P_z}{\partial z}$: $\rho_b = -\frac{\partial}{\partial z} \left[ \frac{q}{S} \left(1 - \frac{\epsilon_0}{\epsilon(z)}\right) \right] = -\frac{q}{S} \frac{\partial}{\partial z} \left(1 - \epsilon_0 \epsilon(z)^{-1}\right)$. $\rho_b = -\frac{q}{S} (-\epsilon_0) (-1) \epsilon(z)^{-2} \frac{d\epsilon}{dz} = -\frac{q \epsilon_0}{S [\epsilon(z)]^2} \frac{d\epsilon}{dz}$. From the linear variation $\epsilon(z) = \epsilon_1 + \frac{\epsilon_2 - \epsilon_1}{d} z$, we have $\frac{d\epsilon}{dz} = \frac{\epsilon_2 - \epsilon_1}{d}$. So, $\rho_b(z) = -\frac{q \epsilon_0}{S [\epsilon(z)]^2} \frac{\epsilon_2 - \epsilon_1}{d}$. We are asked for the space density as a function of $\epsilon$. Since $\epsilon$ varies linearly with z, z can be expressed in terms of $\epsilon$: $z = \frac{d}{\epsilon_2 - \epsilon_1} (\epsilon - \epsilon_1)$. So specifying z is equivalent to specifying $\epsilon$. The expression for $\rho_b$ already has $\epsilon(z)$ in the denominator. We can just write $\epsilon$ instead of $\epsilon(z)$ to indicate the local value of permittivity. $\rho_b(\epsilon) = -\frac{q \epsilon_0 (\epsilon_2 - \epsilon_1)}{S d \epsilon^2}$. The direction of $\mathbf{E}$ is assumed to be from z=0 to z=d. If the field were directed toward the other plate, the sign of $\mathbf{E}$ would be reversed, but the sign of q would also be reversed (assuming q is the charge on the plate where the field originates), leaving the expression for $\rho_b$ unchanged.

Final check: The total bound charge $q_b = \int \rho_b dV = \int_0^d \rho_b(z) S dz = \int_0^d -\frac{q \epsilon_0}{S [\epsilon(z)]^2} \frac{\epsilon_2 - \epsilon_1}{d} S dz$. $q_b = -\frac{q \epsilon_0 (\epsilon_2 - \epsilon_1)}{d} \int_0^d \frac{dz}{[\epsilon_1 + \frac{\epsilon_2 - \epsilon_1}{d} z]^2}$. Let $u = \epsilon_1 + \frac{\epsilon_2 - \epsilon_1}{d} z$, $dz = \frac{d}{\epsilon_2 - \epsilon_1} du$. $q_b = -\frac{q \epsilon_0 (\epsilon_2 - \epsilon_1)}{d} \int_{\epsilon_1}^{\epsilon_2} \frac{1}{u^2} \frac{d}{\epsilon_2 - \epsilon_1} du = -q \epsilon_0 \int_{\epsilon_1}^{\epsilon_2} u^{-2} du$. $q_b = -q \epsilon_0 [-u^{-1}]_{\epsilon_1}^{\epsilon_2} = -q \epsilon_0 (-\frac{1}{\epsilon_2} + \frac{1}{\epsilon_1}) = q \epsilon_0 \left(\frac{1}{\epsilon_2} - \frac{1}{\epsilon_1}\right) = q \epsilon_0 \frac{\epsilon_1 - \epsilon_2}{\epsilon_1 \epsilon_2}$. The bound surface charge density on the plates are $\sigma_{b1} = \mathbf{P}(0) \cdot (-\hat{\mathbf{z}})$ and $\sigma_{b2} = \mathbf{P}(d) \cdot (\hat{\mathbf{z}})$. $\mathbf{P}(z) = \frac{q}{S} (1 - \frac{\epsilon_0}{\epsilon(z)}) \hat{\mathbf{z}}$. $\sigma_{b1} = \frac{q}{S} (1 - \frac{\epsilon_0}{\epsilon_1}) (-1) = -\frac{q}{S} (1 - \frac{\epsilon_0}{\epsilon_1})$. $\sigma_{b2} = \frac{q}{S} (1 - \frac{\epsilon_0}{\epsilon_2})$. The total bound charge is $Q_b = q_b + \sigma_{b1} S + \sigma_{b2} S$. $Q_b = q \epsilon_0 \frac{\epsilon_1 - \epsilon_2}{\epsilon_1 \epsilon_2} - q (1 - \frac{\epsilon_0}{\epsilon_1}) + q (1 - \frac{\epsilon_0}{\epsilon_2})$. $Q_b = q \epsilon_0 \frac{\epsilon_1 - \epsilon_2}{\epsilon_1 \epsilon_2} - q + \frac{q \epsilon_0}{\epsilon_1} + q - \frac{q \epsilon_0}{\epsilon_2} = q \epsilon_0 \frac{\epsilon_1 - \epsilon_2}{\epsilon_1 \epsilon_2} + q \epsilon_0 (\frac{1}{\epsilon_1} - \frac{1}{\epsilon_2})$. $Q_b = q \epsilon_0 \frac{\epsilon_1 - \epsilon_2}{\epsilon_1 \epsilon_2} + q \epsilon_0 \frac{\epsilon_2 - \epsilon_1}{\epsilon_1 \epsilon_2} = 0$. The total bound charge in the volume and on the surfaces must sum to zero for a neutral dielectric. This confirms the expression for $\rho_b$.