##### Boundary Value Problems with Linear Dielectrics
emsm.esm.ld.bvp

In homogeneous linear dielectric, the bound charge density is directly proportional to the free charge:

- Gr (4.39)

\[ \rho_b = -{\boldsymbol \nabla} \cdot {\bf P} = - {\boldsymbol \nabla} \cdot \left(\varepsilon_0 \frac{\chi_e}{\varepsilon} {\bf D}\right) = -\left( \frac{\chi_e}{1 + \chi_e} \right) \rho_f \tag{rhob_ld}\label{rhob_ld} \] If \(\rho = 0\), then any net charge is located on the surface, and the potential then obeys the Laplace equation.

It is convenient to rewrite our boundary conditions in terms of free charge densities: from Dperp_disc,

- Gr (4.40)

\[ \varepsilon_{above} E^{\perp}_{above} - \varepsilon_{below} E^{\perp}_{below} = \sigma_f \tag{Edisc_f}\label{Edisc_f} \]

or in terms of the potential,

- Gr (4.41)

\[ \varepsilon_{above} \frac{\partial \phi_{above}}{\partial n} - \varepsilon_{below} \frac{\partial \phi_{below}}{\partial n} = -\sigma_f \tag{dpdisc_f}\label{dpdisc_f} \]

The potential itself is continuous,

- Gr (4.42)

\[ \phi_{above} = \phi_{below} \tag{pcont}\label{pcont} \]

Example: sphere of homogeneous dielectric material in uniform electric field

Consider a sphere made of a linear isotropic dielectric material and subjected to an external electrostatic field \({\bf E}_0\).

Task: find the electric field inside the sphere.

Solution: resembles the case of the conducting sphere, here cancellation is not total.

Need to solve Laplace's equation for \(\phi(r, \theta)\) with boundary conditions

\begin{align*} &(i)~~\phi_{in} (R,\theta) = \phi_{out} (R, \theta), \nonumber \\ &(ii)~~\varepsilon \frac{\partial \phi_{in} (R,\theta)}{\partial n} = \varepsilon_0 \frac{\partial \phi_{out} (R,\theta)}{\partial n}, \nonumber \\ &(iii)~~\phi_{out} (r) \rightarrow -E_0 r \cos \theta, ~~r \gg R. \end{align*}

Inside and outside sphere: \[ \phi_{in} (r,\theta) = \sum_{l=0}^\infty A_l r^l P_l (\cos \theta), \hspace{1cm} \phi_{out} (r,\theta) = -E_0 r \cos \theta + \sum_{l=0}^\infty \frac{B_l}{r^{l+1}} P_l (\cos \theta). \] Boundary condition \((i)\) imposes \[ \sum_{l=0}^\infty A_l R^l P_l(\cos \theta) = -E_0 R \cos \theta + \sum_{l=0}^\infty \frac{B_l}{R^{l+1}} P_l (\cos \theta) \] so \[ A_l R^l = \frac{B_l}{R^{l+1}}, ~~ l \neq 1, \hspace{1cm} A_1 R = -E_0 R + \frac{B_1}{R^2}. \] Boundary condition \((ii)\): \[ \varepsilon_r \sum_{l=0}^\infty l A_l R^{l-1} P_l (\cos \theta) = - E_0 \cos \theta - \sum_{l=0}^\infty \frac{(l+1)B_l}{R^{l+2}} P_l (\cos \theta) \] so \[ \varepsilon_r l A_l R^{l-1} = -\frac{(l+1)B_l}{R^{l+2}}, ~~ l \neq 1, \hspace{1cm} \varepsilon_r A_1 = -E_0 - \frac{2B_1}{R^3}. \] We then have \[ A_l = 0 = B_l, ~~ l \neq 1, \hspace{1cm} A_l = -\frac{3}{\varepsilon_r + 2} E_0, ~~ B_1 = \frac{\varepsilon_r - 1}{\varepsilon_r + 2} R^3 E_0. \] Thus, \[ \phi_{in} (r, \theta) = -\frac{3E_0}{\varepsilon_r + 2} r \cos \theta = -\frac{3E_0}{\varepsilon_r + 2} z, \hspace{1cm} {\bf E} = \frac{3}{\varepsilon_r + 2} {\bf E}_0. \]

Example: point charge above dielectric plane

Suppose that the whole region below \(z = 0\) is filled with a uniform linear dielectric with susceptibility \(\chi_e\).

Task: calculate the force on a point charge \(q\) situated a distance \(d\) above the origin.

Solution: bound surface charge is of opposite sign, force is attractive. No bound volume charge because of rhob_ld. Using sigmab and PchiE, \[ \sigma_b = {\bf P} \cdot \hat{\bf n} = P_z = \varepsilon_0 \chi_e E_z \] where \(E_z\) is the z-component of total field just below surface of dielectric (due to \(q\) and to bound charge). Contribution from charge \(q\) from Coulomb (careful: \(\theta\) is \(\pi\) -rotated as compared to spherical coord so \(\theta = 0\) represents \(-\hat{z}\)) \[ -\frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + d^2} \cos \theta = -\frac{1}{4\pi\varepsilon_0} \frac{qd}{(r^2 + d^2)^{3/2}}, \hspace{1cm} r = \sqrt{x^2 + y^2}. \] \(z\) component of field from bound charge: \(-\sigma_b/2\varepsilon_0\), so \[ \sigma_b = \varepsilon_0 \chi_e \left[ -\frac{1}{4\pi\varepsilon_0} \frac{qd}{(r^2 + d^2)^{3/2}} - \frac{\sigma_b}{2\varepsilon_0} \right], \] so \[ \sigma_b = -\frac{1}{2\pi} \left(\frac{\chi_e}{\chi_e + 2}\right) \frac{qd}{(r^2 + d^2)^{3/2}}. \label{Gr(4.50)} \] As per conducting plane, except for factor \(\chi_e/(\chi_e + 2)\). Total bound charge: \[ q_b = -\left(\frac{\chi_e}{\chi_e + 2}\right) q. \] Field: by direct integration, or more nicely by method of images: replace dielectric by single point charge \(q_b\) at \((0,0,-d)\): \[ \phi (x,y,z>0) = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{\sqrt{x^2 + y^2 + (z-d)^2}} + \frac{q_b}{\sqrt{x^2 + y^2 + (z+d)^2}}\right] \] A charge \(q + q_b\) at \((0,0,d)\) gives \[ \phi (x,y,z<0) = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q + q_b}{\sqrt{x^2 + y^2 + (z-d)^2}}\right] \] Putting these two together yields a solution to Poisson going to zero at infinity, and is therefore the unique solution. Correct discontinuity at \(z = 0\): \[ -\varepsilon_0 \left( \frac{\partial \phi}{\partial z}|_{z = 0^+} - \frac{\partial \phi}{\partial z}|_{z = 0^-} \right) = -\frac{1}{2\pi} \left(\frac{\chi_e}{\chi_e + 2}\right) \frac{qd}{(r^2 + d^2)^{3/2}}. \] Force on \(q\): \[ {\bf F} = \frac{1}{4\pi\varepsilon_0} \frac{q q_b}{4d^2} \hat{\bf z} = -\frac{1}{4\pi\varepsilon_0} \left( \frac{\chi_e}{\chi_e + 2} \right) \frac{q^2}{4d^2} \hat{\bf z} \label{Gr(4.54)} \]

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42