Pre-Quantum Electrodynamics

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 \(q\): \[ {\bf F} = \frac{1}{4\pi\varepsilon_0} \frac{q q_b}{4d^2} \hat{\bf z} = -\frac{1}{4\pi\varepsilon_0} \left( \frac{\chi_e}{\chi_e + 2} \right) \frac{q^2}{4d^2} \hat{\bf z} \label{Gr(4.54)} \]