Pre-Quantum Electrodynamics

\paragraph{Example 4.5:} metal sphere of radius \(a\) carrying charge \(Q\), surrounded out to radius \(b\) by a linear dielectric material of permittivity \(\varepsilon\). Find potential at center (relative to infinity). \paragraph{Solution:} need to know \({\bf E}\). Could try to locate bound charge: but we don't know \({\bf P}\) ! What we do know: free charge, situation is spherically symmetric, so can calculate \({\bf D}\) using \ref{Gr(4.23)}: \[ {\bf D} = \frac{Q}{4\pi r^2} \hat{\bf r}, \hspace{1cm} r > a. \] Inside sphere, \({\bf E} = {\bf P} = {\bf D} = 0\). Find \({\bf E}\) using \ref{Gr(4.32)}: \[ {\bf E} = \left\{ \begin{array}{cc} \frac{Q}{4\pi \varepsilon r^2} \hat{\bf r}, & a < r < b, \\ \frac{Q}{4\pi \varepsilon_0 r^2} \hat{\bf r}, & r > b. \end{array} \right. \] The potential is thus \[ V = -\int_\infty^0 d{\bf l} \cdot {\bf E} = -\int_\infty^b dr \frac{Q}{4\pi\varepsilon_0 r^2} - \int_b^a dr \frac{Q}{4\pi\varepsilon r^2} = \frac{Q}{4\pi} \left( \frac{1}{\varepsilon_0 b} + \frac{1}{\varepsilon a} - \frac{1}{\varepsilon b} \right). \]

It was thus not necessary to compute the polarization or the bound charge explicitly. This can be done: \[ {\bf P} = \varepsilon_0 \chi_e {\bf E} = \frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon r^2} \hat{\bf r}, \] so \[ \rho_b = -{\boldsymbol \nabla} \cdot {\bf P} = 0, \hspace{1cm} \sigma_b = {\bf P} \cdot \hat{\bf n} = \left\{ \begin{array}{cc} \frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon b^2}, & \mbox{outer surface} \\ -\frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon a^2}, & \mbox{inner surface} \end{array} \right. \]

Dielectric thus like an imperfect conductor: charge \(Q\) not fully screened.

\paragraph{Example 4.6:} parallel-plate capacitor filled with insulating material of dielectric constant \(\varepsilon_r\). What is the effect on the capacitance ? \paragraph{Solution:} field confined between plates, and reduced by factor \(1/\varepsilon_r\). Potential difference \(V\) also reduced by same factor. Since \(Q = C/V\), capacitance is increased by factor of \(\varepsilon_r\), so \[ C = \varepsilon_r C_{vac} \label{Gr(4.37)} \]

\paragraph{Example 4.7:} sphere of homogeneous dielectric material in uniform electric field \({\bf E}_0\). Find electric field inside sphere. \paragraph{Solution:} resembles Example 3.8 (conducting sphere), here cancellation is not total.

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

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

Inside and outside sphere: \[ V_{in} (r,\theta) = \sum_{l=0}^\infty A_l r^l P_l (\cos \theta), \hspace{1cm} V_{out} (r,\theta) = -E_0 r \cos \theta + \sum_{l=0}^\infty \frac{B_l}{r^{l+1}} P_l (\cos \theta). \] 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, \[ V_{in} (r, \theta) = -\frac{3E_0}{\varepsilon_r + 2} r \cos \theta = -\frac{3E_0}{\varepsilon_r + 2} z, \hspace{1cm} {\bf E} = \frac{3}{\varepsilon_r + 2} {\bf E}_0. \]

\paragraph{Example 4.8:} suppose region below \(z = 0\) is filled with uniform linear dielectric with susceptibility \(\chi_e\). Calculate force on point charge \(q\) situated a distance \(d\) above origin. \paragraph{Solution:} bound surface charge is of opposite sign, force is attractive. No bound volume charge because of \ref{Gr(4.39)}. Using \ref{Gr(4.11)} and \ref{Gr(4.30)}, \[ \sigma_b = {\bf P} \cdot \hat{\bf n} = P_z = \varepsilon_0 \chi_e E_z \] 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 $π$-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 \((0,0,-d)\): \[ V (x,y,z>0) = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{\sqrt{x^2 + y^2 + (z-d)^2}} + \frac{q_b}{\sqrt{x^2 + y^2 + (z+d)^2}}\right] \] A charge \(q + q_b\) at \((0,0,d)\) gives \[ V (x,y,z<0) = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q + q_b}{\sqrt{x^2 + y^2 + (z-d)^2}}\right] \] Putting these two together yields a solution to Poisson going to zero at infinity, and is therefore the unique solution. Correct discontinuity at \(z = 0\): \[ -\varepsilon_0 \left( \frac{\partial V}{\partial z}|_{z = 0^+} - \frac{\partial V}{\partial z}|_{z = 0^-} \right) = -\frac{1}{2\pi} \left(\frac{\chi_e}{\chi_e + 2}\right) \frac{qd}{(r^2 + d^2)^{3/2}}. \] Force on \(q\): \[ {\bf F} = \frac{1}{4\pi\varepsilon_0} \frac{q q_b}{4d^2} \hat{\bf z} = -\frac{1}{4\pi\varepsilon_0} \left( \frac{\chi_e}{\chi_e + 2} \right) \frac{q^2}{4d^2} \hat{\bf z} \label{Gr(4.54)} \]