Homework 7 · Displacement Current, Energy Flow, and Angular Momentum

Ch. 14, Ex. 1 · Displacement current in a circular capacitor

Problem. A parallel-plate capacitor has circular plates of radius \(a\), and an AC source imposes \(Q(t)=Q_0\sin\omega t\). Find the displacement current density \(\mathbf j_d\), total displacement current \(i_d\), and magnetic field at radius \(r<a\) between the plates. Ignore edges, assume uniform gap electric field, and use \(\omega a\ll c\).

Method: Displacement current and Ampère–Maxwell law.

Solution. Choose \(+z\) along the gap field for \(Q>0\). Uniform field gives

\[ \mathbf E=\frac{Q(t)}{\epsilon_0\pi a^2}\hat{\mathbf z},\qquad \boxed{\mathbf j_d=\epsilon_0\frac{\partial\mathbf E}{\partial t} =\frac{Q_0\omega}{\pi a^2}\cos\omega t\,\hat{\mathbf z},} \]

\[ \boxed{i_d=\pi a^2j_{d,z}=Q_0\omega\cos\omega t=\dot Q.} \]

For a circular Amperian contour of radius \(r<a\), its enclosed displacement current is \(j_{d,z}\pi r^2\). Thus

\[ 2\pi r B_\phi=\mu_0j_{d,z}\pi r^2, \qquad \boxed{\mathbf B(r,t)=\frac{\mu_0Q_0\omega r}{2\pi a^2}\cos\omega t\,\hat{\boldsymbol\phi}.} \]

The sense follows the right-hand rule for signed \(i_d\). \(\omega a\ll c\) is the quasistatic condition: propagation across the plate is fast compared with the field variation. 位移电流不是电荷穿过真空间隙，却在磁场方程中起相应作用。

Ch. 14, Ex. 3 · Energy entering a resistive wire

Problem. A long cylindrical lead of radius \(a\) and resistivity \(\rho_e\) carries steady current \(i\), uniformly distributed over its cross-section. Its interior electric field is uniform; neglect the return circuit's magnetic field. (a) Find magnitudes and directions of \(\mathbf E,\mathbf B,\mathbf S\) at \(r<a\). (b) Verify \(\nabla\cdot\mathbf S+\partial u/\partial t=-\rho_eJ^2=-\mathbf j\cdot\mathbf E\), where \(J=|\mathbf j|\).

Method: Ohm's law, Ampère's law, and Poynting's theorem.

Solution. (a) Choose the positive current along \(+z\). With \(J=i/(\pi a^2)>0\),

\[ \boxed{\mathbf E=\rho_eJ\hat{\mathbf z},\qquad \mathbf B=\frac{\mu_0Jr}{2}\hat{\boldsymbol\phi}.} \]

Since \(\hat{\mathbf z}\times\hat{\boldsymbol\phi}=-\hat{\mathbf r}\),

\[ \boxed{\mathbf S=\frac{\mathbf E\times\mathbf B}{\mu_0} =-\frac{\rho_eJ^2r}{2}\hat{\mathbf r}.} \]

Thus \(E=\rho_ei/(\pi a^2)\), \(B=\mu_0ir/(2\pi a^2)\), and \(S=\rho_ei^2r/(2\pi^2a^4)\). Energy flows radially inward, even though the current flows along the wire. Reversing current reverses \(\mathbf E\) and \(\mathbf B\) together, leaving \(\mathbf S\) inward.

(b) All fields are steady, so \(\partial u/\partial t=0\). Cylindrical divergence gives

\[ \nabla\cdot\mathbf S=\frac1r\frac{\partial(rS_r)}{\partial r} =\frac1r\frac{\partial}{\partial r}\left(-\frac12\rho_eJ^2r^2\right) =-\rho_eJ^2=-\mathbf j\cdot\mathbf E. \]

This extends continuously to \(r=0\). For a length \(\ell\), the inward surface power is \(-S_r(a)2\pi a\ell=\rho_eJ^2\pi a^2\ell=i^2R_\ell\), where \(R_\ell=\rho_e\ell/(\pi a^2)\). 流入导线的场能恰好转化为焦耳热。

Ch. 14, Ex. 4 · A spherical capacitor starts rotating

Problem. A spherical capacitor has inner radius \(r_1\) and outer radius \(r_2>r_1\). Initially at rest, its insulating shells carry uniform fixed charges \(+Q\) and \(-Q\). They are rigidly joined, free to rotate about \(z\), and have combined moment of inertia \(I\). A surrounding long coaxial solenoid initially produces uniform \(\mathbf B=B_0\hat{\mathbf z}\), \(B_0>0\), which is slowly removed. Neglect the field produced by the shells' own rotation. Find the final angular velocity and its sense.

Method: Induced electric field followed by mechanical torque and angular impulse.

Solution. At cylindrical radius \(s\) inside the solenoid, a circular contour gives

\[ 2\pi s E_\phi=-\pi s^2\dot B, \qquad E_\phi=-\frac{s}{2}\dot B. \]

On a spherical shell of radius \(r\) and signed charge \(Q_s\), \(s=r\sin\theta\). Its axial torque is

\[ \tau_z=\int sE_\phi\,dq =-\frac{\dot B}{2}\int r^2\sin^2\theta\,dq. \]

Uniform surface charge has \(\langle\sin^2\theta\rangle=2/3\), so \(\tau_z=-Q_sr^2\dot B/3\). Adding the inner positive shell and outer negative shell,

\[ \tau_{z,\rm total}=\frac{Q(r_2^2-r_1^2)}{3}\dot B. \]

Integrate \(I\dot\Omega=\tau_z\) while \(B\) decreases from \(B_0\) to zero:

\[ I\Omega_f=\frac{Q(r_2^2-r_1^2)}3\int_{B_0}^{0}dB, \qquad \boxed{\boldsymbol\Omega_f=-\frac{QB_0(r_2^2-r_1^2)}{3I}\hat{\mathbf z}.} \]

The shells turn clockwise seen from \(+z\). The induced electric field is counterclockwise during removal, but the negative outer shell receives a larger clockwise torque because of its larger squared lever arm. 这里的电荷固定在绝缘壳上，不应使用导体的静电平衡重新分布电荷。