2023 Midterm · Worked Solutions

Question paper · Revision plan · Transfer practice

Use \(k=1/(4\pi\epsilon_0)\) and \(V(\infty)=0\). These are worked reference solutions; the supplied images contain no official answer key.

1. Conducting sphere and an external charge

Question 1 · Conductor equilibrium · Spherical images · Related homework

(a) Isolated sphere without external sources

Spherical symmetry and Gauss's law give

\[ \boxed{\mathbf E_1(r)=\begin{cases} 0,&r<R,\\ kQ\hat{\mathbf r}/r^2,&r>R, \end{cases}\qquad V_1(r)=\begin{cases} kQ/R,&r\leq R,\\ kQ/r,&r\geq R. \end{cases}} \]

\(V\) is continuous at \(R\); \(E_r\) jumps from \(0\) to \(Q/(4\pi\epsilon_0R^2)\). The field at a charged ideal surface is specified by its two one-sided limits.

(b) External charge, sphere still isolated

After redistribution the metal remains equipotential and field-free:

\[ \boxed{\mathbf E_2(O)=0.} \]

At the center, every surface element is still distance \(R\) away, regardless of its density. Since the sphere's total charge remains \(Q\),

\[ V_2(O)=k\int_{\rm sphere}\frac{dq_s}{R}+\frac{kq}{d} =\boxed{\frac{kQ}{R}+\frac{kq}{d}.} \]

表面电荷已不均匀；中心电势仍可这样算，是因为所有面元到中心等距。计算电势相加，计算电场则必须相加矢量。

(c) Charge exchanged on grounding

Grounding imposes \(V(O)=0\) while allowing the net sphere charge to change:

\[ 0=\frac{kQ_2}{R}+\frac{kq}{d} \quad\Rightarrow\quad \boxed{Q_2=-\frac{qR}{d},\qquad Q-Q_2=Q+\frac{qR}{d}.} \]

The latter is signed positive charge transferred from sphere to ground. Equivalently, electrons flow from ground to sphere. 接地固定电势，不固定净电荷，也不保证电中性。

(d) Exterior potential by images

Source correction. The grounded sphere's induced charge is nonuniform in the external charge's presence. The uniformity sentence in the question contradicts its image-method hint; solve the stated \(V(R,\theta)=0\) boundary problem.

Put the real \(q\) at \(d\hat{\mathbf z}\) and an image \(-q'\) at \(a\hat{\mathbf z}\), \(0<a<R\). On the sphere,

\[ \frac{q}{\sqrt{R^2+d^2-2Rd\cos\theta}} =\frac{q'}{\sqrt{R^2+a^2-2Ra\cos\theta}}. \]

Matching the constant and \(\cos\theta\) terms after squaring gives the interior image

\[ \boxed{a=\frac{R^2}{d},\qquad q'=\frac{qR}{d}.} \]

Indeed the real-charge distance on the surface is \(d/R\) times the image distance, so cancellation holds for every \(\theta\). Uniqueness fixes the exterior solution:

\[ \boxed{V_3(r,\theta)=k\left[ \frac{q}{\sqrt{r^2+d^2-2rd\cos\theta}} -\frac{qR/d}{\sqrt{r^2+(R^2/d)^2-2r(R^2/d)\cos\theta}} \right],\quad r\geq R.} \]

Exclude the real charge's singular point. The image is a mathematical replacement inside the excluded conductor; the represented total induced charge is \(-q'=Q_2\). Its nonuniform density, if needed, follows from the normal derivative:

\[ \sigma_s(\theta)=-\epsilon_0\left.\frac{\partial V_3}{\partial r}\right|_{R^+} =-\frac{q(d^2-R^2)}{4\pi R(R^2+d^2-2Rd\cos\theta)^{3/2}}. \]

The density depends on \(\theta\), while its integral is \(-qR/d\). If the sphere were isolated instead, a central compensating image would be needed; see transfer practice 2.

2. Lossy spherical capacitor

Question 2 · Capacitance and local Ohm's law · RC discharge

Set \(\epsilon=\epsilon_0\epsilon_r\). Assume a uniform linear medium and free discharge with no connected voltage source. Here \(R\) means electrical resistance.

(a) Capacitance and resistance

At instantaneous inner charge \(Q\), spherical symmetry gives

\[ E_r(r)=\frac{Q}{4\pi\epsilon r^2},\qquad V_a-V_b=\int_a^bE_r\,dr=\frac{Q}{4\pi\epsilon}\left(\frac1a-\frac1b\right). \]

Consequently,

\[ \boxed{C=\frac{4\pi\epsilon}{1/a-1/b} =\frac{4\pi\epsilon ab}{b-a}.} \]

A radial current \(I\) crosses area \(4\pi r^2\), so \(J_r=I/(4\pi r^2)\), \(E_r=J_r/\sigma\), and

\[ V_a-V_b=\frac{I}{4\pi\sigma}\int_a^b\frac{dr}{r^2}, \qquad \boxed{R=\frac1{4\pi\sigma}\left(\frac1a-\frac1b\right).} \]

Do not substitute a constant area into \(R=\ell/(\sigma A)\); the current-carrying area varies with radius.

(b) Charge relaxation

Outward conventional current removes positive inner charge:

\[ -\dot Q=I=\frac{V_a-V_b}{R}=\frac{Q}{RC}. \]

Using \(Q(0)=Q_0\),

\[ \boxed{Q(t)=Q_0e^{-t/\tau},\qquad \tau=RC=\frac\epsilon\sigma.} \]

Geometry cancels in \(RC\) for this uniform medium. The equivalent circuit is a leakage resistor in parallel with the capacitance; without an external connection they still form a discharge path.

(c) Joule power and energy check

\[ \boxed{P(t)=I^2R=\frac{(V_a-V_b)^2}{R} =\frac{Q_0^2}{RC^2}e^{-2t/\tau} =\frac{\sigma Q_0^2}{4\pi\epsilon^2} \left(\frac1a-\frac1b\right)e^{-2\sigma t/\epsilon}.} \]

The local check gives the same result: \(P=\int_a^b\sigma E_r^2\,4\pi r^2dr\). The total heat is

\[ \int_0^\infty P(t)dt=\frac{Q_0^2}{2C}=U(0),\qquad P=-\frac d{dt}\left(\frac{Q^2}{2C}\right). \]

电荷按 \(e^{-t/\tau}\) 衰减，能量和功率按 \(e^{-2t/\tau}\) 衰减。As \(\sigma\to0\), leakage power tends to zero and \(\tau\to\infty\).

3. Rotating charged spherical surface

Question 3 · Biot–Savart law · Loop axial field · Related homework

Take \(\boldsymbol\omega=\omega\hat{\mathbf z}\), \(\omega>0\). The usual surface density is \(\sigma_s=Q/(4\pi R^2)\).

(a) Latitude-ring charge and current

The ring's circumference is \(2\pi R\sin\theta\) and its meridional width is \(R\,d\theta\), so

\[ dq=\sigma_s(2\pi R\sin\theta)(R\,d\theta) =\frac Q2\sin\theta\,d\theta. \]

In the paper's notation,

\[ \boxed{\sigma(\theta)=\frac{dq}{d\theta}=\frac Q2\sin\theta,\qquad J(\theta)=\frac{\sigma(\theta)}{2\pi/\omega} =\frac{Q\omega}{4\pi}\sin\theta.} \]

The physical ring current is \(dI=J(\theta)d\theta\). Here \(\sigma(\theta)\) is charge per polar angle, not C/m²; \(J(\theta)\) is current per polar angle, not A/m². 漏掉面元中的 \(\sin\theta\) 会同时影响后两问。

(b) Magnetic dipole moment

The ring radius is \(s=R\sin\theta\), so its enclosed area is \(A=\pi R^2\sin^2\theta\):

\[ d\boldsymbol\mu=A\,dI\,\hat{\mathbf z} =\frac{Q\omega R^2}{4}\sin^3\theta\,d\theta\,\hat{\mathbf z}. \]

Since \(\int_0^\pi\sin^3\theta\,d\theta=4/3\),

\[ \boxed{\boldsymbol\mu=\frac{Q\omega R^2}{3}\hat{\mathbf z}.} \]

Use the area enclosed by the ring for its moment, not the spherical strip's surface area.

(c) Magnetic field at the center

The ring is offset by \(z'=R\cos\theta\); the center of the sphere is not generally the center of that ring. Its axial field there is

\[ dB_z=\frac{\mu_0(dI)s^2}{2(s^2+z'^2)^{3/2}} =\frac{\mu_0Q\omega}{8\pi R}\sin^3\theta\,d\theta. \]

Integrating,

\[ \boxed{\mathbf B(O)=\frac{\mu_0Q\omega}{6\pi R}\hat{\mathbf z}.} \]

For positive surface charge the field follows \(\boldsymbol\omega\). The far-field dipole formula cannot be applied at the sphere's center.

4. Faraday disk

Question 4 · Motional emf and power · Related homework

Use cylindrical radius \(r\), \(\boldsymbol\omega=\omega\hat{\mathbf z}\) and \(\mathbf B=B\hat{\mathbf z}\), with \(B>0\). Neglect friction and circuit inductance in the resistive-circuit model of (b,c).

(a) Open circuit: electric field and terminal voltage

The material velocity is \(\mathbf u=\omega r\hat{\boldsymbol\phi}\). Charge separation stops the radial current when the force per unit charge vanishes:

\[ \mathbf E+\mathbf u\times\mathbf B=0, \qquad \boxed{\mathbf E=-B\omega r\hat{\mathbf r}.} \]

This is a moving conductor; the static-conductor rule \(\mathbf E=0\) cannot be imposed in the laboratory frame. Integrating the electric field gives

\[ \boxed{V_{\rm rim}-V_O=-\int_0^a\mathbf E\cdot d\boldsymbol\ell =\frac12B\omega a^2.} \]

The rim is at higher potential. If “drop” means \(V_O-V_{\rm rim}\), its signed value is \(-B\omega a^2/2\).

(b) Closed circuit: torque and mechanical power

The motional emf from center to rim is \(\mathcal E=\int_0^a(\mathbf u\times\mathbf B)\cdot d\boldsymbol\ell=B\omega a^2/2\). Hence the current is outward through the disk and returns through the external resistor:

\[ I=\frac{\mathcal E}{R}=\frac{B\omega a^2}{2R}. \]

For radial current, \(d\mathbf F=I\,dr\,\hat{\mathbf r}\times\mathbf B=-IB\,dr\,\hat{\boldsymbol\phi}\). Its lever arm is \(r\), giving

\[ \boldsymbol\tau_B=-\int_0^a rIB\,dr\,\hat{\mathbf z} =-\frac{IBa^2}{2}\hat{\mathbf z}, \qquad \boxed{\boldsymbol\tau_{\rm ext}=\frac{B^2a^4\omega}{4R}\hat{\mathbf z}.} \]

Check the energy flow:

\[ \tau_{\rm ext}\omega=\frac{B^2a^4\omega^2}{4R}=I\mathcal E=I^2R. \]

Current in (b) follows the emf around the complete circuit and its total resistance. For an ideal moving conductor, a zero rest-frame electric field does not imply zero current.

(c) Free magnetic braking

Let \(\mathcal I\) be the disk's moment of inertia about \(z\). With the applied torque removed,

\[ \mathcal I\dot\omega=-\frac{B^2a^4}{4R}\omega, \qquad \boxed{\omega(t)=\omega_0e^{-t/\tau_m},\qquad \tau_m=\frac{4\mathcal I R}{B^2a^4}.} \]

The supplied question gives no mass or inertia, so a numerical decay time is undetermined. For a uniform disk of mass \(M\), \(\mathcal I=Ma^2/2\) and \(\tau_m=2MR/(B^2a^2)\); with mass density \(\rho_m\), \(M=\rho_m\pi a^2h\) and \(\tau_m=2\pi\rho_mhR/B^2\).

Finally,

\[ -\frac d{dt}\left(\frac12\mathcal I\omega^2\right)=I(t)^2R, \qquad\int_0^\infty I(t)^2R\,dt=\frac12\mathcal I\omega_0^2. \]

Thickness \(h\) enters through the moment of inertia if density is specified; it does not enter the ideal motional emf. 若计入电路自感，需联立 \(L\dot I+RI=B a^2\omega/2\) 与 \(\mathcal I\dot\omega=-Ba^2I/2\)；单指数结果对应忽略自感的模型。