Section A · Conceptual Trap Questions (Multiple Choice)
A1.
A parallel-plate capacitor remains permanently connected to a source of constant voltage $U$. The plate separation $d$ is slowly doubled while everything else stays the same. Which of the following are correct? (Select all that apply.)
A. The capacitance $C$ is halved.
B. The charge on the plates $Q$ is halved.
C. The electric field strength $E$ between the plates is halved.
D. The electric field strength $E$ between the plates is unchanged.
Comparison: If the capacitor is instead charged then disconnected from the source ($Q$ fixed) before $d$ is doubled, which answers would change?
Show answer for A1
Answer: ABC Key point:
voltage $U$ is held constant by the source.
• $C=\dfrac{\varepsilon S}{4\pi kd}$, so doubling $d$ halves $C$ (A correct).
• $Q=CU$; $U$ unchanged and $C$ halved, so $Q$ is halved (B correct).
• $E=\dfrac{U}{d}$; $U$ unchanged and $d$ doubled, so $E$ is halved (C correct, D wrong).
Common trap: option B (charge also halves) is most often missed. Contrast: with the source disconnected ($Q$ fixed), use $E=\dfrac{Q}{\varepsilon_0 S}$ → $E$ unchanged, while $U=Ed$ increases. Connected source: $E$ changes with $d$; disconnected source: $E$ unchanged.
A2.
A charged particle moves in uniform circular motion inside a uniform magnetic field. If only its speed is doubled (charge, mass, and magnetic field unchanged), which of the following are correct? (Select all that apply.)
A. The orbital radius doubles.
B. The period doubles.
C. The period is unchanged.
D. The frequency (revolutions per second) is unchanged.
Show answer for A2
Answer: ACD Using $r=\dfrac{mv}{qB}$ and $T=\dfrac{2\pi m}{qB}$:
• Doubling $v$ doubles $r$ (A correct).
• $T=\dfrac{2\pi m}{qB}$ is independent of $v$, so the period is unchanged (C correct, B wrong); frequency $f=\dfrac{1}{T}$ is also unchanged (D correct).
Core idea: the cyclotron period depends only on $\dfrac{m}{qB}$, not on speed — this is the operating principle of the cyclotron.
Section B · Comprehensive Calculation Problems
B1. (Electric-field acceleration + magnetic-field deflection)
A positive ion of mass $m$ and charge $q$ starts from rest, is accelerated through voltage $U$, and then enters a uniform magnetic field of magnitude $B$ (directed into the page) perpendicularly at point $O$.
(1) Find the speed $v$ upon entering the magnetic field.
Work-energy theorem: $qU=\dfrac{1}{2}mv^2\Rightarrow \boxed{v=\sqrt{\dfrac{2qU}{m}}}$
(2) Find the radius $r$ of circular motion.
Lorentz force provides centripetal force: $qvB=\dfrac{mv^2}{r}$, so $\boxed{r=\dfrac{mv}{qB}=\dfrac{1}{B}\sqrt{\dfrac{2mU}{q}}}$
(3) The magnetic field has width $L$ ($L<r$). Find the deflection angle $\theta$ of the velocity upon exit.
Geometry gives $\sin\theta=\dfrac{L}{r}$, so $\boxed{\theta=\arcsin\dfrac{L}{r}=\arcsin\!\left(\dfrac{BL}{\sqrt{2mU/q}}\right)}$
Key points
Work-energy theorem (electric field) → Lorentz-force circular motion (magnetic field) → geometry for deflection angle. Three steps spanning two domains.
B2. (Charged oil drop + variable capacitor voltage + work)
Two horizontal parallel plates are separated by distance $d$, with the upper plate positively charged. An oil drop of mass $m$ and charge $q$ is in static equilibrium exactly midway between the plates.
(1) Determine the sign of the oil drop's charge and find the voltage $U_0$.
Static equilibrium requires the electric force to act upward to balance gravity. Upper plate positive ⇒ field points downward between the plates ⇒ the drop must carry negative charge. Equilibrium: $q\dfrac{U_0}{d}=mg\Rightarrow\boxed{U_0=\dfrac{mgd}{q}}$
(2) The voltage is suddenly increased to $2U_0$. Find the velocity after time $t$.
Electric force becomes $q\dfrac{2U_0}{d}=2mg$ (upward); net force $F_{\text{net}}=2mg-mg=mg$ (upward); acceleration $a=g$. $\boxed{v=gt\ (\text{vertically upward})}$
(3) Find the work $W$ done by the electric force during this time interval.
Displacement $s=\dfrac{1}{2}gt^2$ (upward); electric force $2mg$ is in the same direction as displacement: $\boxed{W=2mg\cdot\dfrac{1}{2}gt^2=mg^2t^2}$
Key points
Force balance determines charge sign → new net force after voltage change → kinematics for displacement → work. Each step builds on the previous.
B3. (Energy analysis of LC oscillations)
An ideal LC circuit has capacitance $C$ and inductance $L$. The maximum charge on the capacitor is $Q_0$.
(1) Use energy conservation to find the maximum current $I_0$.
When charge is maximum, all energy is electric: $\dfrac{Q_0^2}{2C}$. When current is maximum, all energy is magnetic: $\dfrac{LI_0^2}{2}$. Setting them equal: $\boxed{I_0=\dfrac{Q_0}{\sqrt{LC}}}$
(2) Write down the oscillation period $T$.
$\boxed{T=2\pi\sqrt{LC}}$
(3) Find the current $i$ when the charge is $\dfrac{Q_0}{2}$.
Energy conservation: $\dfrac{q^2}{2C}+\dfrac{Li^2}{2}=\dfrac{Q_0^2}{2C}$. Substituting $q=\dfrac{Q_0}{2}$: $\dfrac{Li^2}{2}=\dfrac{3}{4}\dfrac{Q_0^2}{2C}$, so $\boxed{i=\dfrac{\sqrt{3}}{2}I_0\approx0.87I_0}$
Note: the energy method gives only the magnitude. With direction: $i=\pm\dfrac{\sqrt{3}}{2}I_0$ — from $q=Q_0\cos\omega t$, the condition $q=\dfrac{Q_0}{2}$ corresponds to $\omega t=\dfrac{\pi}{3}$ (discharging, $i$ negative) or $\omega t=\dfrac{5\pi}{3}$ (charging, $i$ positive). The sign depends on whether the capacitor is discharging or charging at that instant.
Key points
When the charge is halved, the current is not half the maximum but rather $\dfrac{\sqrt{3}}{2}$ times it — a consequence of the nonlinear (quadratic) relationship in the energy expressions.
B4. (Rail-and-rod problem · electromagnetic induction, advanced)
Two smooth horizontal rails are separated by distance $L$, with a vertical downward magnetic field $B$. The rails have no resistance and are connected at one end by resistance $R$. A conducting rod of mass $m$ is pushed by a constant horizontal force $F$, starting from rest.
(1) Find the maximum speed $v_{\text{max}}$ the rod can reach.
Induced EMF $\varepsilon=BLv$ → current $I=\dfrac{BLv}{R}$ → Ampere force $F_{\text{A}}=\dfrac{B^2L^2v}{R}$. At maximum speed the acceleration is zero, so $F=F_{\text{A}}$: $\boxed{v_{\text{max}}=\dfrac{FR}{B^2L^2}}$
(2) Find the acceleration $a$ when $v=\dfrac{v_{\text{max}}}{2}$.
At this speed $F_{\text{A}}=\dfrac{1}{2}F$, so the net force is $\dfrac{1}{2}F$: $\boxed{a=\dfrac{F}{2m}}$
(3) When the rod has traveled displacement $s$ and reached speed $v$ ($v<v_{\text{max}}$), find the Joule heat $Q_R$ dissipated in $R$.
Energy conservation (smooth rails, rod with no resistance): $Fs=\dfrac{1}{2}mv^2+Q_R$, so $\boxed{Q_R=Fs-\dfrac{1}{2}mv^2}$. (This is exact for any $v<v_{\text{max}}$, using the actual $s$ and $v$ at the same instant.)
Key points
Induced EMF → current → Ampere force → dynamics (for $v_{\text{max}}$ and $a$) → energy conservation (for heat). This is the standard structure of advanced electromagnetic-induction problems.
B5. (Rod + capacitor · elegant variation)
Replace the resistor in B4 with a capacitor of capacitance $C$ (rails and rod still have no resistance), with all other conditions the same, starting from rest.
(1) Describe qualitatively what motion the rod undergoes.
The rod undergoes uniform acceleration (constant acceleration) in a straight line.
(2) Find the acceleration $a$.
The capacitor voltage equals the induced EMF: $U_C=BLv$. Charge on capacitor: $q=CBLv$. Charging current: $I=\dfrac{\Delta q}{\Delta t}=CBLa$. Ampere force: $F_{\text{A}}=BIL=CB^2L^2a$. Newton's second law: $F-CB^2L^2a=ma$, so $\boxed{a=\dfrac{F}{m+CB^2L^2}}$ (constant ⇒ uniform acceleration).
Key points
Contrast with B4: with a resistor the Ampere force is proportional to $v$ → non-uniform acceleration approaching a terminal speed; with a capacitor the Ampere force is proportional to $a$ → constant acceleration, uniform motion. Same apparatus, different component, completely different kinematics.
B6. (Geometry of a bounded magnetic field · challenge)
A positive charge of mass $m$ and charge $q$ enters a uniform magnetic field (directed into the page) of width $d$ perpendicular to the left boundary with speed $v$.
(1) If the particle just barely exits from the right boundary, find $B$.
Radius $r=\dfrac{mv}{qB}$; the maximum horizontal penetration depth equals $r$. The critical condition for just exiting the right boundary is $r=d$: $\boxed{B=\dfrac{mv}{qd}}$
(2) Under this critical condition, find the time $t$ the particle spends inside the field.
When $r=d$ the particle turns exactly $90^\circ$ (one quarter of a full circle): $t=\dfrac{T}{4}=\dfrac{\pi m}{2qB}$. Substituting $B$: $\boxed{t=\dfrac{\pi d}{2v}}$ (independent of $m$, $q$, and $B$).
Key points
Lorentz-force circular motion + critical condition ($r=d$) + geometry ($90^\circ$ turn) + period formula.
B6 Extension. (Bounded magnetic field + gravity · beyond-syllabus challenge)
Same setup as B6, but gravity is not neglected (directed vertically downward, perpendicular to the entry velocity), and $B$ is given while $d$ is unknown. The particle just exits tangentially from the right boundary. Find $d$ and $t$. Hint: drift velocity $v_{\text{drift}}=\dfrac{mg}{qB}$; tangency condition $v_x=0$.
(1) Find the width $d$ of the magnetic field region.
$d=x(t^{*})$; substituting $v_{\text{drift}}=\dfrac{mg}{qB}$ and simplifying: $$\boxed{d=\dfrac{m}{q^{2}B^{2}}\left[mg\arccos\!\left(\dfrac{-mg}{qBv-mg}\right)+\sqrt{qBv(qBv-2mg)}\right]}$$
(2) Find the time $t$ from entry to reaching the right boundary.
Tangency condition $\cos\omega t^{*}=-\dfrac{mg}{qBv-mg}$ (where $\omega=\dfrac{qB}{m}$): $$\boxed{t=\dfrac{m}{qB}\arccos\!\left(\dfrac{-mg}{qBv-mg}\right)}$$
Key points
Gravity deforms the circular trajectory into a cycloid. The tangency condition $v_x=0$ determines $t$, which is then substituted to obtain $d$. Taking $g\to0$ recovers the B6 results $t=\dfrac{T}{4}$ and $d=r$. Finding $B$ given $d$ leads to a transcendental equation — the forward and inverse problems are asymmetric.