Section A · Relationship Analysis (Multiple Choice)
A1.
A parallel-plate capacitor is charged and then disconnected from the power supply (charge $Q$ is fixed). The plate separation $d$ is then doubled while everything else remains unchanged. Which of the following are correct? (Multiple choice)
A. The capacitance $C$ becomes half its original value.
B. The electric field strength $E$ between the plates remains unchanged.
C. The voltage $U$ across the plates doubles.
D. The energy stored in the capacitor becomes half its original value.
Comparison: if the capacitor stays connected to the power supply ($U$ fixed) while $d$ is doubled, which answers would differ?
Show answer for A1
Answer: ABC Disconnected supply → charge $Q$ is clamped.
• A correct: $C=\dfrac{\varepsilon S}{4\pi kd}$, $d$ doubles → $C$ halves.
• B correct: $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0 S}$; since $Q$ and $S$ are fixed, $E$ is independent of $d$ and does not change.
• C correct: $U=Ed$; $E$ unchanged, $d$ doubles → $U$ doubles.
• D incorrect: $W=\dfrac{Q^2}{2C}$; $C$ halves → $W$ doubles. The extra energy comes from the work done by the external force separating the plates.
Comparison: with supply connected ($U$ fixed), $E=\dfrac{U}{d}$ halves, $Q$ halves, $W$ halves. Disconnected: clamp $\sigma$; connected: clamp $U$.
A2.
A charged particle moves in uniform circular motion inside a uniform magnetic field. If only the speed is doubled (charge, mass, and field strength unchanged), which of the following are correct? (Multiple choice)
A. The orbital radius doubles.
B. The period doubles.
C. The kinetic energy becomes 4 times its original value.
D. The number of revolutions per second (frequency) is unchanged.
Show answer for A2
Answer: ACD $r=\dfrac{mv}{qB}$, $T=\dfrac{2\pi m}{qB}$.
• A correct: $r\propto v$ → radius doubles.
• B incorrect / D correct: $T$ and $f$ are independent of $v$ → both unchanged.
• C correct: $E_k=\dfrac{1}{2}mv^2\propto v^2$ → 4 times.
A3.
Mutually perpendicular uniform electric field $E$ and magnetic field $B$ form a velocity selector. A charged particle enters perpendicularly and travels in a straight line at constant velocity. Which of the following are correct? (Multiple choice)
A. The selected speed is $v=\dfrac{E}{B}$, independent of both charge and mass.
B. If $E$ is doubled, the selected speed doubles.
C. If $B$ is doubled, the selected speed is halved.
D. A particle with speed greater than $\dfrac{E}{B}$ experiences a net force in the same direction as the electric force.
Show answer for A3
Answer: ABC Equilibrium: $qE=qvB\Rightarrow v=\dfrac{E}{B}$.
• A correct: $q$ cancels and $m$ does not appear → selects speed only, regardless of particle type.
• B correct: $v\propto E$ → doubles; C correct: $v\propto\dfrac{1}{B}$ → halves.
• D incorrect: when $v>\dfrac{E}{B}$, the magnetic force $qvB>qE$, so the net force is in the direction of the magnetic force, which is opposite to the electric force.
Section B · Quantitative Relationship Problems
B1. (Electric acceleration + circular motion in magnetic field · charge-to-mass ratio)
A positive particle of mass $m$ and charge $q$ starts from rest, is accelerated through voltage $U$, and enters a magnetic field $B$ perpendicularly, following a circular arc of radius $r$.
(1) Prove that the specific charge is $\dfrac{q}{m}=\dfrac{2U}{B^2r^2}$
Acceleration: $qU=\dfrac{1}{2}mv^2$. Circular motion: $r=\dfrac{mv}{qB}$, so $v=\dfrac{qBr}{m}$. Substituting: $qU=\dfrac{q^2B^2r^2}{2m}\Rightarrow\boxed{\dfrac{q}{m}=\dfrac{2U}{B^2r^2}}$
(2) $U$ is doubled (same particle, same $B$). How does $r$ change?
$r^2\propto U$ → $r\propto\sqrt{U}$. Doubling $U$ gives $r\to\boxed{\sqrt{2}}$ times the original.
(3) A different particle with charge $2q$ and mass $4m$ is used (same $U$, same $B$). By what factor does $r$ change?
$r\propto\sqrt{\dfrac{m}{q}}$: $\dfrac{r'}{r}=\sqrt{\dfrac{4m/2q}{m/q}}=\boxed{\sqrt{2}}$ times.
Key points
Acceleration fixes $v$ → circular motion fixes $r$ → eliminate $v$. Core results: $r\propto\sqrt{U}$ and $r\propto\sqrt{m/q}$.
B2. (Parallel-plate capacitor · Full comparison: constant $U$ vs. constant $Q$)
A parallel-plate capacitor with plate area $S$ and separation $d$ is charged to voltage $U_0$. The separation is then doubled. Case 1: the supply remains connected ($U$ fixed); Case 2: the supply is disconnected first ($Q$ fixed). Find how $C$, $Q$, $U$, $E$, and $W$ each change, and explain why $E$ behaves differently in the two cases.
Show comparison table and explanation
$C=\dfrac{\varepsilon S}{4\pi kd}$; in both cases $C$
halves.
| Quantity | Case 1 ($U$ fixed) | Case 2 ($Q$ fixed) |
|---|
| $C$ | halves | halves |
| $Q$ | halves | unchanged |
| $U$ | unchanged | doubles |
| $E$ | $\dfrac{U}{d}$ → halves | $\dfrac{\sigma}{\varepsilon_0}$ → unchanged |
| $W$ | halves | doubles |
Why $E$ differs: In Case 1 the voltage is clamped, so $E=\dfrac{U}{d}$ decreases as $d$ grows. In Case 2 the surface charge density $\sigma$ is clamped, so $E=\dfrac{\sigma}{\varepsilon_0}$ is independent of $d$.
Key points
Identify which quantity is clamped first, then choose the corresponding field-strength formula; everything else follows in sequence.
B3. (Velocity selector + mass spectrometer · relationship chain)
A beam of particles (all with charge $q$) first passes through a velocity selector (crossed fields $E$ and $B_1$) to select speed $v$, then enters a region with only magnetic field $B_2$ and traces a semicircle before hitting a photographic plate.
(1) Find the selected speed $v$
$\boxed{v=\dfrac{E}{B_1}}$
(2) Find the radius $r$ of the semicircle in $B_2$
$r=\dfrac{mv}{qB_2}=\boxed{\dfrac{mE}{qB_1B_2}}$
(3) Two isotopes have masses $m_1$ and $m_2$. Find the separation $\Delta x$ between their impact points.
The impact point is a full diameter $2r$ from the entry slit: $\boxed{\Delta x=|2r_1-2r_2|=\dfrac{2E}{qB_1B_2}|m_1-m_2|}$
Key points
The selector fixes speed (independent of mass) → radius $r\propto m$ → different masses produce different impact points. This is how a mass spectrometer separates isotopes.
B4. (Conducting rail · compound proportionality + power)
A horizontal frictionless rail has track width $L$, a vertical magnetic field $B$, and a terminal resistor $R$. A bar of mass $m$ is pushed from rest by a constant force $F$ and asymptotically approaches a maximum speed.
(1) Find the maximum speed $v_{\max}$
At $a=0$: $F=\dfrac{B^2L^2v_{\max}}{R}$, so $\boxed{v_{\max}=\dfrac{FR}{B^2L^2}}$.
(2) $B$ is doubled. By what factor does $v_{\max}$ change?
$v_{\max}\propto\dfrac{1}{B^2}$ → becomes $\boxed{\dfrac{1}{4}}$ of the original.
(3) Both $R$ and $B$ are doubled simultaneously. By what factor does $v_{\max}$ change?
$v_{\max}\propto\dfrac{R}{B^2}$; $R\times2$, $B^2\times4$ → $\boxed{\dfrac{1}{2}}$ of the original.
(4) Find $i_{\max}$ and power $P$; verify that $P=Fv_{\max}$
$i_{\max}=\dfrac{BLv_{\max}}{R}=\dfrac{F}{BL}$; $P=i_{\max}^2R=\dfrac{F^2R}{B^2L^2}$. Verification: $Fv_{\max}=\dfrac{F^2R}{B^2L^2}$ ✓ (at constant velocity the applied force power equals the Joule dissipation).
Key points
$v_{\max}\propto R/B^2$ is a compound proportionality; $P=Fv_{\max}$ expresses the equivalence of mechanical input power and electrical output power.
B5. (LC oscillation · quantity relationships and scaling)
An ideal LC circuit has inductance $L$, capacitance $C$, and maximum charge $Q_0$.
(1) Write expressions for the maximum current $I_0$ and maximum voltage $U_0$
$\boxed{I_0=\dfrac{Q_0}{\sqrt{LC}},\quad U_0=\dfrac{Q_0}{C}}$
(2) $Q_0$ is unchanged and $C\to4C$. By what factor do $I_0$, $U_0$, and $T$ each change?
$I_0\propto\dfrac{1}{\sqrt{C}}$ → $\dfrac{1}{2}$; $U_0\propto\dfrac{1}{C}$ → $\dfrac{1}{4}$; $T=2\pi\sqrt{LC}\propto\sqrt{C}$ → $2$ times.
(3) When the charge equals $\dfrac{Q_0}{2}$, what is the current?
$i=I_0\sqrt{1-\left(\dfrac{q}{Q_0}\right)^2}=I_0\sqrt{1-\dfrac{1}{4}}=\boxed{\dfrac{\sqrt{3}}{2}I_0}$
Key points
$I_0\propto1/\sqrt{C}$, $U_0\propto1/C$, $T\propto\sqrt{C}$ — a single change in $C$ drives three quantities at different power laws. The $q$–$i$ relation is quadratic (nonlinear).
B6. (Cyclotron · relationship challenge)
A cyclotron accelerates particles of charge $q$ and mass $m$ in a magnetic field $B$. The maximum orbital radius is $R$.
(1) Find the period $T$ and frequency $f$; explain why they are independent of speed and radius
$\boxed{T=\dfrac{2\pi m}{qB},\quad f=\dfrac{qB}{2\pi m}}$. Because $r=\dfrac{mv}{qB}$ grows in proportion to $v$, the time per revolution $T=\dfrac{2\pi r}{v}=\dfrac{2\pi m}{qB}$ remains constant — a faster particle traces a proportionally larger circle in exactly the same time. This isochronous property is what allows a fixed-frequency alternating voltage to keep accelerating the particles.
(2) Find the maximum kinetic energy $E_k$
At maximum radius: $v_{\max}=\dfrac{qBR}{m}$, so $E_k=\dfrac{1}{2}mv_{\max}^2=\boxed{\dfrac{q^2B^2R^2}{2m}}$.
(3) $B$ is doubled, then (separately) the mass is doubled with $q$ unchanged. By what factor does $E_k$ change in each case?
$B$ doubled: $E_k\propto B^2$ → $\boxed{4}$ times; mass doubled: $E_k\propto\dfrac{1}{m}$ → $\boxed{\dfrac{1}{2}}$.
Key points
Period independent of $v$ (enabling fixed-frequency acceleration) → maximum energy $\propto B^2R^2/m$; the field strength and machine size set the energy ceiling.