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Electrostatics and Electric Fields · Solutions静电学与电场 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 18 marksAP 选择题 + 安/卑省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Charge basics电荷基础 · HS-PS2-4 [3 marks][3 分]

A positively charged rod is brought near neutral aluminium foil. What happens, and why?一根带正电的棒靠近中性铝箔。发生什么，为什么？

Answer:答案：  (B)  attracted, by induced charge separation被吸引，因感应电荷分离

(a) Reason from induction in a conductor从导体中的感应现象推理 M1·A1·A1

The aluminium foil is a conductor with mobile electrons. The positive rod attracts the foil's free electrons to the near side, leaving the far side positive. The near (negative) side is closer to the rod than the far (positive) side, so by Coulomb's inverse-square law the attraction on the near side exceeds the repulsion on the far side. The net force is attractive, option (B). No charge is transferred (no contact), so the foil stays neutral overall.铝箔是含可动电子的导体。带正电的棒把铝箔的自由电子吸引到近端，使远端带正电。近端（负）比远端（正）更靠近棒，由库仑平方反比定律，近端受到的吸引大于远端受到的排斥。净力为吸引力，选 (B)。没有电荷转移（无接触），故铝箔整体仍中性。
Why the distractors fail.干扰项分析。
(A): there is no like-charge repulsion; the foil is neutral, not positive.不存在同种电荷排斥；铝箔是中性，并非带正电。
(C): "neutral overall" is true, but a neutral conductor still responds via induction."整体中性"虽对，但中性导体仍会通过感应作出响应。
(D): conduction requires contact and would not give net positive charge here.传导需要接触，且此处不会使其带净正电。
A charged object always attracts a neutral conductor.带电体总会吸引中性导体。 This is the everyday "balloon sticks to the wall" effect. The charged object polarises the neutral object: opposite charge migrates closer, like charge moves away. Because force falls off as $1/r^2$, the closer (opposite) charge wins and the net interaction is attractive, regardless of the rod's sign. This is why a charged comb picks up neutral paper bits. The key distinction is induction (no contact, charge only rearranges) versus conduction (contact, charge actually transfers).这就是日常"气球贴墙"的效应。带电体使中性物体极化：异种电荷迁移到靠近处，同种电荷移到远处。由于力按 $1/r^2$ 衰减，较近的（异种）电荷占优，无论棒的符号如何，净相互作用都是吸引。这正是带电梳子能吸起中性纸屑的原因。关键区别在于感应（无接触，电荷仅重新分布）与传导（有接触，电荷真正转移）。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Conservation of charge电荷守恒 · HS-PS2-4 [3 marks][3 分]

A $+8.0\ \mu\text{C}$ sphere touches an identical neutral sphere, then they separate. Charge on each?一个 $+8.0\ \mu\text{C}$ 的球接触一个相同的中性球后分开。每个球的电荷？

Answer:答案：  (B)  $+4.0\ \mu\text{C}$ each各

(a) Apply conservation of charge with equal sharing应用电荷守恒与等量分配 M1·A1·A1

Total charge is conserved: before contact it is $+8.0\ \mu\text{C} + 0 = +8.0\ \mu\text{C}$. Two identical conductors in contact share charge equally:总电荷守恒：接触前为 $+8.0\ \mu\text{C} + 0 = +8.0\ \mu\text{C}$。两个相同导体接触时等量分配电荷： $$ q_{\text{each}} \;=\; \frac{+8.0\ \mu\text{C}}{2} \;=\; +4.0\ \mu\text{C}. $$ Option (B). Check: $+4.0 + (+4.0) = +8.0\ \mu\text{C}$, conserved. ✓选 (B)。验证：$+4.0 + (+4.0) = +8.0\ \mu\text{C}$，守恒。✓
Why the distractors fail.干扰项分析。
(A): no sharing; ignores that contact between conductors redistributes charge.未分配；忽略了导体接触会重新分布电荷。
(C): would violate conservation of charge (total $+16\ \mu\text{C}$ from nothing).将违反电荷守恒（凭空多出总计 $+16\ \mu\text{C}$）。
(D): charge is never created; the total cannot double.电荷不会被创造；总量不能翻倍。
Identical conductors in contact split the total charge evenly.相同导体接触时平分总电荷。 When two identical conducting spheres touch, mobile charge spreads until both are at the same potential, which (by symmetry) means equal charge on each. The rule "halve for two identical spheres" is just conservation of charge plus the symmetry of identical bodies. If the spheres were different sizes, the split would be unequal (larger sphere holds more). This same idea, applied repeatedly, lets you produce fractional charges like $+8 \to +4 \to +2 \to \ldots$ by successive touches, a standard provincial and Diploma exam set-up.两个相同导体球接触时，可动电荷扩散到两球电位相同为止，由对称性这意味着每球电荷相等。"两个相同球减半"的规则只是电荷守恒加上相同物体的对称性。若两球大小不同，分配将不均（大球持有更多）。反复应用这一思想，可通过连续接触得到 $+8 \to +4 \to +2 \to \ldots$ 等分数电荷，这是省考与毕业考的标准考法。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §1 Conduction & induction传导与感应 · SPH4U D3 [4 marks][4 分]

(a) Difference between conduction and induction. (b) Sign after a negative rod touches a neutral sphere. (c) Conductor or insulator allows free charge flow.(a) 传导与感应的区别。(b) 负电棒接触中性球后的符号。(c) 导体还是绝缘体允许电荷自由流动。

Answer:答案：  (a) conduction needs contact; induction does not传导需接触；感应不需  ·  (b) negative负  ·  (c) a conductor导体

(a) State the key difference说出关键区别 A1·A1

Charging by conduction requires direct contact: electrons transfer between the charged object and the target, and the target keeps the same sign of charge as the source. Charging (or polarising) by induction needs no contact: a nearby charged object rearranges the charges already in the target; if then grounded, the target keeps charge of the opposite sign to the inducing object.通过传导带电需要直接接触：电子在带电体与目标之间转移，目标所带电荷与源同号。通过感应带电（或极化）无需接触：邻近的带电体使目标内部已有的电荷重新分布；若随后接地，目标会保留与施感物体异号的电荷。

(b) Sign after conduction传导后的符号 A1

A negative rod in contact transfers electrons onto the sphere, so the sphere becomes negative (same sign as the rod).负电棒接触时把电子转移到球上，故球变为负（与棒同号）。

(c) Free charge flow电荷自由流动 A1

A conductor (e.g. a metal) has mobile electrons and allows charge to flow freely; an insulator does not.导体（如金属）有可动电子，允许电荷自由流动；绝缘体则不。
Conduction copies the sign; grounded induction flips it.传导带电同号；接地感应带电异号。 A reliable memory hook: touch (conduction) gives the same sign, because charge actually flows across the contact. Induction with grounding gives the opposite sign, because you remove the repelled like-charge to ground and trap the attracted opposite charge. The reason metals conduct is the "sea" of delocalised electrons; insulators hold their electrons tightly so charge stays local. Every charge-transfer problem reduces to: (1) is there contact? (2) is the object a conductor? (3) is conservation of charge respected? Those three checks resolve the whole topic.一个可靠的记忆法：接触（传导）给出同号，因为电荷确实跨过接触面流动。接地感应给出异号，因为你把被排斥的同种电荷导入大地，留住被吸引的异种电荷。金属导电的原因是离域电子的"海洋"；绝缘体的电子被牢牢束缚，故电荷停留在局部。每道电荷转移题都归结为：(1) 是否接触？(2) 物体是否为导体？(3) 是否遵守电荷守恒？这三项检查可解决整个主题。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Coulomb's law (scaling)库仑定律（标度） · HS-PS2-4 [3 marks][3 分]

Force is $F$ at separation $r$. Separation increased to $3r$, charges unchanged. New force?间距 $r$ 时力为 $F$。间距增大到 $3r$，电荷不变。新的力？

Answer:答案：  (C)  $F/9$

(a) Apply the inverse-square dependence应用平方反比关系 M1·A1·A1

Coulomb's law gives $F = k|q_1||q_2|/r^2$, so $F \propto 1/r^2$ when the charges are fixed. Tripling the separation:库仑定律为 $F = k|q_1||q_2|/r^2$，故电荷固定时 $F \propto 1/r^2$。间距变为三倍： $$ F_{\text{new}} \;=\; \frac{k|q_1||q_2|}{(3r)^2} \;=\; \frac{k|q_1||q_2|}{9r^2} \;=\; \frac{F}{9}. $$ Option (C).选 (C)。
Why the distractors fail.干扰项分析。
(A) $3F$: treats $F$ as proportional to $r$, ignoring the square and the inverse.把 $F$ 当作与 $r$ 成正比，忽略了平方与反比。
(B) $F/3$: uses $1/r$ instead of $1/r^2$ (forgets to square).用 $1/r$ 而非 $1/r^2$（忘记平方）。
(D) $9F$: squares but drops the inverse, treating $F \propto r^2$.做了平方但丢掉了反比，当作 $F \propto r^2$。
"Times $n$ on distance" means "divide force by $n^2$.""距离乘以 $n$"意味着"力除以 $n^2$"。 Inverse-square scaling is the single most-tested idea in this unit, and it is shared exactly with Newton's law of gravitation. The fast method: read off the distance factor ($\times 3$), square it ($9$), and because the dependence is inverse, divide ($F/9$). The same logic runs the electric field $E = kQ/r^2$ and gravitation $F = Gm_1m_2/r^2$. Distance dominates: a small change in separation has a large effect because of the square. Watch the direction of the inequality, halving distance gives $\times 4$, doubling gives $\div 4$.平方反比标度是本单元最常考的单一概念，且与牛顿万有引力定律完全一致。快捷法：读出距离因子（$\times 3$），平方它（$9$），由于是反比，故相除（$F/9$）。同样的逻辑适用于电场 $E = kQ/r^2$ 与万有引力 $F = Gm_1m_2/r^2$。距离起主导作用：因为有平方，间距的微小变化会产生很大影响。注意不等式方向：距离减半为 $\times 4$，加倍为 $\div 4$。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §2 Coulomb's law库仑定律 · Physics 12 [5 marks][5 分]

$q_1 = +3.0\ \mu\text{C}$, $q_2 = +5.0\ \mu\text{C}$, $r = 0.20$ m. (a) Force magnitude. (b) Attractive or repulsive. (c) Compare force on each.$q_1 = +3.0\ \mu\text{C}$，$q_2 = +5.0\ \mu\text{C}$，$r = 0.20$ m。(a) 力的大小。(b) 引力还是斥力。(c) 比较各自受力。

Answer:答案：  (a) $F \approx 3.4\ \text{N}$  ·  (b) repulsive斥力  ·  (c) equal magnitudes大小相等

(a) Apply Coulomb's law套用库仑定律 M1·A1

Convert: $|q_1| = 3.0 \times 10^{-6}$ C, $|q_2| = 5.0 \times 10^{-6}$ C, $r = 0.20$ m.换算：$|q_1| = 3.0 \times 10^{-6}$ C，$|q_2| = 5.0 \times 10^{-6}$ C，$r = 0.20$ m。 $$ F \;=\; k\frac{|q_1||q_2|}{r^2} \;=\; (8.99\times10^9)\frac{(3.0\times10^{-6})(5.0\times10^{-6})}{(0.20)^2} \;=\; \frac{0.13485}{0.040} \;\approx\; 3.4\ \text{N}. $$

(b) Determine direction from the signs由符号判断方向 A1·A1

Both charges are positive (same sign), so the force is repulsive: each charge is pushed directly away from the other along the line joining them.两电荷均为正（同号），故力为斥力：每个电荷沿连线被直接推离另一个。

(c) Compare the forces (Newton's third law)比较受力（牛顿第三定律） A1

The forces are an action-reaction pair: the magnitude of the force on $q_1$ equals the magnitude of the force on $q_2$ ($3.4$ N each), directed oppositely.两力是作用-反作用对：$q_1$ 所受力的大小等于 $q_2$ 所受力的大小（各 $3.4$ N），方向相反。
Coulomb's law gives magnitude; signs give direction; Newton's third law guarantees the pair is equal.库仑定律给大小；符号给方向；牛顿第三定律保证成对相等。 Always feed magnitudes into $F = k|q_1||q_2|/r^2$ and decide attraction/repulsion separately from the signs (same sign repels, opposite attracts). A frequent student error is to expect the larger charge to "feel a larger force," but the interaction force is identical on both bodies, by Newton's third law, exactly as Earth and a falling apple pull on each other equally. The charges differ, yet the product $|q_1||q_2|$ is what enters the formula, and that product is symmetric. Convert $\mu\text{C} \to \text{C}$ before substituting, the most common arithmetic slip here.务必把大小代入 $F = k|q_1||q_2|/r^2$，再单独由符号判断引力/斥力（同号排斥，异号吸引）。学生常见错误是以为较大电荷"受到较大的力"，但由牛顿第三定律，两物体所受相互作用力完全相同，正如地球与下落的苹果彼此拉力相等。两电荷不同，但进入公式的是乘积 $|q_1||q_2|$，而该乘积是对称的。代入前先把 $\mu\text{C} \to \text{C}$，这是此处最常见的算术失误。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3 Electric field & force电场与力 · HS-PS2-4 [8 marks][8 分]

$Q = +6.0\ \mu\text{C}$ fixed; $P$ is $0.30$ m away. (a) Field at $P$. (b) Field direction. (c) Force on $q = -2.0\ \mu\text{C}$ at $P$. (d) Force direction and why it differs.$Q = +6.0\ \mu\text{C}$ 固定；$P$ 距 $0.30$ m。(a) $P$ 处场。(b) 场方向。(c) $P$ 处 $q = -2.0\ \mu\text{C}$ 所受力。(d) 力方向及差异原因。

Answer:答案：  (a) $E \approx 6.0\times10^5\ \text{N/C}$  ·  (b) away from $Q$背向 $Q$  ·  (c) $F \approx 1.2\ \text{N}$  ·  (d) toward $Q$指向 $Q$

(a) Field of a point charge点电荷的场 M1·A1

$$ E \;=\; k\frac{|Q|}{r^2} \;=\; (8.99\times10^9)\frac{6.0\times10^{-6}}{(0.30)^2} \;=\; \frac{53940}{0.090} \;\approx\; 6.0\times10^5\ \text{N/C}. $$

(b) Field direction场的方向 A1·A1

$\vec{E}$ points in the direction a positive test charge would be pushed. Since $Q$ is positive, the field at $P$ points radially away from $Q$.$\vec{E}$ 指向正试验电荷被推动的方向。由于 $Q$ 为正，$P$ 处的场径向背向 $Q$。

(c) Force on the placed charge放入电荷所受的力 M1·A1

$$ |F| \;=\; |q|E \;=\; (2.0\times10^{-6})(6.0\times10^5) \;\approx\; 1.2\ \text{N}. $$

(d) Force direction力的方向 A1·A1

Because $q$ is negative, $\vec{F} = q\vec{E}$ points opposite to $\vec{E}$, i.e. toward $Q$ (attractive). The field direction is fixed by the source $Q$ alone; the force direction depends on the sign of the charge placed in it.因为 $q$ 为负，$\vec{F} = q\vec{E}$ 与 $\vec{E}$ 反向，即指向 $Q$（引力）。场的方向仅由源 $Q$ 决定；力的方向取决于放入其中的电荷的符号。
Field is a property of space set by the source; force depends on the test charge's sign.场是由源决定的空间属性；力取决于试验电荷的符号。 The two-step structure here is the heart of the field concept: first find $\vec{E}$ from the source ($E = kQ/r^2$, direction set by the source's sign), then get the force from $\vec{F} = q\vec{E}$. A positive charge feels a force along $\vec{E}$; a negative charge feels a force against $\vec{E}$. This is why the same field can push two charges in opposite directions. Separating "what the field is" from "what it does to a given charge" prevents the most common conceptual error in the unit, and it generalises directly to gravitational and magnetic fields.此处的两步结构是场概念的核心：先由源求 $\vec{E}$（$E = kQ/r^2$，方向由源的符号决定），再由 $\vec{F} = q\vec{E}$ 求力。正电荷受力沿 $\vec{E}$；负电荷受力逆 $\vec{E}$。这就是同一个场能把两种电荷推向相反方向的原因。把"场是什么"与"它对给定电荷做什么"分开，可避免本单元最常见的概念错误，并可直接推广到引力场与磁场。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §5 Electric potential & energy电势与能量 · SPH4U D2 [8 marks][8 分]

$Q = +5.0\ \mu\text{C}$ fixed; $A$ is $0.50$ m away; $V = 0$ at infinity. (a) Potential at $A$. (b) Work to bring $q = +3.0\ \mu\text{C}$ from afar. (c) Sign of work, explained. (d) Scalar or vector.$Q = +5.0\ \mu\text{C}$ 固定；$A$ 距 $0.50$ m；无穷远 $V = 0$。(a) $A$ 处电势。(b) 将 $q = +3.0\ \mu\text{C}$ 从远处带来所需功。(c) 功的符号及解释。(d) 标量还是矢量。

Answer:答案：  (a) $V_A \approx 9.0\times10^4\ \text{V}$  ·  (b) $W \approx 0.27\ \text{J}$  ·  (c) positive正  ·  (d) scalar标量

(a) Potential of a point charge点电荷的电势 M1·A1·A1

$$ V_A \;=\; k\frac{Q}{r} \;=\; (8.99\times10^9)\frac{5.0\times10^{-6}}{0.50} \;=\; \frac{44950}{0.50} \;\approx\; 9.0\times10^4\ \text{V.} $$

(b) Work to bring $q$ in from infinity将 $q$ 从无穷远带入所需的功 M1·A1

The work equals the change in potential energy. With $V_\infty = 0$, $W = q(V_A - V_\infty) = qV_A$:功等于电势能的变化。由于 $V_\infty = 0$，$W = q(V_A - V_\infty) = qV_A$： $$ W \;=\; (3.0\times10^{-6})(9.0\times10^4) \;\approx\; 0.27\ \text{J.} $$

(c) Sign of the work功的符号 A1

The work is positive. Both charges are positive, so they repel; an external agent must push the incoming $+q$ against this repulsion, doing positive work that is stored as electric potential energy in the system.功为正。两电荷均为正，相互排斥；外部施力必须推动进入的 $+q$ 克服此排斥，做正功，并以系统的电势能形式储存。

(d) Scalar or vector标量还是矢量 A1

Electric potential is a scalar: it has magnitude and sign but no direction. Contributions from several charges add algebraically.电势是标量：有大小和符号，但无方向。多个电荷的贡献代数相加。
Potential is the scalar companion to the field; work to assemble a charge configuration equals $q\,\Delta V$.电势是场的标量伴随量；组装电荷构型所做的功等于 $q\,\Delta V$。 Because $V$ is a scalar, it is often easier to work with than the vector field $\vec{E}$: you add signed numbers, not arrows. The work to bring a charge from infinity to a point is $W = qV$, and its sign carries physical meaning, positive when you push against repulsion (energy stored), negative when the field does the work for you (e.g. bringing a positive charge toward a negative source). Setting $V = 0$ at infinity is the standard reference for isolated point charges. This energy view becomes the engine of capacitors (§6) and of every "accelerate a charge through a voltage" problem, where $\Delta KE = |q\,\Delta V|$.由于 $V$ 是标量，处理起来常比矢量场 $\vec{E}$ 更容易：你相加的是带符号的数，而非箭头。把电荷从无穷远带到某点所做的功为 $W = qV$，其符号有物理意义，克服排斥推入时为正（储存能量），场替你做功时为负（如把正电荷带向负源）。取无穷远处 $V = 0$ 是孤立点电荷的标准参考。这种能量观点是电容器（§6）以及所有"电荷经电压加速"问题的引擎，其中 $\Delta KE = |q\,\Delta V|$。
Q8HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 Field lines & superposition电场线与叠加 · Physics 12 [9 marks][9 分]

Dipole: $+5.0\ \mu\text{C}$ at $x=0$, $-5.0\ \mu\text{C}$ at $x=0.40$ m; midpoint $M$ at $x=0.20$ m. (a) Field from each at $M$. (b) Net field by superposition. (c) Why they add. (d) Why field lines never cross.偶极子：$+5.0\ \mu\text{C}$ 在 $x=0$，$-5.0\ \mu\text{C}$ 在 $x=0.40$ m；中点 $M$ 在 $x=0.20$ m。(a) 各电荷在 $M$ 处的场。(b) 叠加求净场。(c) 为何相加。(d) 电场线为何不相交。

Answer:答案：  (a) $E_+ = E_- \approx 1.12\times10^6\ \text{N/C}$  ·  (b) $E_{\text{net}} \approx 2.25\times10^6\ \text{N/C}$, $+x$  ·  (c) same direction方向相同  ·  (d) one $\vec{E}$ per point每点仅一个 $\vec{E}$

(a) Field magnitude from each charge at $M$ ($r = 0.20$ m)各电荷在 $M$ 处的场强（$r = 0.20$ m） M1·A1·A1

$$ E_{\pm} \;=\; k\frac{|Q|}{r^2} \;=\; (8.99\times10^9)\frac{5.0\times10^{-6}}{(0.20)^2} \;=\; \frac{44950}{0.040} \;\approx\; 1.12\times10^6\ \text{N/C}. $$

(b) Net field by superposition叠加求净场 M1·A1·A1

Take $+x$ pointing from $+Q$ toward $-Q$. The field from $+Q$ points away from it ($+x$); the field from $-Q$ points toward it (also $+x$). Both contributions are in $+x$, so they add:取 $+x$ 为从 $+Q$ 指向 $-Q$ 的方向。$+Q$ 的场背向自身（$+x$）；$-Q$ 的场指向自身（也是 $+x$）。两个贡献都在 $+x$ 方向，故相加： $$ E_{\text{net}} \;=\; E_+ + E_- \;=\; 2(1.12\times10^6) \;\approx\; 2.25\times10^6\ \text{N/C, in } +x. $$

(c) Why they add rather than cancel为何相加而非相消 A1·A1

At the midpoint of a dipole, the field of the positive charge points away from it and the field of the negative charge points toward it. Both of these are the same direction (from $+$ to $-$), so the magnitudes add. They would cancel only if both charges had the same sign.在偶极子的中点，正电荷的场背向自身，负电荷的场指向自身。两者方向相同（从 $+$ 指向 $-$），故大小相加。只有当两电荷同号时它们才会相消。

(d) Why field lines never cross电场线为何从不相交 A1

The electric field has a single, well-defined direction at every point. If two field lines crossed, $\vec{E}$ would have two directions there, which is impossible.电场在每一点都有唯一确定的方向。若两条场线相交，$\vec{E}$ 在该点将有两个方向，这是不可能的。
Superposition is vector addition: resolve each field, then sum components, never add magnitudes blindly.叠加是矢量相加：先分解各场，再对分量求和，切勿盲目相加大小。 The dipole midpoint is the one geometry where the two fields happen to be collinear, so the sum is just $E_+ + E_-$. Contrast this with two like charges at the midpoint, where the fields are equal and opposite and cancel to zero. The general rule is always: find each $\vec{E}_i = kQ_i/r_i^2$ with its direction, then add as vectors (components). The fact that the dipole midpoint field points from $+$ to $-$ and is twice a single-charge field is worth memorising; it is the canonical NGSS HS-PS3-5 "two opposite charges" model and the building block of how molecules with permanent dipoles (like water) interact.偶极子中点是两个场恰好共线的特例，故和就是 $E_+ + E_-$。与之对比，两个同号电荷在中点的场等大反向，相消为零。一般规则始终是：求出每个 $\vec{E}_i = kQ_i/r_i^2$ 及其方向，再按矢量（分量）相加。偶极子中点场从 $+$ 指向 $-$ 且是单电荷场的两倍，这一结论值得记住；它是 NGSS HS-PS3-5"两个异号电荷"的典型模型，也是水等永久偶极子分子如何相互作用的基本单元。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3 + §5 Uniform field & energy匀强场与能量 · HS-PS3-5 [10 marks][10 分]

Parallel plates, $d = 0.020$ m, $200$ V supply, uniform field. Electron ($-e$, $m_e$) starts from rest at the negative plate. (a) Field. (b) KE gained across the gap. (c) Speed at the positive plate. (d) Why it accelerates toward the positive plate.平行板，$d = 0.020$ m，$200$ V 电源，匀强场。电子（$-e$，$m_e$）从负板静止出发。(a) 场。(b) 穿越间隙获得的动能。(c) 到达正板时的速率。(d) 为何向正板加速。

Answer:答案：  (a) $E = 1.0\times10^4\ \text{N/C}$  ·  (b) $\Delta KE = 3.2\times10^{-17}\ \text{J}$  ·  (c) $v \approx 8.4\times10^6\ \text{m/s}$  ·  (d) force on $-e$ is opposite $\vec{E}$$-e$ 受力与 $\vec{E}$ 反向

(a) Uniform field between plates ($E = V/d$)两板间匀强场（$E = V/d$） M1·A1

$$ E \;=\; \frac{V}{d} \;=\; \frac{200}{0.020} \;=\; 1.0\times10^4\ \text{N/C.} $$

(b) Kinetic energy gained ($\Delta KE = |q\,\Delta V|$)获得的动能（$\Delta KE = |q\,\Delta V|$） M1·A1·A1

Crossing the full gap, the electron moves through the full $200$ V. The work done by the field equals the kinetic energy gained:穿越整个间隙时，电子经过完整的 $200$ V。场所做的功等于获得的动能： $$ \Delta KE \;=\; |q|\,\Delta V \;=\; (1.60\times10^{-19})(200) \;=\; 3.2\times10^{-17}\ \text{J.} $$

(c) Speed at the positive plate ($\Delta KE = \tfrac{1}{2}m_e v^2$)到达正板时的速率（$\Delta KE = \tfrac{1}{2}m_e v^2$） M1·A1·A1

$$ v \;=\; \sqrt{\frac{2\,\Delta KE}{m_e}} \;=\; \sqrt{\frac{2(3.2\times10^{-17})}{9.11\times10^{-31}}} \;=\; \sqrt{7.03\times10^{13}} \;\approx\; 8.4\times10^6\ \text{m/s.} $$

(d) Why the electron moves toward the positive plate电子为何向正板运动 A1·A1

The field $\vec{E}$ points from the positive plate to the negative plate. The force on a charge is $\vec{F} = q\vec{E}$; because the electron's charge is negative, $\vec{F}$ points opposite to $\vec{E}$, i.e. from the negative plate toward the positive plate. So the electron accelerates toward the positive plate even though the field points the other way.场 $\vec{E}$ 从正板指向负板。电荷所受力为 $\vec{F} = q\vec{E}$；因为电子电荷为负，$\vec{F}$ 与 $\vec{E}$ 反向，即从负板指向正板。所以尽管场指向相反方向，电子仍向正板加速。
Across parallel plates, a charge gains $|q\,\Delta V|$ of kinetic energy; convert to speed with $\tfrac{1}{2}mv^2$.穿越平行板时，电荷获得 $|q\,\Delta V|$ 的动能；用 $\tfrac{1}{2}mv^2$ 换算成速率。 This is the working principle of cathode-ray tubes, mass spectrometers, and particle accelerators, exactly the applications BC Physics 12 names. The energy method is far cleaner than chasing the force, the acceleration, and then SUVAT: the field does work $W = qE d = q\,\Delta V$ on the charge regardless of path detail, and all of it becomes kinetic energy if it starts from rest. The electron's huge final speed ($\sim 8.4 \times 10^6$ m/s, about $3\%$ of light) comes from its tiny mass. Note the unit "electron-volt": an electron through $200$ V gains exactly $200$ eV, which is why eV is the natural energy unit in this regime.这是阴极射线管、质谱仪和粒子加速器的工作原理，正是 BC Physics 12 所列举的应用。能量法比追踪力、加速度再用 SUVAT 干净得多：无论路径细节如何，场对电荷做功 $W = qE d = q\,\Delta V$，若从静止出发，这些功全部变为动能。电子极大的末速率（约 $8.4 \times 10^6$ m/s，约为光速的 $3\%$）源于其极小的质量。注意单位"电子伏"：电子通过 $200$ V 恰好获得 $200$ eV，这正是 eV 在此范畴内成为自然能量单位的原因。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解AB Diploma + Universal · 28 marks阿省毕业考 + 通用题型 · 共 28 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6 Capacitance (applied)电容（应用） · 30-B2.9k [9 marks][9 分]

Flash capacitor: $A = 0.015\ \text{m}^2$, $d = 1.5\times10^{-3}$ m, charged to $90$ V. (a) Capacitance. (b) Charge. (c) Energy. (d) Field.闪光灯电容器：$A = 0.015\ \text{m}^2$，$d = 1.5\times10^{-3}$ m，充电到 $90$ V。(a) 电容。(b) 电荷。(c) 能量。(d) 场。

Answer:答案：  (a) $C \approx 88.5\ \text{pF}$  ·  (b) $Q \approx 8.0\ \text{nC}$  ·  (c) $U \approx 3.6\times10^{-7}\ \text{J}$  ·  (d) $E = 6.0\times10^4\ \text{N/C}$

(a) Capacitance ($C = \varepsilon_0 A/d$)电容（$C = \varepsilon_0 A/d$） M1·A1·A1

$$ C \;=\; \varepsilon_0\frac{A}{d} \;=\; (8.85\times10^{-12})\frac{0.015}{1.5\times10^{-3}} \;=\; 8.85\times10^{-11}\ \text{F} \;\approx\; 88.5\ \text{pF.} $$

(b) Charge stored ($Q = CV$)储存的电荷（$Q = CV$） M1·A1

$$ Q \;=\; CV \;=\; (8.85\times10^{-11})(90) \;\approx\; 7.97\times10^{-9}\ \text{C} \;\approx\; 8.0\ \text{nC.} $$

(c) Energy stored ($U = \tfrac{1}{2}CV^2$)储存的能量（$U = \tfrac{1}{2}CV^2$） M1·A1

$$ U \;=\; \tfrac{1}{2}CV^2 \;=\; \tfrac{1}{2}(8.85\times10^{-11})(90)^2 \;\approx\; 3.6\times10^{-7}\ \text{J.} $$

(d) Uniform field ($E = V/d$)匀强场（$E = V/d$） M1·A1

$$ E \;=\; \frac{V}{d} \;=\; \frac{90}{1.5\times10^{-3}} \;=\; 6.0\times10^4\ \text{N/C.} $$
A parallel-plate capacitor ties together geometry ($C = \varepsilon_0 A/d$), charge ($Q = CV$), energy ($U = \tfrac12 CV^2$), and field ($E = V/d$).平行板电容器把几何（$C = \varepsilon_0 A/d$）、电荷（$Q = CV$）、能量（$U = \tfrac12 CV^2$）与场（$E = V/d$）联系起来。 Capacitance depends only on geometry and the material between the plates, not on the voltage applied. Charge and energy then follow from the voltage. Watch the energy formula: it is $\tfrac{1}{2}CV^2$, not $CV^2$, and the voltage is squared, two of the most common slips. Note also the three equivalent energy forms $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}QV = \tfrac{1}{2}Q^2/C$; pick whichever matches your knowns. A camera flash works exactly this way: charge the capacitor slowly over seconds, then dump the stored energy through the bulb in milliseconds, producing a bright burst.电容只取决于几何形状与极板间的材料，与所加电压无关。电荷与能量随后由电压决定。注意能量公式是 $\tfrac{1}{2}CV^2$ 而非 $CV^2$，且电压要平方，这是两个最常见的失误。还要注意三种等价的能量形式 $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}QV = \tfrac{1}{2}Q^2/C$；按已知量选用。相机闪光灯正是这样工作的：用几秒缓慢给电容充电，再在几毫秒内把储存的能量倾泻通过灯泡，产生明亮的闪光。
Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §2 Coulomb's law (applied)库仑定律（应用） · 30-B1.6k [9 marks][9 分]

Two spheres: $q_1 = +8.0\ \mu\text{C}$, $q_2 = -3.0\ \mu\text{C}$, $0.10$ m apart. (a) Force magnitude. (b) Attractive or repulsive. (c) New force if separation halved. (d) Compare with gravitation.两球：$q_1 = +8.0\ \mu\text{C}$，$q_2 = -3.0\ \mu\text{C}$，相距 $0.10$ m。(a) 力的大小。(b) 引力还是斥力。(c) 间距减半后的新力。(d) 与万有引力比较。

Answer:答案：  (a) $F \approx 21.6\ \text{N}$  ·  (b) attractive引力  ·  (c) $4\times$ larger ($\approx 86\ \text{N}$)（约 $86\ \text{N}$）  ·  (d) both inverse-square同为平方反比

(a) Apply Coulomb's law套用库仑定律 M1·A1·A1

$|q_1| = 8.0\times10^{-6}$ C, $|q_2| = 3.0\times10^{-6}$ C, $r = 0.10$ m:$|q_1| = 8.0\times10^{-6}$ C，$|q_2| = 3.0\times10^{-6}$ C，$r = 0.10$ m： $$ F \;=\; k\frac{|q_1||q_2|}{r^2} \;=\; (8.99\times10^9)\frac{(8.0\times10^{-6})(3.0\times10^{-6})}{(0.10)^2} \;=\; \frac{0.21576}{0.010} \;\approx\; 21.6\ \text{N.} $$

(b) Direction from the signs由符号判断方向 A1·A1

The charges have opposite signs ($+$ and $-$), so the force is attractive: the spheres are pulled toward each other.两电荷符号相反（$+$ 与 $-$），故力为引力：两球相互吸引。

(c) Halving the separation间距减半 M1·A1

$F \propto 1/r^2$. Halving $r$ multiplies $r^2$ by $1/4$, so the force becomes $4\times$ larger: $F_{\text{new}} \approx 4(21.6) \approx 86\ \text{N}$.$F \propto 1/r^2$。$r$ 减半使 $r^2$ 乘以 $1/4$，故力变为原来的 $4$ 倍：$F_{\text{new}} \approx 4(21.6) \approx 86\ \text{N}$。

(d) Compare with Newton's law of gravitation与牛顿万有引力定律比较 A1·A1

Both are inverse-square laws ($F \propto 1/r^2$) with the same mathematical form. The key difference: gravitation uses mass (always positive) and is always attractive, whereas the Coulomb force uses charge (positive or negative) and can be attractive or repulsive.两者都是平方反比定律（$F \propto 1/r^2$），数学形式相同。关键区别：万有引力用质量（恒为正），始终为吸引；而库仑力用电荷（可正可负），可吸引也可排斥。
Coulomb's law and Newton's gravitation share the inverse-square skeleton; signs are what make electricity richer.库仑定律与牛顿万有引力共用平方反比骨架；符号让电学更丰富。 Alberta Physics 30 (30-B1.8k) explicitly asks students to compare the two laws. Structurally identical, $F = k q_1 q_2 / r^2$ versus $F = G m_1 m_2 / r^2$, they differ in two ways: (1) the source quantity (charge can be $\pm$, mass cannot), so electric forces can attract or repel; (2) the strength, the electric force between two protons is about $10^{36}$ times their gravitational attraction. This is why electrostatics, not gravity, governs the structure of atoms and molecules. The scaling logic ($\times 4$ when distance halves) is the same for both laws, making this comparison a powerful unifying idea.阿尔伯塔 Physics 30（30-B1.8k）明确要求学生比较这两条定律。它们结构相同，$F = k q_1 q_2 / r^2$ 对 $F = G m_1 m_2 / r^2$，区别有二：(1) 源量（电荷可为 $\pm$，质量不可），故电力可吸引或排斥；(2) 强度，两个质子间的电力约为其引力的 $10^{36}$ 倍。这正是支配原子与分子结构的是静电力而非引力的原因。标度逻辑（距离减半时 $\times 4$）对两条定律都相同，使这一比较成为有力的统一思想。
Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 Superposition: null point叠加：零点 · HS-PS2-4 (above two-object floor)（超出双物体基准） [10 marks][10 分]

$q_1 = +9.0\ \mu\text{C}$ at $x=0$, $q_2 = +4.0\ \mu\text{C}$ at $x=0.50$ m. Find the null point $P$. (a) Why $P$ lies between them. (b) Set up the equation. (c) Solve for $x$. (d) Closer to larger or smaller charge?$q_1 = +9.0\ \mu\text{C}$ 在 $x=0$，$q_2 = +4.0\ \mu\text{C}$ 在 $x=0.50$ m。求零点 $P$。(a) 为何 $P$ 在两者之间。(b) 列方程。(c) 求 $x$。(d) 更靠近大电荷还是小电荷？

Answer:答案：  (a) fields oppose only between two like charges仅在两同号电荷之间场才反向  ·  (b) $9/x^2 = 4/(0.50-x)^2$  ·  (c) $x = 0.30\ \text{m}$  ·  (d) closer to the smaller charge更靠近小电荷

(a) Why the null point is between the charges零点为何在两电荷之间 A1·A1

Both charges are positive. Between them, $q_1$ pushes its field in the $+x$ direction while $q_2$ pushes its field in the $-x$ direction, so the two fields oppose and can cancel. Outside the pair (to the left of $q_1$ or right of $q_2$), both fields point the same way and can never cancel. So the null point must be between them.两电荷均为正。在它们之间，$q_1$ 的场指向 $+x$ 方向，而 $q_2$ 的场指向 $-x$ 方向，两场反向，可以相消。在两电荷之外（$q_1$ 左侧或 $q_2$ 右侧），两场同向，永不相消。因此零点必在两者之间。

(b) Set the field magnitudes equal令两场强相等 M1·A1·A1

Let $P$ be a distance $x$ from $q_1$, so it is $(0.50 - x)$ from $q_2$. The $k$ and $10^{-6}$ factors cancel:设 $P$ 距 $q_1$ 为 $x$，则距 $q_2$ 为 $(0.50 - x)$。$k$ 与 $10^{-6}$ 因子相消： $$ \frac{k(9.0\times10^{-6})}{x^2} \;=\; \frac{k(4.0\times10^{-6})}{(0.50-x)^2} \;\Longrightarrow\; \frac{9}{x^2} \;=\; \frac{4}{(0.50-x)^2}. $$

(c) Solve for $x$ (take square roots)求 $x$（开平方根） M1·A1·A1

Cross-multiply and take the positive square root:交叉相乘并取正平方根： $$ 3(0.50 - x) \;=\; 2x \;\Longrightarrow\; 1.5 - 3x \;=\; 2x \;\Longrightarrow\; 5x \;=\; 1.5 \;\Longrightarrow\; x \;=\; 0.30\ \text{m.} $$ Check: $9/(0.30)^2 = 100$ and $4/(0.20)^2 = 100$. Equal magnitudes, so $E_{\text{net}} = 0$. ✓验证：$9/(0.30)^2 = 100$，$4/(0.20)^2 = 100$。大小相等，故 $E_{\text{net}} = 0$。✓

(d) Closer to which charge?更靠近哪个电荷？ A1·A1

$P$ is $0.30$ m from $q_1$ ($9\ \mu\text{C}$) but only $0.20$ m from $q_2$ ($4\ \mu\text{C}$), so it lies closer to the smaller charge. To balance the stronger charge, you must stand farther from it.$P$ 距 $q_1$（$9\ \mu\text{C}$）为 $0.30$ m，但距 $q_2$（$4\ \mu\text{C}$）仅 $0.20$ m，故它更靠近小电荷。要平衡较强的电荷，必须离它更远。
A null point between two like charges sits closer to the weaker one; take the positive root only.两同号电荷之间的零点更靠近较弱者；只取正根。 The strategy is always the same: (1) reason out the region where the fields can oppose (between two like charges, or outside two opposite charges), (2) equate magnitudes $kQ_1/r_1^2 = kQ_2/r_2^2$, (3) take the square root to linearise, avoiding a messy quadratic. The square-root step is the key shortcut: $\sqrt{9} : \sqrt{4} = 3 : 2$, so the distances are in ratio $3 : 2$, immediately giving $x = 0.30$ m. Always discard the negative root, which corresponds to a point outside the segment where the fields actually reinforce. This null-point method extends directly to AP Physics and is a favourite extended-response setup.策略始终相同：(1) 推断场可相消的区域（两同号电荷之间，或两异号电荷之外）；(2) 令大小相等 $kQ_1/r_1^2 = kQ_2/r_2^2$；(3) 开平方根使方程线性化，避免繁琐的二次方程。开方这一步是关键捷径：$\sqrt{9} : \sqrt{4} = 3 : 2$，故距离之比为 $3 : 2$，立即得 $x = 0.30$ m。务必舍去负根，它对应于场实际相互增强的区段外的点。这一零点方法可直接推广到 AP 物理，是简答题的常见考法。