High School Physics

Electrostatics and Electric Fields静电学与电场

Electrostatics describes the forces and fields arising from electric charges at rest. This guide covers the full path: the nature and conservation of electric charge, Coulomb's inverse-square force law (the electrostatic parallel to gravity), the electric field as force per unit charge, field-line diagrams and the superposition principle for multiple charges, electric potential and voltage as energy per charge, an introduction to capacitance, and a systematic problem-solving strategy. KaTeX math is used throughout for Coulomb's law $F = k\tfrac{q_1 q_2}{r^2}$, the field definition $E = \tfrac{F}{q}$, and the point-charge potential $V = \tfrac{kq}{r}$.静电学(electrostatics)描述静止电荷(electric charge,电荷)产生的力与场。本指南覆盖完整路径:电荷的本质与守恒、库仑反平方力定律(库仑定律,Coulomb's law,与重力的静电平行)、电场(electric field,电场)作为单位电荷所受的力、多电荷的电场线图与叠加原理、电势(electric potential)与电压(voltage)作为单位电荷的能量、电容(capacitance)入门,以及系统性解题策略。全程使用 KaTeX 数学公式:库仑定律 $F = k\tfrac{q_1 q_2}{r^2}$、场定义 $E = \tfrac{F}{q}$、点电荷电势 $V = \tfrac{kq}{r}$。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB Connects to Unit 9 Circuits · Unit 10 Magnetism衔接第 9 单元电路 · 第 10 单元磁学

How to use this guide如何使用本指南

Electrostatics sits at different grade levels in each curriculum. US NGSS addresses it through two PEs (HS-PS2-4 and HS-PS3-5) that bracket both Coulomb's law and field/energy modelling. In Ontario it appears explicitly in SPH4U Strand D (Grade 12) rather than SPH3U (Grade 11), though SPH3U Strand F builds the charge/current foundation. In BC it is a core Physics 12 topic (Big Idea: "Forces and energy interactions occur within fields"). In Alberta it is Physics 30 Unit B (Grade 12). The table below tells you which sections are core for you now; each row cites the curriculum document it was checked against.静电学在各大纲中位于不同年级。美国 NGSS 通过两个 PE(HS-PS2-4 与 HS-PS3-5)涵盖库仑定律与场/能量建模。在安大略,它明确出现在 SPH4U D 单元(12 年级),而非 SPH3U(11 年级),但 SPH3U F 单元奠定了电荷/电流基础。在 BC,它是 Physics 12 的核心主题(大概念:"力与能量相互作用发生在场中")。在阿尔伯塔,它是 Physics 30 B 单元(12 年级)。下表告诉你当前哪些节属于你的核心;每行都注明所依据的课纲文件。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Physical Science美国 NGSS 物理科学 §1 through §5 (charge, Coulomb, field, field lines, potential) — the core under HS-PS2-4 and HS-PS3-5§1 至 §5(电荷、库仑、电场、电场线、电势)—— HS-PS2-4 与 HS-PS3-5 下的核心 §6 (capacitance intro) and §7 (extended problem-solving): valuable but above the two-object Assessment Boundary§6(电容入门)与 §7(拓展解题):有价值,但高于两个物体的评估边界 ngss_hs_ps_extract.md — HS-PS2-4 (Coulomb/field) + HS-PS3-5 (field/energy modelling)— HS-PS2-4(库仑/场)+ HS-PS3-5(场/能量建模)
🇨🇦 ON Grade 12 — SPH4U安大略 12 年级 — SPH4U §1 through §6 in full. SPH4U Strand D (Gravitational, Electric, and Magnetic Fields) covers all of charge, Coulomb, electric field, and electric potential quantitatively§1 至 §6 完整学习。SPH4U D 单元(引力场、电场与磁场)定量涵盖电荷、库仑、电场与电势的全部内容 Nothing — the full unit is core Grade 12 content in Ontario无 — 全单元均为安大略 12 年级核心内容 science_11-12_physics_extract.md — SPH4U Strand D Overall Expectations D1–D3— SPH4U D 单元总体期望 D1–D3
🇨🇦 BC Grade 12 — Physics 12BC 12 年级 — Physics 12 §1 through §6. Physics 12 Content lists "electric field and Coulomb's law"; "electric potential energy, electric potential, and electric potential difference"; "electrostatic dynamics and energy relationships" — all sections are core§1 至 §6。Physics 12 内容含"电场与库仑定律";"电势能、电势与电势差";"静电动力学与能量关系"——各节均为核心 Nothing — BC Physics 12 Big Idea is explicitly "forces and energy interactions occur within fields"无 — BC Physics 12 大概念明确为"力与能量相互作用发生在场中" physics_11-12_extract.md — Physics 12 Content: electric field, Coulomb's law, electric potential, electrostatic dynamics— Physics 12 内容:电场、库仑定律、电势、静电动力学
🇨🇦 AB Grade 12 — Physics 30阿尔伯塔 12 年级 — Physics 30 §1 through §6 in full. Physics 30 Unit B GO1–GO2 cover Coulomb's law quantitatively, vector fields, electric potential difference, field intensity and direction (30–B1.6k, 30–B2.1k, 30–B2.4k, 30–B2.6k)§1 至 §6 完整学习。Physics 30 B 单元 GO1–GO2 定量涵盖库仑定律、矢量场、电势差、场强与方向(30–B1.6k、30–B2.1k、30–B2.4k、30–B2.6k) Nothing — AB Physics 30 expects quantitative Coulomb and field calculations; Diploma-style problems reward exact field-vector setup无 — AB Physics 30 要求定量库仑与场计算;文凭考风格题奖励精确的场矢量设置 physics_20-30_extract.md — Physics 30 Unit B GO1 (30–B1.1k–30–B1.8k) and GO2 (30–B2.1k–30–B2.10k)— Physics 30 B 单元 GO1(30–B1.1k–30–B1.8k)与 GO2(30–B2.1k–30–B2.10k)
🇺🇸 AP / IB feeder trackAP / IB 衔接轨道 All seven sections plus every going-deeper derivation. AP Physics C: E&M and IB Physics HL Topic D assume fluent Coulomb's law, field superposition, and potential from the outset全部 7 节,并完成每个"深入"推导。AP Physics C: E&M 与 IB Physics HL D 主题从一开始就默认熟练库仑定律、场的叠加与电势 Nothing — this unit is the field-and-potential foundation that Circuits, Magnetism, and AP/IB E&M build on无 — 本单元是电路、磁学和 AP/IB 电磁学所依赖的场与电势基础 ngss_hs_ps_extract.md — HS-PS2-4 maps to both this unit and Unit 5 (Gravitation); the AP/IB feeder reads the full Gauss-law and field-theory depth— HS-PS2-4 同时映射本单元与第 5 单元(引力);AP/IB 衔接阅读完整的高斯定律与场论深度

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后,用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise four things: the sign rule for charge (like repels, unlike attracts); Coulomb's law $F = k\tfrac{q_1 q_2}{r^2}$ with $k = 8.99 \times 10^9\ \mathrm{N\,m^2/C^2}$; the field definition $E = \tfrac{F}{q}$ (force per unit positive test charge); and the point-charge potential $V = \tfrac{kq}{r}$. Read every cram-cheat box. Skip the going-deeper derivations.背熟四件事:电荷符号法则(同斥异吸);库仑定律 $F = k\tfrac{q_1 q_2}{r^2}$,$k = 8.99 \times 10^9\ \mathrm{N\,m^2/C^2}$;电场定义 $E = \tfrac{F}{q}$(单位正试验电荷所受的力);以及点电荷电势 $V = \tfrac{kq}{r}$。读每个速记框,跳过深入推导。

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If you are going for the top mark如果你目标顶分

Always draw a diagram showing charge signs and the direction of the force or field vector before writing any equation. For superposition, add field vectors component-by-component — not magnitudes directly. Distinguish field ($E$, a vector, force per charge) from potential ($V$, a scalar, energy per charge). Know that field lines point from positive to negative and never cross. AB Physics 30 and SPH4U both expect quantitative Coulomb calculations with full vector direction, not just magnitude.写任何方程前先画图,标出电荷符号与力或场矢量的方向。叠加时,逐分量相加场矢量,而非直接相加大小。区分电场($E$,矢量,力除以电荷)与电势($V$,标量,能量除以电荷)。熟知电场线从正指向负、永不交叉。AB Physics 30 与 SPH4U 都要求带完整矢量方向的定量库仑计算,不只是大小。

Grade-level note.年级说明。 This unit sits at Grade 12 in all four curricula (SPH4U, BC Physics 12, AB Physics 30, and the NGSS HS-PS performance expectations). If you are in Grade 11, the charge/Coulomb basics (§1–§2) often appear as a preview in SPH3U Strand F, but the full quantitative treatment is Grade 12. Use the table above to locate your row and focus your effort there. The going-deeper boxes in §3 and §5 extend to AP Physics C: E&M and IB Physics HL depth (Gauss's law, full potential-energy derivations).本单元在四套大纲中均位于 12 年级(SPH4U、BC Physics 12、AB Physics 30 以及 NGSS HS-PS 表现期望)。若你在 11 年级,电荷/库仑基础(§1–§2)通常在 SPH3U F 单元作为预览出现,但完整的定量处理在 12 年级。用上表找到你的行并集中精力。§3 与 §5 的深入框延伸至 AP Physics C: E&M 与 IB Physics HL 深度(高斯定律、完整电势能推导)。

Electric Charge and Conservation of Charge电荷与电荷守恒

Three facts that anchor the whole unit.支撑整个单元的三个基本事实。
  • Two types of charge.两种电荷。 Positive (protons) and negative (electrons). Like charges repel; unlike charges attract. The unit is the coulomb (C); the elementary charge is $e = 1.60 \times 10^{-19}$ C.正电荷(质子)与负电荷(电子)。同种排斥,异种吸引。单位是库仑(C);基本电荷量 $e = 1.60 \times 10^{-19}$ C。
  • Conservation of charge.电荷守恒。 Charge is neither created nor destroyed in any interaction; it is only transferred. The net charge of an isolated system is constant.在任何相互作用中,电荷既不产生也不消失,只是转移。孤立系统的净电荷量恒定不变。
  • Charge transfer methods.电荷转移方式。 Conduction (direct contact, electrons transfer) and induction (no contact; rearrangement by proximity of a charged object). Conductors allow charge to flow freely; insulators do not.传导(直接接触,电子转移)与感应(无接触;因带电体靠近而重新分布)。导体允许电荷自由流动;绝缘体则不。
Alberta 30–B1.1k: "explain electrical interactions in terms of the law of conservation of charge." Alberta 30–B1.3k: "compare the methods of transferring charge (conduction and induction)."阿尔伯塔 30–B1.1k:"用电荷守恒定律解释电相互作用。"30–B1.3k:"比较传导与感应两种转移电荷的方法。"
Worked Example 1 · Charge by conduction例题 1 · 传导带电

A neutral metal sphere A is touched by a negatively charged rod that transfers $-3.2 \times 10^{-8}$ C to it. After contact the rod is removed and sphere A then touches neutral sphere B. Assuming equal sharing, what is the charge on each sphere?一个中性金属球 A 被一根转移了 $-3.2 \times 10^{-8}$ C 的负电荷棒接触。接触后棒被移走,球 A 再接触中性球 B。假设等量分配,每个球的电荷量是多少?

After rod touches A.棒接触 A 后。 Sphere A gains $-3.2 \times 10^{-8}$ C. Conservation of charge: rod loses that charge.球 A 获得 $-3.2 \times 10^{-8}$ C。电荷守恒:棒损失该电荷。

A touches neutral B.A 接触中性球 B。 Total charge shared $= -3.2 \times 10^{-8}$ C. Equal sharing:共享总电荷 $= -3.2 \times 10^{-8}$ C。等量分配:

$$ q_A = q_B = \frac{-3.2 \times 10^{-8}}{2} = -1.6 \times 10^{-8}\ \text{C each.} $$

Key point.关键点。 The total charge $(-3.2 \times 10^{-8}\ \text{C} + 0)$ before equals the total $(-1.6 \times 10^{-8} + (-1.6 \times 10^{-8}))$ after. Charge is conserved. ✓之前总电荷 $(-3.2 \times 10^{-8}\ \text{C} + 0)$ 等于之后总电荷 $(-1.6 \times 10^{-8} + (-1.6 \times 10^{-8}))$。电荷守恒。✓

Two identical neutral conducting spheres are touched by a $+6.0\ \mu\text{C}$ rod and then separated. What is the charge on each?两个相同的中性导体球被一根 $+6.0\ \mu\text{C}$ 的棒接触后分开。每个球的电荷量是多少?
§1 · Q1
$+6.0\ \mu\text{C}$ each各 $+6.0\ \mu\text{C}$
$+3.0\ \mu\text{C}$ each各 $+3.0\ \mu\text{C}$
$0$ each各 $0$
$-3.0\ \mu\text{C}$ each各 $-3.0\ \mu\text{C}$
Total charge $+6.0\ \mu\text{C}$ is conserved and shared equally between two identical spheres: $+3.0\ \mu\text{C}$ each.总电荷 $+6.0\ \mu\text{C}$ 守恒,在两个相同球之间等量分配:各 $+3.0\ \mu\text{C}$。
Charge is conserved; the rod's charge is shared equally, so each sphere gets half.电荷守恒;棒的电荷等量分配,每个球得一半。
Which statement about electric charge is correct?关于电荷,哪项陈述正确?
§1 · Q2
Charge can be created from nothing in a conductor导体中电荷可以无中生有
Like charges attract each other同种电荷相互吸引
The net charge of an isolated system is constant孤立系统的净电荷量恒定
Induction requires direct contact between objects感应需要物体间直接接触
Conservation of charge: charge is never created or destroyed, only transferred. The net charge of an isolated system is constant.电荷守恒:电荷既不产生也不消失,只是转移。孤立系统的净电荷量恒定。
Like charges repel. Induction works without contact. Charge cannot be created. Only conservation of charge is always true.同种电荷相互排斥。感应无需接触。电荷不能凭空产生。只有电荷守恒始终成立。
Going deeper — quantisation of charge and the elementary charge $e$深入 — 电荷量子化与基本电荷 $e$

Every observable charge is an integer multiple of the elementary charge $e = 1.602 \times 10^{-19}$ C. You cannot have $1.5e$ or $2.7e$ of charge — only $\pm 1e, \pm 2e, \ldots$ This is charge quantisation, confirmed by Millikan's oil-drop experiment (30–B2.10k in Alberta Physics 30). In everyday macroscopic objects, the number of excess charges is so large (a microcoulomb is about $6 \times 10^{12}$ elementary charges) that charge appears continuous. But at the atomic level, charge comes in discrete packets — the foundation of the whole charge-and-field picture this unit builds.一切可观测的电荷都是基本电荷 $e = 1.602 \times 10^{-19}$ C 的整数倍。你不可能有 $1.5e$ 或 $2.7e$ 的电荷,只能是 $\pm 1e, \pm 2e, \ldots$ 这就是电荷量子化,由密立根油滴实验证实(阿尔伯塔 Physics 30 30–B2.10k)。在日常宏观物体中,多余电荷的数目极大(一微库仑约含 $6 \times 10^{12}$ 个基本电荷),电荷看上去是连续的。但在原子层面,电荷以离散单元存在——这正是本单元所建立的整个电荷与场图像的基础。


Coulomb's Law: the Electrostatic Force库仑定律:静电力

One formula, four things to remember.一条公式,四点要记。 $$ F = k\frac{|q_1|\,|q_2|}{r^2}, \qquad k = 8.99 \times 10^9\ \frac{\mathrm{N\,m^2}}{\mathrm{C^2}} $$
  • Inverse-square law.平方反比定律。 Double the distance $\Rightarrow$ force drops to one-quarter. Same structure as Newton's law of gravitation.距离加倍 $\Rightarrow$ 力减为四分之一。结构与牛顿万有引力定律相同。
  • Direction from sign.方向由符号决定。 Use the magnitude formula to find $|F|$; the direction is repulsive if signs are the same, attractive if opposite. Draw a force diagram.用大小公式求 $|F|$;方向:同号排斥,异号吸引。画受力图。
  • Newton's third law pair.牛顿第三定律对。 $q_1$ exerts $F$ on $q_2$, and $q_2$ exerts $F$ on $q_1$ — equal magnitudes, opposite directions.$q_1$ 对 $q_2$ 施力 $F$,$q_2$ 对 $q_1$ 施力 $F$ — 大小相等,方向相反。
  • Gravity parallel.与重力的对比。 $F_g = G\frac{m_1 m_2}{r^2}$ uses mass; $F_e = k\frac{q_1 q_2}{r^2}$ uses charge. Coulomb force can be repulsive; gravity is always attractive. (HS-PS2-4, 30–B1.8k)$F_g = G\frac{m_1 m_2}{r^2}$ 用质量;$F_e = k\frac{q_1 q_2}{r^2}$ 用电荷。库仑力可排斥;重力始终吸引。(HS-PS2-430–B1.8k
Worked Example 2 · Force between two point charges例题 2 · 两点电荷间的力

Two point charges $q_1 = +4.0\ \mu\text{C}$ and $q_2 = -2.0\ \mu\text{C}$ are $0.30$ m apart. Find the magnitude of the electric force between them and state whether it is attractive or repulsive.两点电荷 $q_1 = +4.0\ \mu\text{C}$ 与 $q_2 = -2.0\ \mu\text{C}$ 相距 $0.30$ m。求电力大小并说明是引力还是斥力。

Convert units.换算单位。 $|q_1| = 4.0 \times 10^{-6}$ C; $|q_2| = 2.0 \times 10^{-6}$ C; $r = 0.30$ m.$|q_1| = 4.0 \times 10^{-6}$ C;$|q_2| = 2.0 \times 10^{-6}$ C;$r = 0.30$ m。

Apply Coulomb's law.套用库仑定律。

$$ F = k\frac{|q_1|\,|q_2|}{r^2} = (8.99\times10^9)\frac{(4.0\times10^{-6})(2.0\times10^{-6})}{(0.30)^2} $$ $$ F = (8.99\times10^9)\frac{8.0\times10^{-12}}{0.090} \approx 0.80\ \text{N.} $$

Direction.方向。 Opposite signs ($+$ and $-$) $\Rightarrow$ attractive. The $0.80$ N force pulls the charges toward each other.符号相反($+$ 与 $-$)$\Rightarrow$ 引力。$0.80$ N 的力把两电荷相互拉近。

Two charges are $0.50$ m apart and experience a $1.0$ N force. If the distance is doubled to $1.0$ m, what is the new force?两电荷相距 $0.50$ m,受到 $1.0$ N 的力。若距离加倍至 $1.0$ m,新的力是多少?
§2 · Q1
$2.0$ N
$1.0$ N
$0.50$ N
$0.25$ N
Force $\propto 1/r^2$. Doubling $r$ multiplies $r^2$ by $4$, so $F$ drops to $1/4$: $1.0/4 = 0.25$ N.力 $\propto 1/r^2$。$r$ 加倍则 $r^2$ 乘以 $4$,故 $F$ 变为 $1/4$:$1.0/4 = 0.25$ N。
Coulomb's law is an inverse-square law: double the distance, quarter the force.库仑定律是平方反比定律:距离加倍,力减为四分之一。
Two positive charges of equal magnitude are $0.20$ m apart. The force on each is $F$. What happens to the force if each charge is doubled and the distance is halved?两个等量正电荷相距 $0.20$ m,每个所受的力为 $F$。若每个电荷加倍、距离减半,力如何变化?
§2 · Q2
$16F$ (repulsive)$16F$(斥力)
$4F$ (repulsive)$4F$(斥力)
$F$ (repulsive)$F$(斥力)
$F/4$ (attractive)$F/4$(引力)
Charges double $\Rightarrow$ numerator multiplies by $2 \times 2 = 4$. Distance halves $\Rightarrow$ $r^2$ multiplies by $1/4 \Rightarrow$ denominator quarters $\Rightarrow$ that factor multiplies force by $4$. Net: $4 \times 4 = 16F$. Both charges are positive, so still repulsive.电荷加倍 $\Rightarrow$ 分子乘以 $2 \times 2 = 4$。距离减半 $\Rightarrow$ $r^2$ 乘以 $1/4$ $\Rightarrow$ 分母变为四分之一 $\Rightarrow$ 该因子使力再乘以 $4$。总计:$4 \times 4 = 16F$。两电荷均为正,故仍为斥力。
$F \propto q_1 q_2 / r^2$. Both charges double ($\times 4$ numerator) and distance halves ($\times 4$ from $1/r^2$), giving $16F$ total. Direction is still repulsive (same sign charges).$F \propto q_1 q_2 / r^2$。两电荷均加倍(分子 $\times 4$),距离减半($1/r^2$ 再 $\times 4$),共计 $16F$。方向仍为斥力(同号电荷)。
Going deeper — comparing Coulomb's law and Newton's law of gravitation深入 — 比较库仑定律与牛顿万有引力定律

Both laws follow the inverse-square form $F \propto 1/r^2$, so the mathematical structure is identical. The differences are: (1) Gravity uses mass, always positive, so gravity is always attractive; Coulomb uses charge, which can be positive or negative, so the force can be repulsive. (2) The ratio of magnitudes is staggering: for two protons $0.10$ nm apart, $F_e / F_g \approx 10^{36}$ — electrostatics utterly dominates at atomic scales. (3) Gravity has no shielding; electric fields can be screened by conductors (Faraday cage). NGSS HS-PS2-4 deliberately pairs the two laws in one PE precisely because of this structural parallel.两条定律都遵从平方反比形式 $F \propto 1/r^2$,数学结构完全相同。区别在于:(1) 重力用质量,恒为正,故重力始终吸引;库仑力用电荷,可正可负,故力可以排斥。(2) 大小之比令人震惊:两个质子相距 $0.10$ nm 时,$F_e / F_g \approx 10^{36}$ — 在原子尺度上静电力完全主导。(3) 重力无法屏蔽;电场可被导体屏蔽(法拉第笼)。NGSS HS-PS2-4 刻意把两条定律并入同一 PE,正是因为这种结构上的对应。


The Electric Field: Force per Unit Charge电场:单位电荷所受的力

The field concept replaces "action at a distance."场的概念取代了"超距作用"。 $$ \vec{E} = \frac{\vec{F}}{q_0} \qquad \Rightarrow \qquad F = qE $$ $$ E_{\text{point charge}} = k\frac{|Q|}{r^2} \quad (\text{SI: N/C or V/m}) $$
  • Definition.定义。 $\vec{E}$ is defined at a point as the force per unit positive test charge placed there. A positive test charge $q_0$ that is small enough not to disturb the field maps it.$\vec{E}$ 定义为放在该点的单位正试验电荷所受的力。足够小、不会扰动场的正试验电荷 $q_0$ 用来探测场。
  • Direction convention.方向约定。 $\vec{E}$ points in the direction a positive test charge would be pushed. Near a positive source charge, $\vec{E}$ points outward; near a negative source, it points inward.$\vec{E}$ 指向正试验电荷将被推动的方向。靠近正源电荷时,$\vec{E}$ 向外;靠近负源时,$\vec{E}$ 向内。
  • Force on a charge in a field.场中电荷所受的力。 $\vec{F} = q\vec{E}$. If $q$ is positive, force is in the direction of $\vec{E}$; if $q$ is negative, force is opposite to $\vec{E}$.$\vec{F} = q\vec{E}$。若 $q$ 为正,力与 $\vec{E}$ 同向;若 $q$ 为负,力与 $\vec{E}$ 反向。
BC Physics 12: "electric field: vector field; interacts with positive/negative elementary charge; attractive or repulsive." Alberta 30–B2.6k: "explain, quantitatively, electric fields in terms of intensity (strength) and direction."BC Physics 12:"电场:矢量场;与正/负基本电荷相互作用;吸引或排斥。"阿尔伯塔 30–B2.6k:"从强度与方向两方面定量解释电场。"
Worked Example 3 · Field from a point charge, then force on a test charge例题 3 · 点电荷产生的场,再求试验电荷所受的力

A source charge $Q = +5.0\ \mu\text{C}$ is fixed in place. (a) Find the magnitude of $\vec{E}$ at a point $P$ that is $0.40$ m from $Q$. (b) A charge $q = -3.0\ \mu\text{C}$ is placed at $P$. Find the force on $q$ and state its direction.源电荷 $Q = +5.0\ \mu\text{C}$ 固定不动。(a) 求距 $Q$ 为 $0.40$ m 的点 $P$ 处的 $\vec{E}$ 大小。(b) 电荷 $q = -3.0\ \mu\text{C}$ 放在 $P$ 处,求 $q$ 所受的力及其方向。

(a) Field magnitude at P.(a) P 处的场强大小。

$$ E = k\frac{|Q|}{r^2} = (8.99\times10^9)\frac{5.0\times10^{-6}}{(0.40)^2} = \frac{44950}{0.16} \approx 2.8\times10^5\ \text{N/C.} $$

Direction: $Q$ is positive, so $\vec{E}$ at $P$ points away from $Q$ (radially outward).方向:$Q$ 为正,故 $P$ 处的 $\vec{E}$ 指向远离 $Q$ 的方向(径向向外)。

(b) Force on $q$ at P.(b) P 处 $q$ 所受的力。

$$ |F| = |q|E = (3.0\times10^{-6})(2.8\times10^5) \approx 0.84\ \text{N.} $$

Since $q$ is negative, $\vec{F}$ is opposite to $\vec{E}$ — the force on $q$ points toward $Q$ (attractive). ✓因为 $q$ 为负,$\vec{F}$ 与 $\vec{E}$ 反向 — $q$ 所受的力指向 $Q$(引力)。✓

A positive test charge experiences a force of $6.0 \times 10^{-4}$ N in a region where $E = 3.0 \times 10^3$ N/C. What is the value of the test charge?一个正试验电荷在 $E = 3.0 \times 10^3$ N/C 的区域受到 $6.0 \times 10^{-4}$ N 的力。试验电荷的值是多少?
§3 · Q1
$5.0 \times 10^{-9}$ C
$2.0 \times 10^{-7}$ C
$1.8$ C
$0.50$ C
$q_0 = F/E = (6.0\times10^{-4})/(3.0\times10^3) = 2.0\times10^{-7}$ C = $0.20\ \mu\text{C}$.$q_0 = F/E = (6.0\times10^{-4})/(3.0\times10^3) = 2.0\times10^{-7}$ C = $0.20\ \mu\text{C}$。
From $F = qE$, solve $q = F/E$. Divide the force by the field strength.由 $F = qE$,解出 $q = F/E$。用力除以场强。
At a point near a negative source charge $-Q$, in which direction does the electric field vector point?在负源电荷 $-Q$ 附近某点处,电场矢量指向哪个方向?
§3 · Q2
Away from $-Q$远离 $-Q$ 的方向
Perpendicular to the line joining the point and $-Q$垂直于该点与 $-Q$ 连线的方向
Toward $-Q$指向 $-Q$ 的方向
In whatever direction the field is applied externally外部施加场的方向
$\vec{E}$ points in the direction a positive test charge would be pushed. Near a negative source, the positive test charge is attracted toward it, so $\vec{E}$ points inward toward $-Q$.$\vec{E}$ 指向正试验电荷被推动的方向。靠近负源时,正试验电荷被吸引向它,故 $\vec{E}$ 向内指向 $-Q$。
Field direction is defined by the motion of a positive test charge. Toward a negative source means the field points inward.场的方向由正试验电荷的运动方向定义。朝向负源意味着场向内指向它。
Going deeper — the field from a uniform parallel-plate capacitor深入 — 均匀平行板电容器的电场

A parallel-plate capacitor consists of two large conducting plates of area $A$ separated by distance $d$, with equal and opposite charges $\pm Q$. Between the plates the field is uniform: every point has the same $E$ pointing from the positive to the negative plate. Using Gauss's law (the HS-level statement: field lines begin on positive charges and end on negative charges, and the total flux through a closed surface equals $Q_{\text{enc}}/\varepsilon_0$), the field magnitude is $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$, where $\sigma$ is the surface charge density. Equivalently, $E = V/d$ where $V$ is the voltage across the plates (see §5). BC Physics 12 explicitly names the "parallel plates (uniform field)" as one of two key electric-field geometries.平行板电容器由面积为 $A$、间距为 $d$、带等量异号电荷 $\pm Q$ 的两块大导体板组成。两板间的电场均匀:每一点的 $E$ 大小相同,方向从正板指向负板。利用高斯定律(高中水平表述:电场线从正电荷出发,终止于负电荷,穿过闭合曲面的总通量等于 $Q_{\text{enc}}/\varepsilon_0$),场强大小为 $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$,其中 $\sigma$ 是面电荷密度。等价地,$E = V/d$,其中 $V$ 是两板间的电压(见 §5)。BC Physics 12 明确把"平行板(均匀场)"列为两种关键电场几何形状之一。


Electric Field Lines and the Superposition Principle电场线与叠加原理

Field lines are maps of $\vec{E}$; superposition adds fields as vectors.电场线是 $\vec{E}$ 的地图;叠加将电场作为矢量相加。
  • Field-line rules.电场线规则。
    • Lines start on positive charges and end on negative charges.线从正电荷出发,终止于负电荷。
    • The tangent to a field line at any point gives the direction of $\vec{E}$ there.场线在任意点的切线方向即为该点 $\vec{E}$ 的方向。
    • The density (number per unit area) of lines indicates the magnitude of $\vec{E}$: more lines = stronger field.场线的密度(单位面积条数)表示 $\vec{E}$ 的大小:线越密,场越强。
    • Field lines never cross. (If they did, $\vec{E}$ would have two directions at one point, which is impossible.)电场线从不相交。(若相交,$\vec{E}$ 在一点将有两个方向,这是不可能的。)
  • Superposition.叠加。 The total electric field at a point due to multiple charges is the vector sum of the individual fields: $\vec{E}_{\text{total}} = \vec{E}_1 + \vec{E}_2 + \cdots$. Calculate each $\vec{E}_i = k|Q_i|/r_i^2$ in direction, then add components.多个电荷在某点产生的合电场是各自电场的矢量之和:$\vec{E}_{\text{total}} = \vec{E}_1 + \vec{E}_2 + \cdots$。分别计算每个 $\vec{E}_i = k|Q_i|/r_i^2$ 的方向,再对分量求和。
NGSS HS-PS3-5 names "drawings of what happens when two charges of opposite polarity are near each other" as an example of a field model — exactly the dipole field-line diagram.NGSS HS-PS3-5 以"两个极性相反的电荷靠近时的图示"为场模型的例子 —— 正是偶极子的电场线图。
Worked Example 4 · Field at the midpoint of two opposite charges例题 4 · 两异号电荷中点处的电场

Charge $+Q = +3.0\ \mu\text{C}$ is at position $x = 0$, and charge $-Q = -3.0\ \mu\text{C}$ is at $x = 0.40$ m. Find the magnitude and direction of the net electric field at the midpoint $x = 0.20$ m.电荷 $+Q = +3.0\ \mu\text{C}$ 在 $x = 0$,电荷 $-Q = -3.0\ \mu\text{C}$ 在 $x = 0.40$ m。求中点 $x = 0.20$ m 处净电场的大小与方向。

Field from $+Q$ at the midpoint.$+Q$ 在中点产生的电场。 $r = 0.20$ m; field points away from $+Q$, i.e. in the $+x$ direction:$r = 0.20$ m;场背向 $+Q$,即 $+x$ 方向:

$$ E_+ = k\frac{Q}{r^2} = (8.99\times10^9)\frac{3.0\times10^{-6}}{(0.20)^2} \approx 6.74\times10^5\ \text{N/C, in } +x. $$

Field from $-Q$ at the midpoint.$-Q$ 在中点产生的电场。 $r = 0.20$ m; field points toward $-Q$, also in the $+x$ direction:$r = 0.20$ m;场指向 $-Q$,也是 $+x$ 方向:

$$ E_- = k\frac{Q}{r^2} \approx 6.74\times10^5\ \text{N/C, in } +x. $$

Superposition.叠加。 Both contributions are in $+x$, so they add: $E_{\text{total}} = 2 \times 6.74 \times 10^5 \approx 1.35 \times 10^6$ N/C in the $+x$ direction (from $+Q$ toward $-Q$). This is a classic electric dipole geometry: the midpoint field is twice that of a single charge and points from positive to negative.两个贡献均在 $+x$ 方向,直接相加:$E_{\text{total}} = 2 \times 6.74 \times 10^5 \approx 1.35 \times 10^6$ N/C,方向 $+x$(从 $+Q$ 指向 $-Q$)。这是经典电偶极子几何:中点场是单个电荷的两倍,方向从正指向负。

Which of these describes a correct property of electric field lines?下列哪项正确描述了电场线的性质?
§4 · Q1
Field lines never cross each other电场线从不相交
Field lines start on negative charges电场线从负电荷出发
Widely spaced field lines indicate a stronger field间隔宽的电场线表示场更强
Field lines are perpendicular to $\vec{E}$ at every point电场线在每个点都垂直于 $\vec{E}$
Field lines never cross because $\vec{E}$ can only have one direction at each point. Lines start on positive, end on negative; denser lines mean stronger field; the tangent (not perpendicular) to a line gives $\vec{E}$'s direction.电场线从不相交,因为 $\vec{E}$ 在每点只能有一个方向。线从正出发、终止于负;线越密场越强;场线的切线(而非法线)给出 $\vec{E}$ 的方向。
Field lines begin on positive charges, not negative. Dense lines mean stronger field. The tangent to a field line gives the direction of $\vec{E}$. Lines never cross.电场线从正电荷出发,不是负电荷。线密表示场强。场线的切线给出 $\vec{E}$ 的方向。场线从不相交。
Two charges $+2Q$ and $+2Q$ are placed $d$ apart. At the midpoint between them, the net electric field is:两电荷 $+2Q$ 与 $+2Q$ 相距 $d$。在它们的中点,合电场为:
§4 · Q2
$2k(2Q)/(d/2)^2$ in the $+x$ direction沿 $+x$ 方向的 $2k(2Q)/(d/2)^2$
$k(2Q)/(d/2)^2$ in the $+x$ direction沿 $+x$ 方向的 $k(2Q)/(d/2)^2$
$k(2Q)/d^2$ pointing toward one charge指向某个电荷的 $k(2Q)/d^2$
Zero
Each $+2Q$ pushes the midpoint field outward (away from itself). Since the charges are equal and symmetric, the two contributions are equal in magnitude and point in opposite directions, cancelling to zero.每个 $+2Q$ 都把中点的场向外推(远离自身)。因为两电荷相等且对称,两个贡献大小相等、方向相反,相消为零。
By symmetry, two equal same-sign charges push the midpoint field in exactly opposite directions, so they cancel. Net field at the midpoint is zero.由对称性,两个等量同号电荷在中点产生的场恰好方向相反,相消为零。中点合场为零。
Going deeper — the electric dipole field pattern深入 — 电偶极子的场图案

A dipole is a pair of equal and opposite charges $+q$ and $-q$ separated by distance $d$. Its field has a characteristic pattern: along the axis between the charges, $\vec{E}$ points from $+q$ to $-q$ (reinforcing at the midpoint as shown above). Perpendicular to the axis, the field curves around from the positive end to the negative end. Far from a dipole, the field falls off as $1/r^3$ rather than the $1/r^2$ of a monopole — cancellation between the two charges makes the far field weaker than a single charge of the same magnitude would produce. NGSS HS-PS3-5 uses the dipole as the canonical "two charges of opposite polarity" model. Many molecules (like water) are permanent dipoles, and the interaction of molecular dipoles underlies chemistry, from hydrogen bonding to protein folding.偶极子是一对相距 $d$ 的等量异号电荷 $+q$ 与 $-q$。其场有特征图案:沿两电荷之间的轴,$\vec{E}$ 从 $+q$ 指向 $-q$(如上所示在中点叠加增强)。垂直于轴的方向,场线从正端弯向负端。在远离偶极子处,场按 $1/r^3$ 衰减,而非单极子的 $1/r^2$ — 两电荷之间的相消使远处的场弱于同大小单个电荷所产生的场。NGSS HS-PS3-5 以偶极子作为"两个极性相反的电荷"的典型模型。许多分子(如水)是永久偶极子,分子偶极子的相互作用是化学的基础,从氢键到蛋白质折叠皆如此。


Electric Potential and Voltage: Energy per Charge电势与电压:单位电荷的能量

Potential is the scalar companion to the vector field.电势是矢量场的标量伴随量。 $$ V = \frac{U}{q} = k\frac{Q}{r} \quad \text{(point charge, SI: volts V = J/C)} $$ $$ \Delta V = V_B - V_A = -\frac{W_{A\to B}}{q} = \frac{\Delta U}{q} \qquad E = \frac{V}{d}\ \text{(uniform field)} $$
  • $V$ is a scalar.$V$ 是标量。 Unlike $\vec{E}$, potential has no direction. Contributions from multiple charges add algebraically (with signs, not as vectors): $V_{\text{total}} = V_1 + V_2 + \cdots$与 $\vec{E}$ 不同,电势没有方向。多个电荷的贡献代数相加(带符号,不是矢量):$V_{\text{total}} = V_1 + V_2 + \cdots$
  • Voltage $\Delta V$.电压 $\Delta V$。 The potential difference (voltage) between two points is the work done per unit charge by the electric field in moving a positive test charge from $A$ to $B$. A positive charge moves spontaneously from high $V$ to low $V$ (like rolling downhill).两点间的电势差(电压)是电场把正试验电荷从 $A$ 移到 $B$ 所做的功除以电荷量。正电荷自发地从高 $V$ 移向低 $V$(如同下坡滚动)。
  • Energy gained by a charge.电荷获得的能量。 $\Delta U = q\,\Delta V$. An electron (charge $-e$) accelerated through $\Delta V = 100$ V gains kinetic energy $|q\,\Delta V| = 100$ eV.$\Delta U = q\,\Delta V$。通过 $\Delta V = 100$ V 加速的电子(电荷 $-e$)获得动能 $|q\,\Delta V| = 100$ eV。
Alberta 30–B2.4k: "define electric potential difference as a change in electric potential energy per unit of charge." BC Physics 12: "electric potential energy, electric potential, and electric potential difference."阿尔伯塔 30–B2.4k:"定义电势差为单位电荷电势能的变化。" BC Physics 12:"电势能、电势与电势差。"
Worked Example 5 · Potential from a point charge; energy to move a charge例题 5 · 点电荷的电势;移动电荷所需的能量

A point charge $Q = +8.0\ \mu\text{C}$ is fixed. (a) Find the electric potential at a point $A$ that is $0.60$ m from $Q$. (b) How much work must an external agent do to move a charge $q = +2.0\ \mu\text{C}$ from very far away (where $V \approx 0$) to point $A$?点电荷 $Q = +8.0\ \mu\text{C}$ 固定不动。(a) 求距 $Q$ 为 $0.60$ m 的点 $A$ 处的电势。(b) 外部施力将电荷 $q = +2.0\ \mu\text{C}$ 从远处($V \approx 0$)移到点 $A$,需做多少功?

(a) Potential at A.(a) A 处的电势。

$$ V_A = k\frac{Q}{r} = (8.99\times10^9)\frac{8.0\times10^{-6}}{0.60} \approx 1.20\times10^5\ \text{V.} $$

(b) Work done to bring $q$ to A.(b) 将 $q$ 带到 A 所做的功。 The work equals the change in potential energy: $W = \Delta U = q(V_A - V_\infty) = q\,V_A$ (since $V_\infty = 0$):功等于电势能的变化:$W = \Delta U = q(V_A - V_\infty) = q\,V_A$(因为 $V_\infty = 0$):

$$ W = (2.0\times10^{-6})(1.20\times10^5) = 0.24\ \text{J.} $$

Physical meaning.物理意义。 Bringing a positive charge toward another positive charge requires work against the repulsive force. The $0.24$ J is stored as electric potential energy in the two-charge system.把一个正电荷带向另一个正电荷需要克服斥力做功。这 $0.24$ J 以两电荷系统的电势能形式储存。

An electron (charge $-1.60\times10^{-19}$ C) is accelerated from rest through a potential difference of $\Delta V = 500$ V. What kinetic energy does it gain?一个电子(电荷 $-1.60\times10^{-19}$ C)从静止开始经过 $\Delta V = 500$ V 的电势差加速。它获得多少动能?
§5 · Q1
$-8.0\times10^{-17}$ J (loses energy)
$500$ J
$8.0\times10^{-17}$ J
$3.2\times10^{-16}$ J
$|\Delta KE| = |q||\Delta V| = (1.60\times10^{-19})(500) = 8.0\times10^{-17}$ J. The electron gains this as kinetic energy (it moves opposite to $\vec{E}$, which is the direction that lowers its potential energy since it is negative).$|\Delta KE| = |q||\Delta V| = (1.60\times10^{-19})(500) = 8.0\times10^{-17}$ J。电子获得这些动能(它向 $\vec{E}$ 相反的方向运动,由于它是负电荷,这会降低其电势能)。
$\Delta KE = |q\,\Delta V|$. Multiply the magnitude of the charge by the voltage. The electron gains kinetic energy regardless of its sign.$\Delta KE = |q\,\Delta V|$。用电荷大小乘以电压。电子无论符号如何都会获得动能。
The electric potential at two points in a uniform field are $V_A = 200$ V and $V_B = 80$ V, and the points are $0.030$ m apart. What is the magnitude of the electric field between them?匀强场中两点的电势分别为 $V_A = 200$ V 与 $V_B = 80$ V,两点相距 $0.030$ m。两点间的电场大小是多少?
§5 · Q2
$2400$ N/C
$4000$ N/C
$6.7$ N/C
$280$ N/C
$E = |\Delta V|/d = |200 - 80|/0.030 = 120/0.030 = 4000$ N/C. The field points from high $V$ (A) to low $V$ (B).$E = |\Delta V|/d = |200 - 80|/0.030 = 120/0.030 = 4000$ N/C。场从高电势(A)指向低电势(B)。
In a uniform field $E = \Delta V / d$. Divide the potential difference by the distance.在匀强场中 $E = \Delta V / d$。用电势差除以距离。
Going deeper — equipotential surfaces and the relation $E = -dV/dr$深入 — 等势面与关系 $E = -dV/dr$

An equipotential surface is a surface on which $V$ is constant. No work is done in moving a charge along an equipotential surface; all the work is done when crossing between equipotentials. Key facts: (1) equipotential surfaces are always perpendicular to field lines, (2) for a point charge they are spheres, (3) for a uniform field (parallel plates) they are planes parallel to the plates. The exact relationship between field and potential is $\vec{E} = -\nabla V$, which in one dimension is $E = -dV/dx$. This means a steep potential gradient (closely spaced equipotentials) means a strong field — the same idea as a steep hill meaning a large gravitational force. AB Physics 30 GO2 (30–B2.5k) requires calculating potential difference in a uniform field; the gradient relationship appears at the AP Physics C and IB Physics HL level.等势面是电势 $V$ 恒定的曲面。在等势面上移动电荷不做功;所有的功都在跨越不同等势面时完成。关键事实:(1) 等势面始终垂直于电场线;(2) 对点电荷,等势面是球面;(3) 对匀强场(平行板),等势面是平行于极板的平面。场与电势的精确关系是 $\vec{E} = -\nabla V$,在一维中为 $E = -dV/dx$。这意味着陡峭的电势梯度(密集的等势面)对应强场 — 与陡坡对应大重力分量的概念相同。AB Physics 30 GO2(30–B2.5k)要求在匀强场中计算电势差;梯度关系出现在 AP Physics C 与 IB Physics HL 水平。


Capacitance: Storing Charge and Energy电容:储存电荷与能量

A capacitor stores charge at a given voltage.电容器在给定电压下储存电荷。 $$ C = \frac{Q}{V} \quad \text{(farads, F = C/V)} \qquad C_{\text{parallel plates}} = \varepsilon_0\frac{A}{d} $$ $$ U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}\frac{Q^2}{C} = \tfrac{1}{2}QV \quad \text{(energy stored)} $$
  • Parallel-plate capacitor.平行板电容器。 Two conducting plates of area $A$ separated by distance $d$. The field between them is uniform ($E = V/d$) and the capacitance is $C = \varepsilon_0 A/d$, where $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m.面积为 $A$、间距为 $d$ 的两块导体板。两板间场均匀($E = V/d$),电容为 $C = \varepsilon_0 A/d$,其中 $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m。
  • Larger $C$ means more charge at the same voltage.$C$ 越大,在相同电压下储存的电荷越多。 Capacitance increases with larger plate area, smaller gap, and higher-permittivity dielectric between the plates.极板面积越大、间距越小、极板间介质的介电常数越高,电容越大。
  • Energy stored.储存的能量。 The electric potential energy stored in a capacitor is $U = \tfrac{1}{2}CV^2$. This energy can be released rapidly (flash lamp) or slowly (filter capacitor in an amplifier).电容器储存的电势能为 $U = \tfrac{1}{2}CV^2$。这些能量可以快速释放(闪光灯)或缓慢释放(放大器中的滤波电容)。
BC Physics 12 "electrostatic dynamics and energy relationships" includes the "cathode ray tube, mass spectrometer, particle accelerator" as applications of potential and capacitance. Alberta 30–B2.9k: "explain, quantitatively, electrical interactions using the law of conservation of energy."BC Physics 12"静电动力学与能量关系"将"阴极射线管、质谱仪、粒子加速器"列为电势与电容的应用。阿尔伯塔 30–B2.9k:"定量地用能量守恒定律解释电相互作用。"
Worked Example 6 · Charging a parallel-plate capacitor例题 6 · 平行板电容器充电

A parallel-plate capacitor has plates of area $A = 0.020\ \text{m}^2$ separated by $d = 2.0\ \text{mm} = 2.0 \times 10^{-3}$ m. (a) Find its capacitance. (b) If connected to a $120$ V source, find the charge stored and the energy stored.一个平行板电容器的极板面积 $A = 0.020\ \text{m}^2$,间距 $d = 2.0\ \text{mm} = 2.0 \times 10^{-3}$ m。(a) 求其电容。(b) 接到 $120$ V 电源后,求储存的电荷量与能量。

(a) Capacitance.(a) 电容。

$$ C = \varepsilon_0\frac{A}{d} = (8.85\times10^{-12})\frac{0.020}{2.0\times10^{-3}} = 8.85\times10^{-11}\ \text{F} \approx 88.5\ \text{pF.} $$

(b) Charge and energy at $V = 120$ V.(b) $V = 120$ V 时的电荷量与能量。

$$ Q = CV = (8.85\times10^{-11})(120) \approx 1.06\times10^{-8}\ \text{C} = 10.6\ \text{nC.} $$ $$ U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(8.85\times10^{-11})(120)^2 \approx 6.4\times10^{-7}\ \text{J.} $$

Note.注意。 The field between the plates is $E = V/d = 120/(2.0\times10^{-3}) = 6.0\times10^4$ N/C, uniform everywhere between the plates. This uniform-field geometry is the bridge between §3, §5, and the capacitance formula.两板间的场强 $E = V/d = 120/(2.0\times10^{-3}) = 6.0\times10^4$ N/C,在两板间处处均匀。这种均匀场几何是 §3、§5 与电容公式之间的桥梁。

A capacitor stores $3.0 \times 10^{-4}$ C of charge when connected to a $150$ V battery. What is its capacitance?一个电容器接到 $150$ V 电池时储存了 $3.0 \times 10^{-4}$ C 的电荷。其电容是多少?
§6 · Q1
$45\ \mu\text{F}$
$500\ \mu\text{F}$
$2.0\ \mu\text{F}$
$0.050\ \mu\text{F}$
$C = Q/V = (3.0\times10^{-4})/(150) = 2.0\times10^{-6}$ F $= 2.0\ \mu\text{F}$.$C = Q/V = (3.0\times10^{-4})/(150) = 2.0\times10^{-6}$ F $= 2.0\ \mu\text{F}$。
$C = Q/V$. Divide the charge by the voltage.$C = Q/V$。用电荷量除以电压。
A $10\ \mu\text{F}$ capacitor is charged to $50$ V. How much energy is stored?一个 $10\ \mu\text{F}$ 的电容器充电到 $50$ V。储存了多少能量?
§6 · Q2
$1.25 \times 10^{-2}$ J
$2.5 \times 10^{-2}$ J
$5.0 \times 10^{-4}$ J
$1.0 \times 10^{-3}$ J
$U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(10\times10^{-6})(50)^2 = \tfrac{1}{2}(10^{-5})(2500) = 1.25\times10^{-2}$ J.$U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(10\times10^{-6})(50)^2 = \tfrac{1}{2}(10^{-5})(2500) = 1.25\times10^{-2}$ J。
Energy stored is $\tfrac{1}{2}CV^2$. Don't forget the factor of $\tfrac{1}{2}$ and to square the voltage.储存的能量为 $\tfrac{1}{2}CV^2$。不要忘记 $\tfrac{1}{2}$ 系数,也要把电压平方。

Problem-Solving Strategy for Electrostatics静电学解题策略

A five-step framework that works for every electrostatics problem.适用于所有静电学题目的五步框架。
  1. Draw and label the diagram.画图并标注。 Mark all charge values with signs, distances, and the point(s) where you need $\vec{E}$, $V$, or $F$.标出所有电荷值(含符号)、距离,以及需要求 $\vec{E}$、$V$ 或 $F$ 的点。
  2. Identify what is asked: force, field, potential, or energy.明确所求:力、场、电势还是能量。 Choose the right formula: $F = kq_1q_2/r^2$; $E = kQ/r^2$; $V = kQ/r$; $U = qV$; $\Delta U = q\,\Delta V$.选择正确公式:$F = kq_1q_2/r^2$;$E = kQ/r^2$;$V = kQ/r$;$U = qV$;$\Delta U = q\,\Delta V$。
  3. If multiple charges contribute, use superposition.若多个电荷有贡献,用叠加原理。 For $\vec{E}$: add components ($x$ and $y$) of each contribution separately, then combine. For $V$: add the signed scalar values directly ($V$ is a scalar).对 $\vec{E}$:分别对每个贡献的分量($x$ 与 $y$)求和,再合成。对 $V$:直接代数相加各带符号的标量值($V$ 是标量)。
  4. Assign directions carefully.仔细确定方向。 For Coulomb force: repulsive if same sign, attractive if opposite. For field at a test point: field from a positive source points away; from a negative source, toward it.对库仑力:同号排斥,异号吸引。对测试点处的场:正源电荷产生的场背向源;负源产生的场指向源。
  5. Check units and reasonableness.检查单位与合理性。 Charge in C, distance in m, force in N, field in N/C or V/m, potential in V, energy in J. Sanity-check: force should decrease if charges move farther apart; potential from a positive charge should be positive and decrease with distance.电荷单位 C、距离 m、力 N、场 N/C 或 V/m、电势 V、能量 J。合理性核验:电荷移远时力减小;正电荷产生的电势为正且随距离减小。
Worked Example 7 · Three-charge force and field例题 7 · 三电荷的力与场

Three charges lie on the $x$-axis: $q_1 = +3.0\ \mu\text{C}$ at $x = 0$, $q_2 = -2.0\ \mu\text{C}$ at $x = 0.30$ m, and a test point $P$ at $x = 0.60$ m. Find the magnitude and direction of the net electric field at $P$.三个电荷在 $x$ 轴上:$q_1 = +3.0\ \mu\text{C}$ 在 $x = 0$,$q_2 = -2.0\ \mu\text{C}$ 在 $x = 0.30$ m,测试点 $P$ 在 $x = 0.60$ m。求 $P$ 处合电场的大小与方向。

Field from $q_1$ at P ($r_1 = 0.60$ m).$q_1$ 在 P 处的场($r_1 = 0.60$ m)。 $q_1$ is positive, so $\vec{E}_1$ points away from $q_1$, i.e. in the $+x$ direction:$q_1$ 为正,故 $\vec{E}_1$ 背向 $q_1$,即 $+x$ 方向:

$$ E_1 = k\frac{|q_1|}{r_1^2} = (8.99\times10^9)\frac{3.0\times10^{-6}}{(0.60)^2} \approx 7.49\times10^4\ \text{N/C in } +x. $$

Field from $q_2$ at P ($r_2 = 0.30$ m).$q_2$ 在 P 处的场($r_2 = 0.30$ m)。 $q_2$ is negative, so $\vec{E}_2$ points toward $q_2$, i.e. in the $-x$ direction:$q_2$ 为负,故 $\vec{E}_2$ 指向 $q_2$,即 $-x$ 方向:

$$ E_2 = k\frac{|q_2|}{r_2^2} = (8.99\times10^9)\frac{2.0\times10^{-6}}{(0.30)^2} \approx 2.00\times10^5\ \text{N/C in } -x. $$

Superposition (components along $x$).叠加(沿 $x$ 的分量)。

$$ E_{\text{net}} = E_1 - E_2 = 7.49\times10^4 - 2.00\times10^5 = -1.25\times10^5\ \text{N/C.} $$

The negative sign means the net field at $P$ points in the $-x$ direction (toward $q_2$ and away from $P$), magnitude $1.25 \times 10^5$ N/C.负号表示 $P$ 处的合场指向 $-x$ 方向(指向 $q_2$),大小为 $1.25 \times 10^5$ N/C。

A proton (charge $+e = +1.60\times10^{-19}$ C, mass $1.67\times10^{-27}$ kg) is placed in a uniform electric field of $E = 5.0\times10^4$ N/C. What is its acceleration?一个质子(电荷 $+e = +1.60\times10^{-19}$ C,质量 $1.67\times10^{-27}$ kg)置于 $E = 5.0\times10^4$ N/C 的匀强电场中。其加速度是多少?
§7 · Q1
$8.0\times10^{-15}\ \text{m/s}^2$
$4.8\times10^{12}\ \text{m/s}^2$
$3.2\times10^{-14}\ \text{m/s}^2$
$1.6\times10^{-14}\ \text{m/s}^2$
$F = qE = (1.60\times10^{-19})(5.0\times10^4) = 8.0\times10^{-15}$ N. Then $a = F/m = 8.0\times10^{-15} / 1.67\times10^{-27} \approx 4.8\times10^{12}\ \text{m/s}^2$.$F = qE = (1.60\times10^{-19})(5.0\times10^4) = 8.0\times10^{-15}$ N。则 $a = F/m = 8.0\times10^{-15} / 1.67\times10^{-27} \approx 4.8\times10^{12}\ \text{m/s}^2$。
Step 1: find the force $F = qE$. Step 2: apply Newton's second law $a = F/m$.第一步:求力 $F = qE$。第二步:用牛顿第二定律 $a = F/m$。
The electric potential at a point due to charge $A$ alone is $+120$ V and due to charge $B$ alone is $-80$ V. What is the total electric potential at that point?由电荷 $A$ 单独在某点产生的电势为 $+120$ V,由电荷 $B$ 单独产生的为 $-80$ V。该点的总电势是多少?
§7 · Q2
$200$ V
$\sqrt{120^2 + 80^2} \approx 144$ V
$+40$ V
$-40$ V
Electric potential is a scalar: contributions add algebraically. $V_{\text{total}} = +120 + (-80) = +40$ V. No vector addition needed.电势是标量:贡献代数相加。$V_{\text{total}} = +120 + (-80) = +40$ V。不需要矢量相加。
Potential is a scalar, not a vector. Add with signs directly: $+120 + (-80) = +40$ V.电势是标量,不是矢量。直接带符号相加:$+120 + (-80) = +40$ V。
Going deeper — a charged particle in a uniform electric field: linking electrostatics to kinematics深入 — 匀强场中的带电粒子:静电学与运动学的联系

A charged particle of mass $m$ and charge $q$ in a uniform field $E$ experiences a constant force $F = qE$, giving it a constant acceleration $a = qE/m$ (Newton's second law). This is exactly the constant-acceleration motion covered in HS Physics Unit 1 (Kinematics) — all the SUVAT equations apply directly. For a particle launched perpendicular to the field (like an electron entering between horizontal plates), the problem splits into two independent 1D motions: constant velocity parallel to the plates and constant acceleration perpendicular to them — a "charged-particle projectile" problem. This is the operating principle of the cathode ray tube (BC Physics 12 elaboration) and the mass spectrometer (AB Physics 30). At AP Physics C and IB Physics HL level, the analysis extends to non-uniform fields using $\vec{F} = q\vec{E}$ and the work-energy theorem $W = q\,\Delta V = \Delta KE$.质量为 $m$、电荷为 $q$ 的带电粒子在匀强场 $E$ 中受到恒力 $F = qE$,获得恒定加速度 $a = qE/m$(牛顿第二定律)。这正是 HS Physics 第 1 单元(运动学)所覆盖的匀加速运动 — 所有 SUVAT 方程都直接适用。对于垂直于场方向射入(例如电子进入水平极板之间)的粒子,问题分解为两道独立的一维运动:平行于极板的匀速运动与垂直于极板的匀加速运动 — 一道"带电粒子抛体"题。这是阴极射线管(BC Physics 12 细化内容)和质谱仪(AB Physics 30)的工作原理。在 AP Physics C 和 IB Physics HL 水平,分析通过 $\vec{F} = q\vec{E}$ 与功能定理 $W = q\,\Delta V = \Delta KE$ 延伸至非均匀场。


Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline解题前的纪律
  • Draw the diagram first.先画图。 Label all charges with signs, all distances, and the target point. Missing a sign on a charge is the single most common error in this unit.标出所有电荷的符号、所有距离,以及目标点。电荷符号遗漏是本单元最常见的错误。
  • Identify the quantity: $F$, $\vec{E}$, $V$, or $U$?明确所求量:$F$、$\vec{E}$、$V$ 还是 $U$? $F$ and $\vec{E}$ are vectors (need direction); $V$ and $U$ are scalars (add algebraically). Confusing vector and scalar superposition is a frequent mistake.$F$ 与 $\vec{E}$ 是矢量(需要方向);$V$ 与 $U$ 是标量(代数相加)。混淆矢量与标量的叠加是常见错误。
  • Convert units early.尽早换算单位。 Distances in m, charges in C ($1\ \mu\text{C} = 10^{-6}$ C, $1\ \text{nC} = 10^{-9}$ C). Never plug $\mu\text{C}$ into Coulomb's law without converting.距离单位 m,电荷单位 C($1\ \mu\text{C} = 10^{-6}$ C,$1\ \text{nC} = 10^{-9}$ C)。代入库仑定律前务必换算,切勿直接用 $\mu\text{C}$。
Key pitfalls (§1–§5)关键陷阱(§1–§5)
  • $\vec{E}$ points toward or away?$\vec{E}$ 指向还是背向? Away from a positive source, toward a negative source. For a negative test charge $q$, the force $\vec{F} = q\vec{E}$ is opposite to $\vec{E}$.背向正源,指向负源。对负试验电荷 $q$,力 $\vec{F} = q\vec{E}$ 与 $\vec{E}$ 反向。
  • Potential is a scalar — add with signs, not vectors.电势是标量 — 带符号相加,不是矢量相加。 $V = kQ/r$ uses the sign of $Q$ directly. A positive charge gives $+V$; a negative charge gives $-V$. Add all contributions numerically.$V = kQ/r$ 直接使用 $Q$ 的符号。正电荷给出 $+V$;负电荷给出 $-V$。将所有贡献数值相加。
  • Inverse-square law: double the distance, quarter the field/force.平方反比定律:距离加倍,场/力变为四分之一。 Do not halve it. $F \propto 1/r^2$ and $E \propto 1/r^2$; only $V \propto 1/r$ (linear in $1/r$).不是减半。$F \propto 1/r^2$,$E \propto 1/r^2$;只有 $V \propto 1/r$(对 $1/r$ 是线性的)。
Superposition in practice (§4 & §7)叠加的实际运用(§4 与 §7)
  • For $\vec{E}$: find $x$ and $y$ components of each contribution, then sum each component separately.对 $\vec{E}$:分别求每个贡献的 $x$ 与 $y$ 分量,再对各分量分别求和。 Only combine into a resultant magnitude and direction at the very end.只在最后一步才合成为合矢量的大小与方向。
  • For $V$: add numbers with signs, no component decomposition needed.对 $V$:带符号相加数值,无需分量分解。 This makes potential calculations faster than field calculations when there are multiple charges.这使得有多个电荷时电势计算比场计算更快。
  • Symmetry cancellation: equal charges of the same sign at symmetric positions produce zero net field at the midpoint (fields cancel), but non-zero total potential (potentials add).对称抵消:等量同号电荷在对称位置时,中点合场为零(场相消),但总电势不为零(电势相加)。
Answer hygiene作答规范
  • Round at the very end.最后一步再四舍五入。 Carry extra digits through intermediate steps; round only the final number to the precision the question asks.中间步骤多留几位;仅在最终答案处按题目要求精度四舍五入。
  • State direction for vector quantities.矢量量要说明方向。 For $\vec{E}$ and $\vec{F}$, a magnitude alone is incomplete. State direction relative to a clearly labelled axis or reference.对 $\vec{E}$ 与 $\vec{F}$,仅给大小是不完整的。说明相对于清晰标注的轴或参考方向的方向。
  • Sanity-check magnitudes.用数量级核验。 Coulomb forces between microcoulomb-scale charges at cm-to-m distances are typically $10^{-3}$ to $10$ N. Field from a $\mu\text{C}$ charge at $0.1$ m is $\sim 9 \times 10^5$ N/C. If your answer is wildly off, check unit conversions first.微库仑量级电荷在厘米至米距离的库仑力通常在 $10^{-3}$ 至 $10$ N 之间。$0.1$ m 处 $\mu\text{C}$ 电荷的场强约为 $9 \times 10^5$ N/C。若答案偏差很大,先检查单位换算。

Flashcards闪卡

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Conservation of charge?电荷守恒?
Charge is never created or destroyed — only transferred. Net charge of an isolated system is constant.电荷既不产生也不消失——只被转移。孤立系统净电荷量恒定。
Coulomb's Law?库仑定律?
$$F = k\frac{|q_1||q_2|}{r^2}$$ $$k = 8.99\times10^9\ \mathrm{N\,m^2/C^2}$$
Direction of Coulomb force?库仑力的方向?
Same sign charges: repulsive. Opposite sign: attractive. Newton's 3rd law pair.同号电荷:排斥。异号电荷:吸引。牛顿第三定律对。
Electric field definition?电场定义?
$$\vec{E} = \frac{\vec{F}}{q_0}$$ force per unit positive test charge; SI: N/C or V/m单位正试验电荷所受的力;单位:N/C 或 V/m
Field from a point charge?点电荷的电场?
$$E = k\frac{|Q|}{r^2}$$ outward from $+Q$, inward toward $-Q$背向 $+Q$,指向 $-Q$
Four field-line rules?电场线四条规则?
Start on $+$, end on $-$; tangent gives $\vec{E}$ direction; density $\propto$ field strength; lines never cross.从 $+$ 出发,终止于 $-$;切线方向即 $\vec{E}$ 方向;密度 $\propto$ 场强;场线从不相交。
Superposition of $\vec{E}$?$\vec{E}$ 的叠加?
Add component-by-component: $\vec{E}_{\text{tot}} = \vec{E}_1 + \vec{E}_2 + \cdots$ (vector sum).逐分量相加:$\vec{E}_{\text{tot}} = \vec{E}_1 + \vec{E}_2 + \cdots$(矢量和)。
Electric potential from a point charge?点电荷的电势?
$$V = k\frac{Q}{r}$$ scalar; $+$ for positive $Q$, $-$ for negative $Q$标量;$Q$ 为正则 $+$,$Q$ 为负则 $-$
Superposition of $V$?$V$ 的叠加?
$V_{\text{tot}} = V_1 + V_2 + \cdots$ — algebraic sum of signed scalars. No components needed.$V_{\text{tot}} = V_1 + V_2 + \cdots$ — 有符号标量的代数和。不需要分量。
Voltage and energy?电压与能量?
$$\Delta U = q\,\Delta V$$ Positive charge gains energy moving to higher $V$; negative charge gains energy moving to lower $V$.正电荷移向高 $V$ 时获得能量;负电荷移向低 $V$ 时获得能量。
Uniform field and voltage?匀强场与电压?
$$E = \frac{V}{d} \quad (\text{SI: N/C or V/m})$$ field points from high $V$ to low $V$场从高 $V$ 指向低 $V$
Capacitance?电容?
$$C = \frac{Q}{V}\ \text{(F = C/V)} \qquad C = \varepsilon_0\frac{A}{d}$$
Energy stored in a capacitor?电容器储存的能量?
$$U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}\frac{Q^2}{C} = \tfrac{1}{2}QV$$
Force on a charge in a field?场中电荷所受的力?
$$F = qE$$ direction: same as $\vec{E}$ for $+q$; opposite for $-q$方向:$+q$ 与 $\vec{E}$ 同向;$-q$ 与 $\vec{E}$ 反向

Practice Quiz综合测验

Two charges $q_1 = +5.0\ \mu\text{C}$ and $q_2 = +5.0\ \mu\text{C}$ are $0.20$ m apart. What is the magnitude of the electric force between them?两电荷 $q_1 = +5.0\ \mu\text{C}$ 与 $q_2 = +5.0\ \mu\text{C}$ 相距 $0.20$ m。它们之间的电力大小是多少?
Q1
$0.56$ N
$5.6$ N
$11.2$ N
$1.1$ N
$F = k|q_1||q_2|/r^2 = (8.99\times10^9)(5\times10^{-6})(5\times10^{-6})/(0.20)^2 = (8.99\times10^9)(25\times10^{-12})/(0.04) \approx 5.6$ N (repulsive).$F = k|q_1||q_2|/r^2 = (8.99\times10^9)(5\times10^{-6})(5\times10^{-6})/(0.20)^2 = (8.99\times10^9)(25\times10^{-12})/(0.04) \approx 5.6$ N(斥力)。
Use $F = k|q_1||q_2|/r^2$. Convert $\mu\text{C}$ to C first, then square the distance.用 $F = k|q_1||q_2|/r^2$。先将 $\mu\text{C}$ 转换为 C,再将距离平方。
A charge $q = -4.0\ \mu\text{C}$ is placed in a region where the electric field is $\vec{E} = 3.0\times10^4$ N/C in the $+x$ direction. What is the force on $q$?电荷 $q = -4.0\ \mu\text{C}$ 置于电场 $\vec{E} = 3.0\times10^4$ N/C($+x$ 方向)的区域中。$q$ 所受的力是多少?
Q2
$0.12$ N in the $-x$ direction沿 $-x$ 方向 $0.12$ N
$0.12$ N in the $+x$ direction沿 $+x$ 方向 $0.12$ N
$1.2\times10^{10}$ N in the $-x$ direction沿 $-x$ 方向 $1.2\times10^{10}$ N
zero
$|F| = |q|E = (4.0\times10^{-6})(3.0\times10^4) = 0.12$ N. Since $q$ is negative, the force is opposite to $\vec{E}$: in the $-x$ direction.$|F| = |q|E = (4.0\times10^{-6})(3.0\times10^4) = 0.12$ N。因为 $q$ 为负,力与 $\vec{E}$ 反向:$-x$ 方向。
$F = qE$. The magnitude is $|q|E$. The direction for a negative charge is opposite to $\vec{E}$.$F = qE$。大小为 $|q|E$。负电荷的方向与 $\vec{E}$ 相反。
A charge $Q = +6.0\ \mu\text{C}$ is fixed. What is the electric potential at a point $0.30$ m from it? 🇨🇦 AB 30–B2.4k电荷 $Q = +6.0\ \mu\text{C}$ 固定不动。距它 $0.30$ m 的点处的电势是多少?🇨🇦 AB 30–B2.4k
Q3
$5.99\times10^5$ N/C
$-1.8\times10^5$ V
$2.0\times10^5$ V
$1.8\times10^5$ V
$V = kQ/r = (8.99\times10^9)(6.0\times10^{-6})/(0.30) = 53940/0.30 \approx 1.80\times10^5$ V. Positive, because $Q$ is positive.$V = kQ/r = (8.99\times10^9)(6.0\times10^{-6})/(0.30) = 53940/0.30 \approx 1.80\times10^5$ V。为正,因为 $Q$ 为正。
$V = kQ/r$, a scalar. Use the signed value of $Q$. The result carries the sign of the source charge.$V = kQ/r$,是标量。使用 $Q$ 的带符号值。结果带源电荷的符号。
Two equal positive charges $+Q$ are $d$ apart. At a point on the perpendicular bisector midway between them, the net electric field is: 🇺🇸 NGSS HS-PS2-4两个等量正电荷 $+Q$ 相距 $d$。在两者之间连线的垂直平分线上的中点,合电场为:🇺🇸 NGSS HS-PS2-4
Q4
$2kQ/(d/2)^2$ along the line joining the charges沿两电荷连线方向的 $2kQ/(d/2)^2$
$kQ/(d/2)^2$ perpendicular to the line垂直于连线方向的 $kQ/(d/2)^2$
Zero
$kQ/d^2$ toward one of the charges指向某个电荷的 $kQ/d^2$
Each $+Q$ pushes the field along the perpendicular bisector. The components along the line joining the charges cancel by symmetry; the components perpendicular to it also cancel since both charges are equal. Net field is zero at the midpoint of the perpendicular bisector.每个 $+Q$ 在垂直平分线上产生的场:沿连线方向的分量由对称性相消;垂直方向的分量也由于两电荷相等而相消。垂直平分线中点处合场为零。
By symmetry, the two equal charges produce fields at the midpoint of the bisector that cancel each other. Net field is zero.由对称性,两等量电荷在平分线中点产生的场相互抵消。合场为零。
How much work is done by the electric field in moving a $+2.0\ \mu\text{C}$ charge from a point at $V = 800$ V to a point at $V = 200$ V?电场将 $+2.0\ \mu\text{C}$ 的电荷从 $V = 800$ V 处移到 $V = 200$ V 处,做了多少功?
Q5
$+1.2\times10^{-3}$ J (field does positive work)
$-1.2\times10^{-3}$ J
$+2.0\times10^{-3}$ J
$0$ J
$W = q\,\Delta V_{\text{initial}\to\text{final}} = q(V_i - V_f) = (2.0\times10^{-6})(800 - 200) = (2.0\times10^{-6})(600) = 1.2\times10^{-3}$ J. The field does positive work on a positive charge moving to lower potential.$W = q(V_i - V_f) = (2.0\times10^{-6})(800 - 200) = (2.0\times10^{-6})(600) = 1.2\times10^{-3}$ J。电场对正电荷移向低电势时做正功。
Work done by the field $= q(V_i - V_f)$. A positive charge moving from high $V$ to low $V$ loses potential energy, so the field does positive work.电场做的功 $= q(V_i - V_f)$。正电荷从高 $V$ 移向低 $V$ 时损失电势能,故电场做正功。
A parallel-plate capacitor has capacitance $C = 50\ \text{pF}$ and is charged to $V = 120$ V. How much charge is on each plate? 🇨🇦 BC Physics 12一平行板电容器电容 $C = 50\ \text{pF}$,充电至 $V = 120$ V。每块极板上的电荷量是多少?🇨🇦 BC Physics 12
Q6
$2.4\ \text{nC}$
$0.42\ \text{nC}$
$6.0\ \text{nC}$
$60\ \text{nC}$
$Q = CV = (50\times10^{-12})(120) = 6.0\times10^{-9}$ C $= 6.0$ nC.$Q = CV = (50\times10^{-12})(120) = 6.0\times10^{-9}$ C $= 6.0$ nC。
$Q = CV$. Convert pF to F first: $50\ \text{pF} = 50\times10^{-12}$ F.$Q = CV$。先将 pF 转换为 F:$50\ \text{pF} = 50\times10^{-12}$ F。
Which of these is correct about electric field lines? 🇺🇸 NGSS HS-PS3-5关于电场线,下列哪项正确?🇺🇸 NGSS HS-PS3-5
Q7
They begin on negative charges and end on positive charges从负电荷出发,终止于正电荷
They can cross where the field is very strong在场很强的地方可以相交
Widely spaced field lines indicate a stronger field间隔宽的场线表示场更强
The tangent at any point gives the direction of $\vec{E}$ at that point任意点处的切线方向给出该点 $\vec{E}$ 的方向
Field lines begin on positive charges and end on negative. They never cross. Dense lines mean stronger field. The tangent at any point gives the direction of $\vec{E}$ — this is true by definition.场线从正电荷出发,终止于负电荷。场线从不相交。线越密场越强。任意点的切线方向给出该点 $\vec{E}$ 的方向——这是定义。
Field lines start on positive, end on negative. They never cross. Dense means strong. Only the tangent-gives-direction statement is always correct.场线从正出发,终止于负。从不相交。线密为强。只有切线给方向这句话总是正确的。
A uniform electric field of $E = 2.0\times10^4$ V/m exists between two parallel plates that are $0.050$ m apart. What is the potential difference between the plates? 🇨🇦 AB 30–B2.5k两平行板间距 $0.050$ m,存在 $E = 2.0\times10^4$ V/m 的匀强场。两板间的电势差是多少?🇨🇦 AB 30–B2.5k
Q8
$4.0\times10^5$ V
$1000$ V
$4.0\times10^2$ V
$2.5\times10^{-6}$ V
$\Delta V = E \times d = (2.0\times10^4)(0.050) = 1000$ V.$\Delta V = E \times d = (2.0\times10^4)(0.050) = 1000$ V。
$E = \Delta V/d$, so $\Delta V = Ed$. Multiply field strength by the plate separation.$E = \Delta V/d$,故 $\Delta V = Ed$。用场强乘以板间距。
A $20\ \mu\text{F}$ capacitor is charged to $100$ V. How much energy is stored? 🇨🇦 BC Physics 12 / AB 30–B2.9k一个 $20\ \mu\text{F}$ 的电容器充电到 $100$ V。储存了多少能量?🇨🇦 BC Physics 12 / AB 30–B2.9k
Q9
$0.10$ J
$0.20$ J
$2.0\times10^{-3}$ J
$1.0\times10^{-3}$ J
$U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(20\times10^{-6})(100)^2 = \tfrac{1}{2}(20\times10^{-6})(10^4) = \tfrac{1}{2}(0.20) = 0.10$ J.$U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(20\times10^{-6})(100)^2 = \tfrac{1}{2}(20\times10^{-6})(10^4) = \tfrac{1}{2}(0.20) = 0.10$ J。
$U = \tfrac{1}{2}CV^2$. Square the voltage before multiplying, and don't forget the $\tfrac{1}{2}$.$U = \tfrac{1}{2}CV^2$。先将电压平方再相乘,不要忘记 $\tfrac{1}{2}$。
Compare the electrostatic force (Coulomb) to the gravitational force for two protons $0.10$ nm apart. The ratio $F_e/F_g$ is approximately: 🇨🇦 AB 30–B1.8k / NGSS HS-PS2-4比较相距 $0.10$ nm 的两质子所受的静电力(库仑力)与引力。比值 $F_e/F_g$ 约为:🇨🇦 AB 30–B1.8k / NGSS HS-PS2-4
Q10
about $1$ (they are similar in magnitude)
about $10^{3}$
about $10^{18}$
about $10^{36}$
$F_e/F_g = (ke^2)/(Gm_p^2) \approx (8.99\times10^9)(1.60\times10^{-19})^2 / [(6.67\times10^{-11})(1.67\times10^{-27})^2] \approx 10^{36}$. Electrostatics utterly dominates at atomic scales.$F_e/F_g = (ke^2)/(Gm_p^2) \approx (8.99\times10^9)(1.60\times10^{-19})^2 / [(6.67\times10^{-11})(1.67\times10^{-27})^2] \approx 10^{36}$。在原子尺度上静电力完全主导。
Both force laws have the same $1/r^2$ dependence so they cancel in the ratio. The ratio reduces to constants: $ke^2/(Gm_p^2) \approx 10^{36}$. Electrostatics is enormously stronger than gravity at atomic scales.两条力的定律有相同的 $1/r^2$ 关系,因此在比值中抵消。比值归结为常数:$ke^2/(Gm_p^2) \approx 10^{36}$。在原子尺度上静电力远强于重力。

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成,再勾选每一项。

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What This Feeds Into本单元的去向

Electrostatics is the field-and-energy foundation for every later electricity topic. The charge/field/potential framework you built here reappears immediately in Unit 9 (Current Electricity and Circuits), where current is the rate of charge flow and voltage is the potential difference driving it. Unit 10 (Magnetism and Electromagnetic Induction) extends the field concept to magnetic fields. No AP E&M unit is currently in this repo, but the IB Physics HL and AP Physics C: E&M feeder paths are described below.静电学是后续所有电学主题的场与能量基础。你在此建立的电荷/场/电势框架立即出现在第 9 单元(电流与电路)中,那里电流是电荷流动的速率,电压是驱动它的电势差。第 10 单元(磁学与电磁感应)将场的概念扩展到磁场。本仓库目前没有 AP E&M 单元,但下面描述了 IB Physics HL 与 AP Physics C: E&M 的衔接路径。

Within High School Physics — next units.在 HS Physics 内部 — 后续单元。

Unit 9 (Current Electricity and Circuits) directly reuses the charge/voltage/energy framework: current $I = \Delta Q / \Delta t$ is flow of the charge you defined here; voltage $V$ drives it; Ohm's law $V = IR$ and power $P = IV$ are built from the same quantities. Unit 10 (Magnetism) introduces a second type of field created by moving charges; the parallel between electric and magnetic fields is an important cross-unit theme. The capacitor of §6 reappears in RC circuits in Unit 9.第 9 单元(电流与电路)直接复用电荷/电压/能量框架:电流 $I = \Delta Q / \Delta t$ 是此处所定义电荷的流动;电压 $V$ 驱动它;欧姆定律 $V = IR$ 与功率 $P = IV$ 由同样的量构建。第 10 单元(磁学)引入由运动电荷产生的第二种场;电场与磁场的平行是重要的跨单元主题。§6 的电容器在第 9 单元的 RC 电路中再次出现。

Feeder note — AP E&M and IB Physics HL.衔接说明 — AP E&M 与 IB Physics HL。

No AP Physics C: Electricity & Magnetism unit is currently in this repo. When it is built, Coulomb's law, electric field, potential, and capacitance from this guide will be the assumed prior knowledge from the first lecture. AP adds Gauss's law (relating total flux through a closed surface to enclosed charge), conductors in electrostatic equilibrium, and full treatment of capacitors with dielectrics. For IB Physics HL, Topic D (Fields) covers exactly this scope and extends it to gravitational-electric analogies and field-line quantification. The going-deeper boxes in §3 and §5 introduce the Gauss-law and gradient language that these advanced courses use.本仓库目前没有 AP Physics C: E&M 单元。待建成后,本指南的库仑定律、电场、电势与电容将是第一堂课的默认先修知识。AP 在此基础上增加高斯定律(将穿过闭合曲面的总通量与被包围电荷相联系)、静电平衡中的导体,以及含介质电容器的完整处理。对于 IB Physics HL,D 主题(场)正好涵盖此范围,并延伸至引力-电力类比与电场线量化。§3 与 §5 的深入框引入了这些高级课程所使用的高斯定律与梯度语言。