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Practice练习题

Electrostatics and Electric Fields静电学与电场

Practice Questions · AP-Feeder · ON / BC / AB Provincial & Diploma Styles练习题集 · AP 衔接 · 安 / 卑 / 阿省考与毕业考风格

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
Name:姓名：Date:日期：
PART I  ·  SHORT RESPONSE第一部分  ·  短答题AP-style MCQ + ON/BC short answer · 18 marksAP 风格选择题 + 安/卑省考短答 · 共 18 分

Section A · Short ResponseA 部分 · 短答题

Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. For short-answer items, state units in every answer. Use $k = 8.99 \times 10^9\ \text{N m}^2/\text{C}^2$ throughout. No calculator on Q1-Q2; calculator permitted on Q3-Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。短答题每道都要写出单位。全卷取 $k = 8.99 \times 10^9\ \text{N m}^2/\text{C}^2$。Q1-Q2 不可使用计算器；Q3-Q5 可用计算器。

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 glass rod with a net positive charge is brought near a small piece of neutral aluminium foil. Which statement best describes what happens, and why?一根带净正电荷的玻璃棒靠近一小片中性铝箔。下列哪项最能描述发生的现象及其原因？

  1. (A) The foil is repelled, because like charges repel.铝箔被排斥，因为同种电荷相斥。
  2. (B) The foil is attracted, because the rod induces a charge separation in the foil.铝箔被吸引，因为棒在铝箔中感应出电荷分离。
  3. (C) Nothing happens, because the foil is neutral overall.没有任何现象，因为铝箔整体呈中性。
  4. (D) The foil gains net positive charge by conduction without contact.铝箔无接触地通过传导获得净正电荷。
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 conducting sphere carrying $+8.0\ \mu\text{C}$ is touched to an identical neutral conducting sphere, then the two are separated. What is the charge on each sphere afterward?一个带 $+8.0\ \mu\text{C}$ 的导体球接触一个相同的中性导体球，随后两球分开。之后每个球上的电荷量是多少？

  1. (A) $+8.0\ \mu\text{C}$ and与 $0$
  2. (B) $+4.0\ \mu\text{C}$ each各
  3. (C) $+8.0\ \mu\text{C}$ each各
  4. (D) $+16\ \mu\text{C}$ total, doubled总计，翻倍
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]

Charge can be transferred to an object by conduction or by induction.电荷可以通过传导或感应转移到物体上。

(a) State the key difference between charging by conduction and charging by induction.说出传导带电与感应带电之间的关键区别。 [2]
(b) A negatively charged rod is touched to a neutral metal sphere. State the sign of the sphere's final charge.一根带负电的棒接触一个中性金属球。写出该球最终电荷的符号。 [1]
(c) State whether a conductor or an insulator allows charge to move freely through it.说明导体还是绝缘体允许电荷在其内部自由移动。 [1]
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]

Two point charges separated by a distance $r$ experience a Coulomb force of magnitude $F$. If the separation is increased to $3r$ with the charges unchanged, what is the new force magnitude?两个点电荷相距 $r$ 时受到大小为 $F$ 的库仑力。若电荷不变而间距增大到 $3r$，新的力的大小是多少？

  1. (A) $3F$
  2. (B) $F/3$
  3. (C) $F/9$
  4. (D) $9F$
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]

Two point charges, $q_1 = +3.0\ \mu\text{C}$ and $q_2 = +5.0\ \mu\text{C}$, are held $0.20$ m apart in air.两个点电荷 $q_1 = +3.0\ \mu\text{C}$ 与 $q_2 = +5.0\ \mu\text{C}$ 在空气中相距 $0.20$ m。

(a) Calculate the magnitude of the electrostatic force between them.计算它们之间静电力的大小。 [2]
(b) State whether the force is attractive or repulsive, with a reason.说明该力是引力还是斥力，并给出理由。 [2]
(c) State how the magnitude of the force on $q_1$ compares with that on $q_2$.说明 $q_1$ 所受力的大小与 $q_2$ 所受力的大小如何比较。 [1]
PART II  ·  EXTENDED RESPONSE第二部分  ·  简答题AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分

Section B · Extended ResponseB 部分 · 简答题

Show every step of reasoning. Identify the formula used before substituting values. State units in every final answer. Use $k = 8.99 \times 10^9\ \text{N m}^2/\text{C}^2$, $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$, $e = 1.60 \times 10^{-19}\ \text{C}$, $m_e = 9.11 \times 10^{-31}\ \text{kg}$. Choose and state your sign convention (positive direction) at the start of each question. Calculator permitted on Q6-Q9.每一步推理都要写出。在代入数值前先注明所用公式。每个最终答案都要写单位。取 $k = 8.99 \times 10^9\ \text{N m}^2/\text{C}^2$，$\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$，$e = 1.60 \times 10^{-19}\ \text{C}$，$m_e = 9.11 \times 10^{-31}\ \text{kg}$。每题开头请选择并说明正方向约定。Q6-Q9 可用计算器。

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]

A source charge $Q = +6.0\ \mu\text{C}$ is fixed in place. Point $P$ lies $0.30$ m from $Q$.源电荷 $Q = +6.0\ \mu\text{C}$ 固定不动。点 $P$ 距 $Q$ 为 $0.30$ m。

(a) Calculate the magnitude of the electric field at $P$.计算 $P$ 处电场的大小。 [2]
(b) State the direction of the field at $P$, with a reason.说明 $P$ 处场的方向，并给出理由。 [2]
(c) A charge $q = -2.0\ \mu\text{C}$ is now placed at $P$. Calculate the magnitude of the force on it.现在把电荷 $q = -2.0\ \mu\text{C}$ 放在 $P$ 处。计算它所受力的大小。 [2]
(d) State the direction of the force on $q$, and explain why it differs from the field direction.说明 $q$ 所受力的方向，并解释为何它与场的方向不同。 [2]
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]

A point charge $Q = +5.0\ \mu\text{C}$ is fixed. Point $A$ lies $0.50$ m from $Q$. Take the potential to be zero infinitely far away.点电荷 $Q = +5.0\ \mu\text{C}$ 固定不动。点 $A$ 距 $Q$ 为 $0.50$ m。取无穷远处电势为零。

(a) Calculate the electric potential at $A$.计算 $A$ 处的电势。 [3]
(b) Calculate the work an external agent must do to bring a charge $q = +3.0\ \mu\text{C}$ from far away to $A$.计算外部施力将电荷 $q = +3.0\ \mu\text{C}$ 从远处带到 $A$ 所需做的功。 [2]
(c) State whether this work is positive or negative, and explain physically.说明该功为正还是为负，并从物理上解释。 [2]
(d) State whether electric potential is a scalar or a vector.说明电势是标量还是矢量。 [1]
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]

Charge $+Q = +5.0\ \mu\text{C}$ is at position $x = 0$, and charge $-Q = -5.0\ \mu\text{C}$ is at $x = 0.40$ m (a dipole). Consider the midpoint $M$ at $x = 0.20$ m.电荷 $+Q = +5.0\ \mu\text{C}$ 在 $x = 0$，电荷 $-Q = -5.0\ \mu\text{C}$ 在 $x = 0.40$ m（构成偶极子）。考虑中点 $M$（$x = 0.20$ m）。

(a) Calculate the magnitude of the field at $M$ due to each charge separately.分别计算每个电荷在 $M$ 处产生的场的大小。 [3]
(b) Using superposition, find the magnitude and direction of the net field at $M$.用叠加原理求 $M$ 处净场的大小与方向。 [3]
(c) Explain why the two contributions add rather than cancel at $M$.解释为何两个贡献在 $M$ 处相加而非相消。 [2]
(d) State one rule that explains why field lines never cross.写出一条解释电场线为何从不相交的规则。 [1]
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]

Two parallel plates are separated by $d = 0.020$ m and connected to a $200$ V supply, producing a uniform field. An electron (charge $-e$, mass $m_e$) starts from rest at the negative plate and accelerates to the positive plate.两块平行板相距 $d = 0.020$ m，接到 $200$ V 电源，产生匀强电场。一个电子（电荷 $-e$，质量 $m_e$）从负板由静止出发，加速到正板。

(a) Calculate the magnitude of the uniform field between the plates.计算两板间匀强场的大小。 [2]
(b) Calculate the kinetic energy the electron gains crossing the full gap.计算电子穿越整个间隙获得的动能。 [3]
(c) Calculate the speed of the electron as it reaches the positive plate.计算电子到达正板时的速率。 [3]
(d) Explain why the electron accelerates toward the positive plate even though the field points from positive to negative.解释为何电子向正板加速，尽管场是从正板指向负板的。 [2]
PART III  ·  MODELING / APPLIED第三部分  ·  建模与应用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

Define symbols (with units) at the start of each question. State the formula used before substituting. Use $k = 8.99 \times 10^9\ \text{N m}^2/\text{C}^2$ and $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义符号（含单位）。代入数值前先写出所用公式。取 $k = 8.99 \times 10^9\ \text{N m}^2/\text{C}^2$ 与 $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。

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]

A camera flash uses a parallel-plate capacitor with plates of area $A = 0.015\ \text{m}^2$ separated by $d = 1.5\ \text{mm} = 1.5 \times 10^{-3}$ m, charged by a $90$ V supply.一台相机闪光灯使用平行板电容器，极板面积 $A = 0.015\ \text{m}^2$，间距 $d = 1.5\ \text{mm} = 1.5 \times 10^{-3}$ m，由 $90$ V 电源充电。

(a) Calculate the capacitance.计算电容。 [3]
(b) Calculate the charge stored at $90$ V.计算 $90$ V 时储存的电荷量。 [2]
(c) Calculate the energy stored.计算储存的能量。 [2]
(d) Calculate the magnitude of the uniform field between the plates.计算两板间匀强场的大小。 [2]
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]

In a lab demonstration, two small charged spheres carry $q_1 = +8.0\ \mu\text{C}$ and $q_2 = -3.0\ \mu\text{C}$ and are held $0.10$ m apart.在一次实验演示中，两个小带电球分别带 $q_1 = +8.0\ \mu\text{C}$ 与 $q_2 = -3.0\ \mu\text{C}$，相距 $0.10$ m。

(a) Calculate the magnitude of the force between the spheres.计算两球之间力的大小。 [3]
(b) State whether the force is attractive or repulsive.说明该力是引力还是斥力。 [2]
(c) If the separation is halved to $0.050$ m, state the new force magnitude as a multiple of the original.若间距减半至 $0.050$ m，写出新的力的大小为原来的几倍。 [2]
(d) Compare Coulomb's law with Newton's law of gravitation in one respect.从一个方面比较库仑定律与牛顿万有引力定律。 [2]
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]

Charge $q_1 = +9.0\ \mu\text{C}$ is at $x = 0$ and charge $q_2 = +4.0\ \mu\text{C}$ is at $x = 0.50$ m. A point $P$ on the line between them has zero net electric field.电荷 $q_1 = +9.0\ \mu\text{C}$ 在 $x = 0$，电荷 $q_2 = +4.0\ \mu\text{C}$ 在 $x = 0.50$ m。它们连线之间存在一点 $P$，净电场为零。

(a) Explain why a null point must lie between the two charges, not outside them.解释为何零点必定位于两电荷之间，而非两侧外部。 [2]
(b) Set up the equation that locates $P$ by equating the two field magnitudes.通过令两个场强相等，列出确定 $P$ 位置的方程。 [3]
(c) Solve for the position $x$ of $P$ measured from $q_1$.求出 $P$ 距 $q_1$ 的位置坐标 $x$。 [3]
(d) State whether the null point is closer to the larger or smaller charge, and why.说明零点更靠近较大电荷还是较小电荷，并说明原因。 [2]
🇺🇸 US NGSS美国 NGSSHS-PS2-4 · HS-PS3-5
🇨🇦 Ontario安大略SPH4U Strand D · D2 · D3
🇨🇦 British Columbia不列颠哥伦比亚Physics 12: electric field, Coulomb's law, potential, energy物理 12：电场、库仑定律、电势、能量
🇨🇦 Alberta阿尔伯塔Physics 30 Unit B · 30-B1.6k · 30-B2.4k · 30-B2.6k

Full Syllabus Map lives in ../Study Guides/Unit_8_Electrostatics_and_Electric_Fields.html. Note: NGSS HS-PS2-4 limits assessment to two-object systems; the three-charge null-point question (Q12) sits above the NGSS-assessed floor but is core for AP-feeder / BC Physics 12 / AB Physics 30.完整大纲对照表见 ../Study Guides/Unit_8_Electrostatics_and_Electric_Fields.html。注：NGSS HS-PS2-4 仅考查双物体系统；三电荷零点题（Q12）超出 NGSS 考查范围，但为 AP 衔接 / 卑诗物理 12 / 阿省物理 30 的核心内容。