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Electric Fields — 题型分析

Question Type 1: Definitions (Electric Field / Potential)​

如何识别​

题目包含 "Define electric field strength" 或 "Define electric potential at a point"。

标准解题方法​

解题要点
  • Electric field strength: force per unit positive charge
  • Electric potential: work done per unit positive charge in bringing a small test charge from infinity to the point

评分标准​

评分模式
  • B1/B1: 每题 2 分，包含两个关键要素
  • 电场强度：force / per unit charge / positive charge
  • 电势：work done / per unit charge / from infinity

完整原题​

Example 1 — 9702/41/O/N/20 Q5(a) (2 marks):

Define electric potential at a point.

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MS:

  • M1: work done per unit charge
  • A1: (work done on charge) moving positive charge from infinity

Example 2 — 9702/41/M/J/23 Q1(a)(ii) (1 mark):

Define electric field.

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MS:

  • B1: force per unit positive charge

常见陷阱​

常见陷阱
  • 电场强度定义必须提到 "positive charge"
  • 电势必须提到 "from infinity"（参考点在无穷远）
  • 不要只说 "work done per unit charge"——必须说明是 "from infinity"

Question Type 2: Coulomb's Law Calculations​

如何识别​

题目给出两个点电荷的电量和距离，求电场力。

标准解题方法​

解题步骤
  1. 写 Coulomb's law: F=Q1Q24πϵ0r2F = \frac{Q_1 Q_2}{4\pi \epsilon_0 r^2}
  2. 代入数值（注意 rr 是距离，单位 m）
  3. 计算 FF
  4. 方向判断：同号相斥，异号相吸

评分标准​

评分模式
  • C1: 写出 F=Q1Q2/(4πϵ0r2)F = Q_1 Q_2 / (4\pi \epsilon_0 r^2)
  • C1: 正确代入（包括 14πϵ0=8.99×109\frac{1}{4\pi \epsilon_0} = 8.99 \times 10^9）
  • A1: 正确答案（含单位）

完整原题​

Example 1 — 9702/41/O/N/20 Q5(b)(i) (3 marks):

Two point charges A (+2.0 × 10−9^{-9} C) and B are separated by 12.0 cm. Point P lies on the line joining them. The variation of electric potential V with distance x from charge A is shown in Fig. 5.2. Use Fig. 5.2 to determine the charge of B.

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MS:

  • C1: 2.0×10−94πϵ0(4.0×10−2)+Q4πϵ0(8.0×10−2)=0\frac{2.0 \times 10^{-9}}{4\pi \epsilon_0 (4.0 \times 10^{-2})} + \frac{Q}{4\pi \epsilon_0 (8.0 \times 10^{-2})} = 0
  • A1: Q=4.0×10−9Q = 4.0 \times 10^{-9} C
  • B1: QQ given with negative sign

常见陷阱​

常见陷阱
  • 距离必须用米（m），不是 cm
  • 不要忘记 14πϵ0=8.99×109\frac{1}{4\pi \epsilon_0} = 8.99 \times 10^9 m F−1^{-1}
  • 同号电荷 FF 为正（斥力），异号为负（引力）——但在只求大小时取绝对值

Question Type 3: Electric Field and Potential of Point Charges​

如何识别​

题目要求计算点电荷周围的电场强度或电势，或画 EE-rr / VV-rr 图。

标准解题方法​

解题步骤
  1. 电场强度：E=Q/(4πϵ0r2)E = Q / (4\pi \epsilon_0 r^2)
    • EE 是矢量，方向：正电荷向外，负电荷向内
  2. 电势：V=Q/(4πϵ0r)V = Q / (4\pi \epsilon_0 r)
    • VV 是标量，正电荷为正，负电荷为负
  3. 多个电荷的叠加：
    • EE 是矢量叠加（平行四边形法则）
    • VV 是代数叠加
  4. 对于球形导体，外部场和电势等同于全部电荷集中在球心的点电荷

完整原题​

Example 1 — 9702/41/M/J/20 Q5 (7 marks):

(a) State one similarity and one difference between the fields produced by an isolated point charge and an isolated point mass. (b) An isolated solid metal sphere A of radius R has charge +Q. Point P is distance 2R from the surface. Determine an expression for E at P. (c) A second sphere B with charge –Q is placed near A, centres separated by 6R. Point P lies midway. Explain why: (i) magnitude of E at P is given by the sum of magnitudes (ii) E at P is not equal to 2E (from part b) in practice.

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MS:

  • 5(a) B1: both obey inverse square law / both radial
  • 5(a) B1: electric field can be attractive or repulsive / gravitational field only attractive
  • 5(b) A1: distance from centre =3R= 3R
  • 5(b) A1: E=Q/(4πϵ0(3R)2)E = Q / (4\pi \epsilon_0 (3R)^2)
  • 5(c)(i) B1: fields are in the same direction at P (so magnitudes add)
  • 5(c)(ii) B1: charge distribution on each sphere is affected by the other
  • 5(c)(ii) B1: they are not point charges / charges not uniformly distributed

Example 2 — 9702/41/M/J/21 Q6(a) (3 marks):

An isolated metal sphere of radius r is charged so that E at its surface is E0E_0. On Fig. 6.1, sketch the variation of E with distance x from the centre from x=0x = 0 to x=3rx = 3r.

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MS:

  • B1: E=0E = 0 for x<rx < r
  • B1: E∝1/x2E \propto 1/x^2 for x>rx > r
  • B1: curve starts at E=E0E = E_0 when x=rx = r

常见陷阱​

常见陷阱
  • E=Q/(4πϵ0r2)E = Q/(4\pi \epsilon_0 r^2) 中的 rr 是从球心算起（不是从表面）
  • 金属球内部电场强度为零（导体静电平衡）
  • EE-rr 图：0<r<R0 < r < R 为 0，r>Rr > R 为 1/r21/r^2 衰减
  • VV-rr 图：0<r<R0 < r < R 为常数（等势体），r>Rr > R 为 1/r1/r 衰减

Question Type 4: Uniform Electric Field (Parallel Plates)​

如何识别​

题目涉及平行板电容器、匀强电场、带电粒子在板间运动。

标准解题方法​

解题步骤
  1. E=ΔV/ΔdE = \Delta V / \Delta d（匀强电场）
  2. 带电粒子在电场中受力：F=qEF = qE
  3. 粒子的加速度：a=F/m=qE/ma = F/m = qE/m
  4. 运动分析：
    • 沿电场方向：匀加速直线运动
    • 垂直电场方向：匀速运动
    • 轨迹：抛物线

完整原题​

Example — 9702/41/O/N/22 Q4 (12 marks):

(a) State what is indicated by the direction of an electric field line. (b) Parallel metal plates with p.d. 2400 V, separation 4.6 cm. (i) Draw five lines to represent the electric field between the plates. (ii) Calculate the electric field strength. (c) A proton enters the region between the plates from the left. (i) Draw the path of the proton. (ii) A helium nucleus enters along the same path at same speed. Compare final speeds.

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MS:

  • 4(a) B1: direction of force on a positive charge
  • 4(a) B1: placed at that point in the field
  • 4(b)(i) B1: equally spaced parallel lines (between plates)
  • 4(b)(i) B1: arrows from + to –
  • 4(b)(i) B1: lines straight and not touching plates
  • 4(b)(ii) C1: E=V/d=2400/0.046E = V/d = 2400 / 0.046
  • 4(b)(ii) A1: =5.2×104= 5.2 \times 10^4 N C−1^{-1}
  • 4(c)(i) B1: curved path (parabolic) towards negative plate
  • 4(c)(i) B1: straight line after leaving field
  • 4(c)(ii) B1: helium nucleus has smaller final speed
  • 4(c)(ii) B1: helium has larger mass (so smaller acceleration)
  • 4(c)(ii) B1: charge/mass ratio is smaller

常见陷阱​

常见陷阱
  • 匀强电场中 E=V/dE = V/d，dd 是板间距离（不是沿路径距离）
  • 带电粒子在匀强电场中的轨迹是抛物线（不是直线或圆）
  • α\alpha 粒子（24^4_2He）的 q/mq/m 是质子的 1/21/2（不是相同）