Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus D2.1 to D2.6考纲 D2.1 至 D2.6PHYSICS HL
Charges $q_1 = +2.0\ \mathrm{nC}$, $q_2 = +8.0\ \mathrm{nC}$, separation $3.0\ \mathrm{cm}$. (a) force on $q_1$ and its nature; (b) new force when separation is halved to $1.5\ \mathrm{cm}$.电荷 $q_1 = +2.0\ \mathrm{nC}$、$q_2 = +8.0\ \mathrm{nC}$，间距 $3.0\ \mathrm{cm}$。(a) $q_1$ 受力及性质；(b) 间距减半到 $1.5\ \mathrm{cm}$ 时的新力。
Use Coulomb's law $F = \dfrac{k q_1 q_2}{r^2}$ with $r = 0.030\ \mathrm{m}$: (M1)
$$ F = \frac{(8.99\times 10^{9})(2.0\times 10^{-9})(8.0\times 10^{-9})}{(0.030)^2} = \frac{1.438\times 10^{-7}}{9.0\times 10^{-4}} \approx 1.6\times 10^{-4}\ \mathrm{N}. $$Both charges are positive, so the product $q_1 q_2 > 0$: the force is repulsive. (A1)
Coulomb's law is inverse-square, $F \propto 1/r^2$. Halving $r$ multiplies $1/r^2$ by $2^2 = 4$. (R1)
$$ F_{\text{new}} = 4 \times 1.6\times 10^{-4} \approx 6.4\times 10^{-4}\ \mathrm{N}. $$(A1)
用库仑定律 $F = \dfrac{k q_1 q_2}{r^2}$，$r = 0.030\ \mathrm{m}$：(M1)
$$ F = \frac{(8.99\times 10^{9})(2.0\times 10^{-9})(8.0\times 10^{-9})}{(0.030)^2} = \frac{1.438\times 10^{-7}}{9.0\times 10^{-4}} \approx 1.6\times 10^{-4}\ \mathrm{N}. $$两电荷均为正，故乘积 $q_1 q_2 > 0$：力为排斥。(A1)
库仑定律为平方反比，$F \propto 1/r^2$。$r$ 减半使 $1/r^2$ 变为 $2^2 = 4$ 倍。(R1)
$$ F_{\text{new}} = 4 \times 1.6\times 10^{-4} \approx 6.4\times 10^{-4}\ \mathrm{N}. $$(A1)
Long straight wire, current west to east, compass directly above. (a) shape of the field lines and direction of $B$ directly above; (b) one feature of a solenoid's interior field, and why magnetic lines form closed loops.长直导线，电流由西向东，磁针在正上方。(a) 磁感线形状及正上方 $B$ 的方向；(b) 螺线管内部场的一个特征，及磁感线为何构成闭合回路。
The field lines of a long straight wire are concentric circles centred on the wire, lying in planes perpendicular to it. (A1)
Apply the right-hand grip rule: the thumb points east (conventional current), so the fingers curl over the top of the wire toward the north. Directly above the wire the field therefore points due north. (A1)
Inside a solenoid the field is approximately uniform (straight, parallel, equally spaced lines along the axis), like the field of a bar magnet's exterior. (B1)
Magnetic field lines always form closed loops because there are no magnetic monopoles for the lines to start or stop on, unlike electric lines which begin and end on charges. (R1)
长直导线的磁感线是以导线为中心的同心圆，位于与导线垂直的平面内。(A1)
用右手螺旋定则：拇指指向东（常规电流），四指便在导线上方向北弯绕。故导线正上方的场指向正北。(A1)
螺线管内部场近似匀强（沿轴线的笔直、平行、等间距的线），如条形磁铁外部的场。(B1)
磁感线总是构成闭合回路，因为没有可供线起止的磁单极子；这与起止于电荷的电场线不同。(R1)
Point charge $Q = +5.0\ \mathrm{nC}$. (a) field strength at $r = 0.10\ \mathrm{m}$ and its direction; (b) force on a $-3.0\ \mathrm{nC}$ test charge there; (c) field at the midpoint between two identical $+5.0\ \mathrm{nC}$ charges.点电荷 $Q = +5.0\ \mathrm{nC}$。(a) $r = 0.10\ \mathrm{m}$ 处的场强及方向；(b) 该处 $-3.0\ \mathrm{nC}$ 检验电荷所受力；(c) 两个相同 $+5.0\ \mathrm{nC}$ 电荷中点处的场。
Use $E = \dfrac{kQ}{r^2}$ with $r = 0.10\ \mathrm{m}$: (M1)
$$ E = \frac{(8.99\times 10^{9})(5.0\times 10^{-9})}{(0.10)^2} = \frac{44.95}{0.010} \approx 4.5\times 10^{3}\ \mathrm{N\,C^{-1}}. $$The field points radially outward, away from the positive charge. (A1)
Use $F = qE$ with magnitudes: (M1)
$$ F = (3.0\times 10^{-9})(4.495\times 10^{3}) \approx 1.3\times 10^{-5}\ \mathrm{N}. $$The test charge is negative, so the force is opposite to $E$: directed radially inward, toward $Q$ (attraction). (A1)
The midpoint is equidistant from two equal positive charges. Each charge produces a field of the same magnitude at that point, but the two fields point in opposite directions (each away from its source). (R1)
The two contributions cancel exactly, so the resultant field strength is $E = 0$. (A1)
用 $E = \dfrac{kQ}{r^2}$，$r = 0.10\ \mathrm{m}$：(M1)
$$ E = \frac{(8.99\times 10^{9})(5.0\times 10^{-9})}{(0.10)^2} = \frac{44.95}{0.010} \approx 4.5\times 10^{3}\ \mathrm{N\,C^{-1}}. $$场沿径向向外，背离正电荷。(A1)
用 $F = qE$（取大小）：(M1)
$$ F = (3.0\times 10^{-9})(4.495\times 10^{3}) \approx 1.3\times 10^{-5}\ \mathrm{N}. $$检验电荷为负，受力与 $E$ 相反：沿径向向内，指向 $Q$（吸引）。(A1)
中点到两个相等正电荷等距。每个电荷在该点产生大小相同的场，但两场方向相反（各自背离其源）。(R1)
两份贡献恰好抵消，故合场强为 $E = 0$。(A1)
Parallel plates $d = 5.0\ \mathrm{cm}$, $V = 250\ \mathrm{V}$; electron between them. (a) field strength and two field-line features; (b) electric force; (c) acceleration and why it is constant everywhere.平行板 $d = 5.0\ \mathrm{cm}$、$V = 250\ \mathrm{V}$；电子在板间。(a) 场强及电场线两个特征；(b) 电力；(c) 加速度及其为何处处恒定。
For the uniform field between plates, $E = \dfrac{V}{d}$ with $d = 0.050\ \mathrm{m}$:
$$ E = \frac{250}{0.050} = 5.0\times 10^{3}\ \mathrm{V\,m^{-1}}. $$(A1)
The field lines are straight, parallel and equally spaced, running from the positive plate to the negative plate. (B1 for any two features)
Use $F = qE$ with $q = e = 1.60\times 10^{-19}\ \mathrm{C}$: (M1)
$$ F = (1.60\times 10^{-19})(5.0\times 10^{3}) = 8.0\times 10^{-16}\ \mathrm{N}. $$(A1)
Use $a = \dfrac{F}{m_e}$ with $m_e = 9.11\times 10^{-31}\ \mathrm{kg}$: (M1)
$$ a = \frac{8.0\times 10^{-16}}{9.11\times 10^{-31}} \approx 8.8\times 10^{14}\ \mathrm{m\,s^{-2}}. $$The field between parallel plates is uniform, so $F = qE$ has the same value everywhere; therefore the acceleration is the same everywhere between the plates. (A1)
板间匀强场用 $E = \dfrac{V}{d}$，$d = 0.050\ \mathrm{m}$：
$$ E = \frac{250}{0.050} = 5.0\times 10^{3}\ \mathrm{V\,m^{-1}}. $$(A1)
电场线笔直、平行、等间距，从正板指向负板。(任写两个特征得 B1)
用 $F = qE$，$q = e = 1.60\times 10^{-19}\ \mathrm{C}$：(M1)
$$ F = (1.60\times 10^{-19})(5.0\times 10^{3}) = 8.0\times 10^{-16}\ \mathrm{N}. $$(A1)
用 $a = \dfrac{F}{m_e}$，$m_e = 9.11\times 10^{-31}\ \mathrm{kg}$：(M1)
$$ a = \frac{8.0\times 10^{-16}}{9.11\times 10^{-31}} \approx 8.8\times 10^{14}\ \mathrm{m\,s^{-2}}. $$板间为匀强场，故 $F = qE$ 处处相同；因此加速度在板间各处相同。(A1)
$Q = +6.0\ \mathrm{nC}$; $A$ at $0.20\ \mathrm{m}$, $B$ at $0.60\ \mathrm{m}$. (a) potentials at $A$ and $B$; (b) work to move $q = +4.0\ \mathrm{nC}$ from $B$ to $A$; (c) sign of the work and the work along an equipotential.$Q = +6.0\ \mathrm{nC}$；$A$ 在 $0.20\ \mathrm{m}$，$B$ 在 $0.60\ \mathrm{m}$。(a) $A$、$B$ 处电势；(b) 把 $q = +4.0\ \mathrm{nC}$ 从 $B$ 移到 $A$ 的功；(c) 功的符号及沿等势面的功。
Use $V_e = \dfrac{kQ}{r}$: (M1)
$$ V_A = \frac{(8.99\times 10^{9})(6.0\times 10^{-9})}{0.20} \approx 270\ \mathrm{V}, \qquad V_B = \frac{(8.99\times 10^{9})(6.0\times 10^{-9})}{0.60} \approx 90\ \mathrm{V}. $$(A1 for both)
Use $W = q\,\Delta V_e = q(V_A - V_B)$: (M1)
$$ W = (4.0\times 10^{-9})(270 - 90) = (4.0\times 10^{-9})(180) \approx 7.2\times 10^{-7}\ \mathrm{J}. $$(A1)
The work is positive: moving a positive charge to a region of higher potential (closer to the positive source) raises its electric potential energy, so an external agent must do positive work against the repulsion. (R1)
Along an equipotential surface $\Delta V_e = 0$, so $W = q\,\Delta V_e = 0$: no work is done. (B1)
用 $V_e = \dfrac{kQ}{r}$：(M1)
$$ V_A = \frac{(8.99\times 10^{9})(6.0\times 10^{-9})}{0.20} \approx 270\ \mathrm{V}, \qquad V_B = \frac{(8.99\times 10^{9})(6.0\times 10^{-9})}{0.60} \approx 90\ \mathrm{V}. $$(两者皆对得 A1)
用 $W = q\,\Delta V_e = q(V_A - V_B)$：(M1)
$$ W = (4.0\times 10^{-9})(270 - 90) = (4.0\times 10^{-9})(180) \approx 7.2\times 10^{-7}\ \mathrm{J}. $$(A1)
功为正：把正电荷移到电势更高处（更靠近正源）会升高其电势能，故外力须逆着斥力做正功。(R1)
沿等势面 $\Delta V_e = 0$，故 $W = q\,\Delta V_e = 0$：不做功。(B1)
Comparing fields. (a) one similarity in the force laws of gravitational and electric fields, and one fundamental difference in the nature of the forces; (b) one difference in the field-line pattern of a magnetic field versus an electric field, linked to the absence of magnetic monopoles.对比各种场。(a) 引力场与电场力定律的一个相似点，及力的性质的一个根本区别；(b) 磁场场线图样与电场的一处不同，并联系磁单极子的缺失。
Similarity: both are inverse-square laws, $F = \dfrac{G m_1 m_2}{r^2}$ and $F = \dfrac{k q_1 q_2}{r^2}$, so each force falls off as $1/r^2$. (B1)
Difference: the gravitational force is always attractive (mass has one sign), whereas the electric force can be attractive or repulsive (charge has two signs). (B1)
Magnetic field lines form continuous closed loops, whereas electric field lines start on positive charges and end on negative charges. (B1)
The loops never start or stop because there are no isolated magnetic poles (no monopoles) for them to begin or end on; cutting a magnet only produces more dipoles. (R1)
相似点：两者都是平方反比律，$F = \dfrac{G m_1 m_2}{r^2}$ 与 $F = \dfrac{k q_1 q_2}{r^2}$，故各力都按 $1/r^2$ 衰减。(B1)
区别：引力恒为吸引（质量只有一种符号），而电力可吸可斥（电荷有两种符号）。(B1)
磁感线构成连续的闭合回路，而电场线起于正电荷、止于负电荷。(B1)
磁感线既不起也不止，因为没有孤立磁极（无磁单极子）供其起止；切开磁铁只会产生更多偶极子。(R1)
$E$ vs $1/r^2$ data for a charged sphere. (a) show $E$ vs $1/r^2$ is linear through the origin and state the gradient; (b) gradient and charge $Q$; (c) percentage uncertainty in $E$ at $1/r^2 = 100$; (d) why $E$ vs $1/r^2$ beats $E$ vs $r$.带电球的 $E$ 对 $1/r^2$ 数据。(a) 证明 $E$ 对 $1/r^2$ 为过原点直线并说明斜率；(b) 斜率与电荷 $Q$；(c) $1/r^2 = 100$ 处 $E$ 的百分比不确定度；(d) 为何 $E$ 对 $1/r^2$ 优于 $E$ 对 $r$。
The radial field of a point charge is $E = \dfrac{kQ}{r^2}$. Rewrite it as $E = kQ\left(\dfrac{1}{r^2}\right)$. (M1)
This has the form $y = mx$ with $y = E$ and $x = 1/r^2$ and no intercept, so a plot of $E$ against $1/r^2$ is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient is $kQ$. (A1)
Read the gradient from two well-separated points, $(25,\,90)$ and $(100,\,360)$: (M1)
$$ \text{gradient} = \frac{360 - 90}{100 - 25} = \frac{270}{75} = 3.6\ \mathrm{N\,m^{2}\,C^{-1}}. $$(A1)
Since gradient $= kQ$: $Q = \dfrac{3.6}{8.99\times 10^{9}} \approx 4.0\times 10^{-10}\ \mathrm{C} = 0.40\ \mathrm{nC}$. (A1)
There $E = 360\ \mathrm{N\,C^{-1}}$ with absolute uncertainty $\pm 10\ \mathrm{N\,C^{-1}}$: (M1)
$$ \frac{10}{360}\times 100\% \approx 2.8\%. $$(A1)
Plotting $E$ against $1/r^2$ gives a straight line, whereas $E$ against $r$ is a curve. (B1)
A straight-line best fit lets the gradient be read across all points, averaging out random scatter, and the through-origin form makes a systematic error show as a non-zero intercept; both are far harder to judge from a curve. (R1)
点电荷的径向场为 $E = \dfrac{kQ}{r^2}$。改写为 $E = kQ\left(\dfrac{1}{r^2}\right)$。(M1)
此式形如 $y = mx$，其中 $y = E$、$x = 1/r^2$、无截距，故 $E$ 对 $1/r^2$ 作图为过原点的直线。(A1)
与 $y = mx$ 比较，斜率为 $kQ$。(A1)
用相距较远的两点 $(25,\,90)$ 与 $(100,\,360)$ 读斜率：(M1)
$$ \text{斜率} = \frac{360 - 90}{100 - 25} = \frac{270}{75} = 3.6\ \mathrm{N\,m^{2}\,C^{-1}}. $$(A1)
因斜率 $= kQ$：$Q = \dfrac{3.6}{8.99\times 10^{9}} \approx 4.0\times 10^{-10}\ \mathrm{C} = 0.40\ \mathrm{nC}$。(A1)
该处 $E = 360\ \mathrm{N\,C^{-1}}$，绝对不确定度 $\pm 10\ \mathrm{N\,C^{-1}}$：(M1)
$$ \frac{10}{360}\times 100\% \approx 2.8\%. $$(A1)
作 $E$ 对 $1/r^2$ 的图得到直线，而 $E$ 对 $r$ 是曲线。(B1)
直线最佳拟合可在所有点上读取斜率、平均掉随机散布，且过原点的形式使系统误差表现为非零截距；这两点在曲线上都更难判断。(R1)
$V_e$ vs $1/r$ data for a charged sphere. (a) show $V_e$ vs $1/r$ is linear through the origin and state the gradient; (b) gradient and charge $Q$; (c) work to move $q = +2.0\ \mathrm{nC}$ from $1/r = 4.0$ to $1/r = 10.0$; (d) how equipotentials sit relative to field lines and what their spacing shows.带电球的 $V_e$ 对 $1/r$ 数据。(a) 证明 $V_e$ 对 $1/r$ 为过原点直线并说明斜率；(b) 斜率与电荷 $Q$；(c) 把 $q = +2.0\ \mathrm{nC}$ 从 $1/r = 4.0$ 移到 $1/r = 10.0$ 的功；(d) 等势面相对电场线的排布及其疏密含义。
The potential of a point charge is $V_e = \dfrac{kQ}{r}$. Rewrite as $V_e = kQ\left(\dfrac{1}{r}\right)$. (M1)
This is $y = mx$ with $y = V_e$ and $x = 1/r$ and no intercept, so a plot of $V_e$ against $1/r$ is a straight line through the origin. (A1)
The gradient is $kQ$. (A1)
Read the gradient from $(4.0,\,36)$ and $(10.0,\,90)$: (M1)
$$ \text{gradient} = \frac{90 - 36}{10.0 - 4.0} = \frac{54}{6.0} = 9.0\ \mathrm{V\,m}. $$(A1)
Since gradient $= kQ$: $Q = \dfrac{9.0}{8.99\times 10^{9}} \approx 1.0\times 10^{-9}\ \mathrm{C} = 1.0\ \mathrm{nC}$. (A1)
From the table, $1/r = 4.0$ gives $V_e = 36\ \mathrm{V}$ and $1/r = 10.0$ gives $V_e = 90\ \mathrm{V}$. Use $W = q\,\Delta V_e$: (M1)
$$ W = (2.0\times 10^{-9})(90 - 36) = (2.0\times 10^{-9})(54) \approx 1.1\times 10^{-7}\ \mathrm{J}. $$(A1)
Around a point charge the equipotentials are concentric spheres (drawn as circles), each crossing the radial field lines at right angles. (B1)
Where the equipotentials (for equal potential steps) are more closely spaced, the potential changes more rapidly over a small distance, so the field is stronger; near the charge they crowd together, far away they spread out. (R1)
点电荷的电势为 $V_e = \dfrac{kQ}{r}$。改写为 $V_e = kQ\left(\dfrac{1}{r}\right)$。(M1)
此为 $y = mx$，其中 $y = V_e$、$x = 1/r$、无截距，故 $V_e$ 对 $1/r$ 作图为过原点的直线。(A1)
斜率为 $kQ$。(A1)
用 $(4.0,\,36)$ 与 $(10.0,\,90)$ 读斜率：(M1)
$$ \text{斜率} = \frac{90 - 36}{10.0 - 4.0} = \frac{54}{6.0} = 9.0\ \mathrm{V\,m}. $$(A1)
因斜率 $= kQ$：$Q = \dfrac{9.0}{8.99\times 10^{9}} \approx 1.0\times 10^{-9}\ \mathrm{C} = 1.0\ \mathrm{nC}$。(A1)
由表，$1/r = 4.0$ 处 $V_e = 36\ \mathrm{V}$，$1/r = 10.0$ 处 $V_e = 90\ \mathrm{V}$。用 $W = q\,\Delta V_e$：(M1)
$$ W = (2.0\times 10^{-9})(90 - 36) = (2.0\times 10^{-9})(54) \approx 1.1\times 10^{-7}\ \mathrm{J}. $$(A1)
点电荷周围的等势面是同心球面（画作圆），各自与径向电场线处处垂直。(B1)
（按等电势间隔画的）等势面越密之处，小距离内电势变化越快，故场越强；近电荷处它们密集，远处则疏散。(R1)
Plates $d = 4.0\ \mathrm{cm}$; droplet $m = 4.0\times 10^{-15}\ \mathrm{kg}$, $q = +8.0\times 10^{-19}\ \mathrm{C}$ floats motionless, upper plate positive. (a) number of excess/missing electrons; (b) field strength from force balance; (c) the pd $V$; (d) initial acceleration after gaining one more electron, and whether suvat applies.板 $d = 4.0\ \mathrm{cm}$；液滴 $m = 4.0\times 10^{-15}\ \mathrm{kg}$、$q = +8.0\times 10^{-19}\ \mathrm{C}$ 静止悬浮，上板为正。(a) 多余/缺失电子数；(b) 由受力平衡求场强；(c) 电势差 $V$；(d) 再获一个电子后的初始加速度及 suvat 是否适用。
The number of elementary charges is $n = \dfrac{q}{e} = \dfrac{8.0\times 10^{-19}}{1.60\times 10^{-19}} = 5$. (M1)
The droplet is positive, so it has lost electrons: it is missing 5 electrons. (A1)
The droplet floats, so the upward electric force balances the downward weight: $qE = mg$. (M1)
$$ E = \frac{mg}{q} = \frac{(4.0\times 10^{-15})(9.81)}{8.0\times 10^{-19}}. $$(M1 for substitution)
$$ E = \frac{3.924\times 10^{-14}}{8.0\times 10^{-19}} \approx 4.9\times 10^{4}\ \mathrm{V\,m^{-1}}. $$(A1)
For the uniform field, $E = \dfrac{V}{d}$, so $V = Ed$ with $d = 0.040\ \mathrm{m}$: (M1)
$$ V = (4.905\times 10^{4})(0.040) \approx 1.96\times 10^{3}\ \mathrm{V} \approx 2.0\times 10^{3}\ \mathrm{V}. $$(A1)
For the upward electric force on a positive charge the field must point upward, so the lower plate is at the higher potential; the stated upper-plate-positive arrangement is consistent only if the droplet sign or field direction is read carefully, and the magnitude of $V$ is $\approx 2.0\ \mathrm{kV}$. (A1)
One more electron makes the charge less positive: the new charge is $q' = q - e = 8.0\times 10^{-19} - 1.60\times 10^{-19} = 6.4\times 10^{-19}\ \mathrm{C}$. (M1)
The field $E$ is unchanged ($V$ fixed), so the new upward electric force is $q'E$, while the weight is still $mg$. The resultant is downward because the upward force has shrunk: (M1)
$$ F_{\text{net}} = mg - q'E = 3.924\times 10^{-14} - (6.4\times 10^{-19})(4.905\times 10^{4}) = 3.924\times 10^{-14} - 3.139\times 10^{-14} = 7.85\times 10^{-15}\ \mathrm{N}. $$$$ a = \frac{F_{\text{net}}}{m} = \frac{7.85\times 10^{-15}}{4.0\times 10^{-15}} \approx 2.0\ \mathrm{m\,s^{-2}} \ \text{(downward)}. $$ (A1)
Both forces are constant (uniform field, fixed charge and mass), so the acceleration is constant and the suvat equations do apply to the subsequent motion. (R1)
元电荷数 $n = \dfrac{q}{e} = \dfrac{8.0\times 10^{-19}}{1.60\times 10^{-19}} = 5$。(M1)
液滴带正电，故失去了电子：缺失 5 个电子。(A1)
液滴悬浮，故向上的电力平衡向下的重力：$qE = mg$。(M1)
$$ E = \frac{mg}{q} = \frac{(4.0\times 10^{-15})(9.81)}{8.0\times 10^{-19}}. $$(代入得 M1)
$$ E = \frac{3.924\times 10^{-14}}{8.0\times 10^{-19}} \approx 4.9\times 10^{4}\ \mathrm{V\,m^{-1}}. $$(A1)
对匀强场 $E = \dfrac{V}{d}$，故 $V = Ed$，$d = 0.040\ \mathrm{m}$：(M1)
$$ V = (4.905\times 10^{4})(0.040) \approx 1.96\times 10^{3}\ \mathrm{V} \approx 2.0\times 10^{3}\ \mathrm{V}. $$(A1)
正电荷受向上电力要求场向上，故下板电势更高；题述上板为正的布置须结合液滴符号与场方向仔细判读，而 $V$ 的大小约为 $2.0\ \mathrm{kV}$。(A1)
多一个电子使电荷正性减弱：新电荷 $q' = q - e = 8.0\times 10^{-19} - 1.60\times 10^{-19} = 6.4\times 10^{-19}\ \mathrm{C}$。(M1)
场 $E$ 不变（$V$ 固定），故新的向上电力为 $q'E$，重力仍为 $mg$。因向上力减小，合力向下：(M1)
$$ F_{\text{net}} = mg - q'E = 3.924\times 10^{-14} - (6.4\times 10^{-19})(4.905\times 10^{4}) = 3.924\times 10^{-14} - 3.139\times 10^{-14} = 7.85\times 10^{-15}\ \mathrm{N}. $$$$ a = \frac{F_{\text{net}}}{m} = \frac{7.85\times 10^{-15}}{4.0\times 10^{-15}} \approx 2.0\ \mathrm{m\,s^{-2}} \ \text{（向下）}. $$ (A1)
两个力都恒定（匀强场、电荷与质量固定），故加速度恒定，suvat 方程适用于其后续运动。(R1)
$q_1 = +3.0\ \mathrm{nC}$, $q_2 = +5.0\ \mathrm{nC}$, separation $r = 6.0\ \mathrm{cm}$. (a) electric PE of the pair and whether external work was needed to assemble it; (b) resultant potential at the midpoint; (c) work by the field around a closed path; (d) new PE when the separation is doubled.$q_1 = +3.0\ \mathrm{nC}$、$q_2 = +5.0\ \mathrm{nC}$，间距 $r = 6.0\ \mathrm{cm}$。(a) 这对电荷的电势能及组装是否需外力做功；(b) 中点处合电势；(c) 场沿闭合路径所做的功；(d) 间距加倍后的新电势能。
Use $E_p = \dfrac{k q_1 q_2}{r}$ with $r = 0.060\ \mathrm{m}$: (M1)
$$ E_p = \frac{(8.99\times 10^{9})(3.0\times 10^{-9})(5.0\times 10^{-9})}{0.060} = \frac{1.3485\times 10^{-7}}{0.060} \approx 2.2\times 10^{-6}\ \mathrm{J}. $$(A1)
The energy is positive, and both charges are positive (they repel), so positive external work was needed to bring them together from infinity against the repulsion. (R1)
The midpoint is $0.030\ \mathrm{m}$ from each charge. Potential is a scalar, so add the contributions: (M1)
$$ V = \frac{k q_1}{0.030} + \frac{k q_2}{0.030} = \frac{k(q_1 + q_2)}{0.030} = \frac{(8.99\times 10^{9})(8.0\times 10^{-9})}{0.030}. $$(M1)
$$ V = \frac{71.92}{0.030} \approx 2.4\times 10^{3}\ \mathrm{V}. $$(A1)
The work done by the electric field over any closed path is zero. (A1)
Work depends only on the change in potential, $W = q\,\Delta V_e$; a closed path returns to its start, so $\Delta V_e = 0$ and hence $W = 0$. The electric field is conservative. (R1)
$E_p = \dfrac{k q_1 q_2}{r} \propto \dfrac{1}{r}$, so doubling $r$ halves the energy: (R1)
$$ E_p' = \tfrac{1}{2}(2.2\times 10^{-6}) \approx 1.1\times 10^{-6}\ \mathrm{J}. $$(A1)
用 $E_p = \dfrac{k q_1 q_2}{r}$，$r = 0.060\ \mathrm{m}$：(M1)
$$ E_p = \frac{(8.99\times 10^{9})(3.0\times 10^{-9})(5.0\times 10^{-9})}{0.060} = \frac{1.3485\times 10^{-7}}{0.060} \approx 2.2\times 10^{-6}\ \mathrm{J}. $$(A1)
能量为正，且两电荷皆正（相斥），故需外力做正功才能克服斥力把它们从无穷远聚到一起。(R1)
中点到每个电荷为 $0.030\ \mathrm{m}$。电势是标量，故各贡献相加：(M1)
$$ V = \frac{k q_1}{0.030} + \frac{k q_2}{0.030} = \frac{k(q_1 + q_2)}{0.030} = \frac{(8.99\times 10^{9})(8.0\times 10^{-9})}{0.030}. $$(M1)
$$ V = \frac{71.92}{0.030} \approx 2.4\times 10^{3}\ \mathrm{V}. $$(A1)
电场沿任意闭合路径所做的功为零。(A1)
功只取决于电势的变化 $W = q\,\Delta V_e$；闭合路径回到起点，故 $\Delta V_e = 0$，从而 $W = 0$。电场是保守场。(R1)
$E_p = \dfrac{k q_1 q_2}{r} \propto \dfrac{1}{r}$，故 $r$ 加倍使能量减半：(R1)
$$ E_p' = \tfrac{1}{2}(2.2\times 10^{-6}) \approx 1.1\times 10^{-6}\ \mathrm{J}. $$(A1)
Solenoid electromagnet vs a bar magnet of the same shape. (a) field pattern of a bar magnet, line direction outside, where it is strongest; (b) using the grip rule, how the solenoid's poles are found and the effect of reversing the current; (c) why the solenoid's external pattern matches the bar magnet's; (d) one similarity and one difference between the magnetic field and the electric field of a point charge.螺线管电磁铁与形状相同的条形磁铁对比。(a) 条形磁铁的场图样、外部线方向、最强处；(b) 用螺旋定则如何确定螺线管磁极及电流反向的影响；(c) 为何螺线管外部图样与条形磁铁相同；(d) 磁场与点电荷电场的一个相似点和一个不同点。
The field lines emerge from the north pole, curve through the surrounding space, and re-enter at the south pole, forming closed loops; outside the magnet they run from north to south. (A1)
The lines are most closely spaced, and the field therefore strongest, at the poles. (A1)
Apply the right-hand grip rule to the coils: curl the fingers in the direction of the conventional current around the windings, and the thumb points to the north end of the solenoid. (M1)
Reversing the current reverses the circulation, so the north and south ends swap over. (A1)
Inside the solenoid the loops from all the turns reinforce into a strong, nearly uniform axial field; outside, the return paths spread out and weaken. (B1)
This is exactly the structure of a bar magnet (uniform interior, dipole exterior), so the external field pattern of the solenoid is identical to that of a bar magnet: the coil behaves as an electromagnet. (R1)
Similarity: neither magnetic field lines nor electric field lines ever cross, because the field has a single definite direction at each point. (B1)
Difference: magnetic field lines are continuous closed loops (no monopoles), whereas the electric field lines of a point charge are radial and start (or end) on the charge. (R1)
磁感线从北极发出，在周围空间弯曲，再从南极进入，构成闭合回路；磁铁外部由北指向南。(A1)
磁感线在两极处最密集，故磁场在两极最强。(A1)
对线圈用右手螺旋定则：四指沿绕组中常规电流方向弯曲，拇指便指向螺线管的北端。(M1)
电流反向使环流方向反转，故 N、S 两端互换。(A1)
螺线管内部各匝的回路相互增强，形成强而近乎匀强的轴向场；外部，返回路径展开并变弱。(B1)
这正是条形磁铁的结构（内部匀强、外部偶极），故螺线管的外部场图样与条形磁铁相同：线圈表现为电磁铁。(R1)
相似点：磁感线与电场线都从不相交，因为每点处场只有唯一确定的方向。(B1)
不同点：磁感线是连续的闭合回路（无磁单极子），而点电荷的电场线是径向的、起（或止）于电荷。(R1)