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电势

AP 物理 C:电磁学 · 第 9 主题

训练
讲义 词汇表
9.1

电势能

大纲
Learning ObjectiveEssential Knowledge

9.1.A
Describe the electric potential energy of a system.

  • 9.1.A.1 The electric potential energy of a system of two point charges equals the amount of work required for an external force to bring the point charges to their current positions from infinitely far away.
  • 9.1.A.2 The general form for the electric potential energy between two charged objects is given by the equation
    • Equation: $U_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r} = k\dfrac{q_1 q_2}{r}$.
  • 9.1.A.3 The total electric potential energy of a system can be determined by finding the sum of the electric potential energies of the individual interactions between each pair of charged objects in the system.

来源:美国大学理事会 AP 课程与考试说明

当你把两个同种电荷推到一起时,你逆着电力做(work)。那个功被这一对储存为电势能(electric potential energy):

$$U_E=\frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}=k\frac{q_1 q_2}{r}.$$

$U_E$ 被定义为一个外力必须做的功,把电荷从无限远带到相距 $r$。符号携带物理:

  • 同种电荷: $U_E>0$。需要一个外部推力把它们带到一起。若释放,它们飞开,而储存的能量变成动能(kinetic energy)。
  • 相反电荷: $U_E<0$。这一对被束缚。你必须供应 $|U_E|$ 的功把它们拉开到无穷。

电力是一个保守力(conservative force):它做的功只取决于起点和终点,不取决于路径,而等于 $-\Delta U_E$

对于几个电荷的一个系统(system),把每个不同对的势能相加:

$$U_{\text{total}}=k\!\left(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right).$$

Worked example. 三个 $+2.0\ \mu\text{C}$ 电荷坐在一个边长 $0.30\ \text{m}$ 的等边三角形的角上。每一对储存 $U=\dfrac{kq^2}{r}=\dfrac{(9.0\times10^{9})(2.0\times10^{-6})^2}{0.30}=0.12\ \text{J}$。有三对,所以 $U_{\text{total}}=3\times0.12\ \text{J}=0.36\ \text{J}$。这是从无穷组装这个三角形所需的功——以及若所有三个都被释放这些电荷会共享的动能。

词汇表 训练
英文 中文 拼音
work gōng
electric potential energy 电势能 diàn shì néng
kinetic energy 动能 dòng néng
conservative force 保守力 bǎo shǒu lì
system 系统 xì tǒng
Electric potential 电势 diàn shì
9.2

电势

大纲
Learning ObjectiveEssential Knowledge

9.2.A
Describe the electric potential due to a configuration of charged objects.

  • 9.2.A.1 Electric potential describes the electric potential energy per unit charge at a point in space.
  • 9.2.A.2 Expressions for the electric potential of charge distributions can be found using integration and the principle of superposition.
    • Equation: $V = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\int \dfrac{dq}{r}$
    • 9.2.A.2.i The electric potential for single point charge is
      • Equation: $V = \dfrac{q}{4\pi\varepsilon_0 r}$.
    • 9.2.A.2.ii The electric potential due to multiple point charges can be determined by the principle of scalar superposition of the electric potential due to each of the point charges.
      • Equation: $V = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum_i \dfrac{q_i}{r_i}$
  • 9.2.A.3 The electric potential difference between two points is the change in electric potential energy per unit charge when a test charge is moved between the two points.
    • Equation: $\Delta V = \dfrac{\Delta U_E}{q}$
  • 9.2.A.4 Electric potential difference may also result from chemical processes that cause positive and negative charges to separate, such as in a battery.

Boundary statement: AP Physics C: Electricity & Magnetism only expects students to use calculus to find the electric potential resulting from the following charge distributions and locations: an infinitely long, uniformly charged wire or cylinder at a distance from its central axis, a thin ring of charge at a location along the axis of the ring, a semicircular arc or part of a semicircular arc at its center, and a finite wire or line charge at a point collinear with the line charge or at a location along its perpendicular bisector.

9.2.B
Describe the relationship between electric potential and electric field.

  • 9.2.B.1 The value of an electric field component in any direction at a given location is equal to the negative of the spatial rate of change in electric potential at that location.
    • Equation: $E_x = -\dfrac{dV}{dx}$
  • 9.2.B.2 The change in electric potential between two points can be determined by integrating the dot product of the electric field and the displacement along the path connecting the points.
    • Equation: $\Delta V = V_b - V_a = -\displaystyle\int_a^b \vec{E}\cdot d\vec{r}$
  • 9.2.B.3 Electric field vector maps and equipotential lines are tools to describe the field produced by a charge or configuration of charges and can be used to predict the motion of charged objects in the field.
    • 9.2.B.3.i Equipotential lines represent lines of equal electric potential. These lines are also referred to as isolines of electric potential.
    • 9.2.B.3.ii Isolines are perpendicular to electric field vectors. An isoline map of electric potential can be constructed from an electric field vector map, and an electric field map may be constructed from an isoline map.
    • 9.2.B.3.iii An electric field vector points in the direction of decreasing potential.
    • 9.2.B.3.iv There is no component of an electric field along an isoline.

来源:美国大学理事会 AP 课程与考试说明

电势(electric potential)是空间里一点处每单位电荷的电势能。它以伏特测量($1\ \text{V}=1\ \text{J/C}$):

$$V=\frac{U_E}{q},\qquad V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}\ \text{(point charge)}.$$

电势是一个标量(scalar):它有一个符号但没有方向。这使 $V$ 比场 $\vec{E}$ 容易处理得多。对于几个点电荷(point charges),用标量叠加(superposition)——把电势连同它们的符号相加:

$$V=\frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{r_i}.$$

Worked example. 距一个 $+3.0\ \text{nC}$ 点电荷 $0.10\ \text{m}$ 的电势是 $V=\dfrac{kq}{r}=\dfrac{9.0\times10^{9}(3.0\times10^{-9})}{0.10}=270\ \text{V}$。现在加一个距同一点 $0.30\ \text{m}$$-3.0\ \text{nC}$ 电荷:它贡献 $\dfrac{9.0\times10^{9}(-3.0\times10^{-9})}{0.30}=-90\ \text{V}$,所以总量是 $270-90=180\ \text{V}$。只是一个有符号的和——没有矢量分量要分解。

Continuous charge distributions

对于一个连续电荷分布(continuous charge distribution),把它切成小片 $dq$、把每片当作一个点电荷,并积分(integrate):

$$V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{r}.$$

这是一个标量积分,所以它通常比场积分容易得多。AP 期望你用微积分处理四个形状:一个薄环(在它轴上的一点)、一段弧(在它的中心)、一条有限的线或导线(共线,或在它的垂直平分线上),和一根无限长的导线或圆柱(距它的轴一个距离)。

一个电荷环:每个元素 dq 距一个轴上的点相同的距离
一个电荷环:每个元素 dq 距一个轴上的点相同的距离

Worked example (ring of charge). 一个半径 $R$ 的薄环携带总电荷 $Q$。对于轴上距中心一个距离 $z$ 的一点 $P$,每个元素 $dq$$P$ 坐在相同的距离 $r=\sqrt{R^2+z^2}$。这个距离恒定,所以它从积分里出来:

$$V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{\sqrt{R^2+z^2}}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{\sqrt{R^2+z^2}}.$$

在一个 FRQ 上, $r$ 对每个元素都相同——那个陈述是评分的步骤。同样的思想给出任何弧的中心处 $V=kQ/R$,无论它的角度。

From potential to field

电势和场包含相同的信息,由每个方向的一个导数和一个路径积分联系:

$$E_x=-\frac{dV}{dx},\qquad \Delta V=V_b-V_a=-\int_a^b \vec{E}\cdot d\vec{r}.$$

场是电势的负梯度(gradient):$\vec{E}$ 从高电势指向低电势,"下坡"。$V$ 变化快的地方,场强。

Worked example. 沿 $x$ 轴,$V(x)=3x^{2}-2x$(伏特,$x$ 以米)。那么 $E_x=-\dfrac{dV}{dx}=2-6x\ \text{V/m}$。在 $x=0.50\ \text{m}$,$E_x=-1.0\ \text{V/m}$:那里的场指向 $-x$ 方向。

在一个均匀场里电势随距离稳定地下降
在一个均匀场里电势随距离稳定地下降

两点之间的电势差(potential difference)是它们之间移动每单位电荷的势能的变化:$\Delta V=\Delta U_E/q$。一个电池化学地制造一个电势差:里面的反应把正电荷从负电荷分开并把端子保持一个固定的 $\Delta V$

一个点电荷附近的电势随 1/r 变化
一个点电荷附近的电势随 1/r 变化

Equipotential maps

一条等势面(equipotential)线(也叫一条等值线(isoline))连接共享相同电势的点。四条规则让你能读任何图:

  • 等值线在它们相交的每个地方都垂直(perpendicular)于电场线(field lines)。
  • $\vec{E}$ 从高 $V$ 指向低 $V$ ——从不沿一条等值线。沿一条等值线移动一个电荷不需要功。
  • 紧密间隔的等值线意味着一个强的场:$E\approx-\Delta V/\Delta x$
  • 你能从一张等值线图草绘场图,并从一张场图草绘等值线图。
一个偶极子周围的等势线以直角穿过场线
一个偶极子周围的等势线以直角穿过场线

Exam skill. 给定一张每 $10\ \text{V}$、相距约 $2\ \text{cm}$ 有等值线的图,估计 $E\approx\dfrac{10}{0.02}=500\ \text{V/m}$,从较高值的线指向较低的。一个外部作用者慢慢地把一个电荷 $q$$A$ 移到 $B$ 所做的功是 $W=q(V_B-V_A)$ ——所走的路径不重要。

High-voltage power lines: electric potential energy is converted and transmitted as current at high voltage
High-voltage power lines: electric potential energy is converted and transmitted as current at high voltage
探索

Field lines and the potential around a charge

Electric potential is the energy per unit charge. It falls off with distance from a positive charge; the field lines point from high potential to low.

词汇表 训练
英文 中文 拼音
scalar 标量 biāo liàng
point charges 点电荷 diǎn diàn hè
superposition 叠加 dié jiā
continuous charge distribution 连续电荷分布 lián xù diàn hè fēn bù
integrate 积分 jī fēn
gradient 梯度 tī dù
potential difference 电势差 diàn shì chà
equipotential 等势面 děng shì miàn
isoline 等值线 děng zhí xiàn
perpendicular 垂直 chuí zhí
field lines 电场线 diàn chǎng xiàn
9.3

电能守恒

大纲
Learning ObjectiveEssential Knowledge

9.3.A
Describe changes in a system due to a difference in electric potential between two locations.

  • 9.3.A.1 When a charged object moves between two locations with different electric potentials, the resulting change in the electric potential energy of the object-field system is given by the following equation.
    • Equation: $\Delta U_E = q\Delta V$
  • 9.3.A.2 The movement of a charged object between two points with different electric potentials results in a change in kinetic energy of the object consistent with the conservation of energy.

来源:美国大学理事会 AP 课程与考试说明

当一个电荷 $q$ 在两个电势相差 $\Delta V$ 的点之间移动时,电荷-场系统的势能变化

$$\Delta U_E=q\,\Delta V.$$

若只有电力作用,总能量守恒,所以动能变化相反的量:

$$\Delta K=-\Delta U_E=-q\,\Delta V.$$

把两个符号一起注意:一个正电荷在它移向较低电势时加速,而一个负电荷在它移向较高电势时加速。两者都只是系统把势能换成动能。这就是粒子加速器如何给带电粒子它们的能量,而它是电路里能量分析的基础。

Worked example. 一个质子($q=1.6\times10^{-19}\ \text{C}$,$m=1.67\times10^{-27}\ \text{kg}$)从静止开始并通过一个 $500\ \text{V}$ 的电势降落加速(它移向较低电势,所以 $\Delta V=-500\ \text{V}$$\Delta K=-q\,\Delta V=+8.0\times10^{-17}\ \text{J}$)。令 $\Delta K=\tfrac12 mv^2$:

$$v=\sqrt{\frac{2(8.0\times10^{-17})}{1.67\times10^{-27}}}=3.1\times10^{5}\ \text{m/s}.$$

Exam skill. 每当场非均匀或路径弯曲时选择能量方法、不是运动学:只有端点电势重要。一个典型的 FRQ 链是 $q\,\Delta V \to \Delta K \to v$,带一行论证:"电力是保守的,所以能量守恒。"

9.3

考试技巧

  • $V=-\int \vec E\cdot d\vec l$$\vec E=-\nabla V$(在一维,$E_x=-\tfrac{dV}{dx}$)关联场和电势。
  • 电势是一个标量——连同符号相加贡献,不需要矢量分量。
  • 一对的势能是 $U=\tfrac{1}{4\pi\varepsilon_0}\tfrac{q_1 q_2}{r}$;对一个电荷的速率用能量守恒。
  • 知道沿一条等势面移动不做功,它垂直于 $\vec E$
  • 选择电势的零(通常是无穷)并陈述它。

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