Electric field of a point charge · 点电荷的电场
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| point charge/pɔɪnt tʃɑːdʒ/ | 点电荷 | diǎn diàn hè |
| inverse-square/ɪnˈvɜːs skweə/ | 平方反比 | píng fāng fǎn bǐ |
| vector sum/ˈvektə sʌm/ | 矢量和 | shǐ liàng hé |
Take away the second charge
- Coulomb's law needs two charges. But a charge sitting alone still changes the space around it, whether or not anything is there to feel it.
- Divide the Coulomb force by the test charge and the test charge cancels out. What is left describes the source charge alone: its field.
- That is the whole move from force to field, and it is the reason a field is worth defining at all.
- This lesson is the field of a point charge 点电荷, its inverse-square fall, and how fields from several charges combine.
把第二个电荷拿走
- 库仑定律需要两个电荷。但一个孤零零待着的电荷仍然改变着它周围的空间,不管那里有没有东西去感受它。
- 用检验电荷去除库仑力,检验电荷就被约掉了。剩下的东西单独描述那个源电荷:它的场。
- 这就是从力到场的全部转变,也是场值得被定义的理由。
- 这一课讲点电荷(point charge)的场、它的平方反比衰减,以及多个电荷的场怎样合成。
The field of a point charge
- Take Coulomb's law and divide by the test charge, $E = F/q$:
- It points outwards from a positive $Q$ and inwards towards a negative one, and it falls as $\dfrac{1}{r^2}$.
- Only the source charge appears. That is the point: the field exists whether or not another charge is there to feel it.
Radial lines, and a curve that drops away fast
点电荷的场
- 取库仑定律,除以检验电荷,$E = F/q$:
- 它由正的 $Q$ 向外、朝负的 $Q$ 向内,并按 $\dfrac{1}{r^2}$ 衰减。
- 式中只出现源电荷。这正是要点:不管有没有另一个电荷去感受它,场都存在。
辐射状的线,以及一条迅速下降的曲线
Field of a point charge · 点电荷的电场
E ∝ Q/r²
The electric field of a point charge spreads out radially and obeys the inverse-square law. · 点电荷的电场呈径向发散，并遵循平方反比定律。
The field strength at distance $r$ from a point charge $Q$ is: · 在距离点电荷 $Q$ 为 $r$ 处的场强是:
From $E = F/q$ with Coulomb's force, the test charge cancels, leaving $E = \dfrac{Q}{4\pi\varepsilon_0 r^{2}}$. · 由 $E = F/q$ 和库仑力,检验电荷抵消,剩下 $E = \dfrac{Q}{4\pi\varepsilon_0 r^{2}}$。
The field is $100\ \dfrac{\text{N}}{\text{C}}$ at distance $r$ from a charge. What is it at $2r$? · 在距离一个电荷 $r$ 处,场是 $100\ \dfrac{\text{N}}{\text{C}}$。在 $2r$ 处是多少?
Inverse-square: $\dfrac{100}{2^{2}} = 25\ \dfrac{\text{N}}{\text{C}}$. · 平方反比:$\dfrac{100}{2^{2}} = 25\ \dfrac{\text{N}}{\text{C}}$。
The field points ____ from a positive point charge. · 场从一个正点电荷指向 ____。
A positive charge pushes a positive test charge away, so its field points outward. · 正电荷把正检验电荷推开,所以它的场指向外。
Exactly like gravity
- Compare $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$ with $g = \dfrac{GM}{r^2}$. The same inverse-square 平方反比 shape, with $Q$ in place of $M$ and $\dfrac{1}{4\pi\varepsilon_0}$ in place of $G$.
- The one real difference: charge has two signs, so an electric field can push or pull, while a gravitational field only ever pulls.
- Because the shapes match, every result transfers. Halve the distance and both fields quadruple.
与引力完全同形
- 把 $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$ 与 $g = \dfrac{GM}{r^2}$ 比较。同样的平方反比(inverse-square)形状,$Q$ 取代 $M$,$\dfrac{1}{4\pi\varepsilon_0}$ 取代 $G$。
- 唯一真正的区别:电荷有两种符号,所以电场能推也能拉,而引力场永远只拉。
- 因为形状相同,所有结论都可迁移。距离减半,两种场都变为四倍。
The electric field of a point charge falls off with distance just like gravity (as 1/r²). · 点电荷的电场随距离减小,就像引力一样(按 1/r²)。
Both are inverse-square: $E = \dfrac{Q}{4\pi\varepsilon_0 r^{2}}$ and $g = \dfrac{GM}{r^{2}}$. · 两者都是平方反比:$E = \dfrac{Q}{4\pi\varepsilon_0 r^{2}}$ 和 $g = \dfrac{GM}{r^{2}}$。
How do the electric field and the electric potential of a point charge depend on distance? · 点电荷的电场和电势怎样依赖于距离?
Confusing these two is the commonest error in the topic. The field is the negative gradient of the potential, which is exactly why one power lower appears in V. · 混淆这两个是本单元最常见的错误。场是电势的负梯度,这正是 V 中幂次低一阶的原因。
Worked example: the charged Earth
- Treat the Earth as a conducting sphere of radius $6.37 \times 10^6\ \text{m}$ carrying $-4.80 \times 10^5\ \text{C}$. Find the field at the surface and compare it with $g$.
- Outside a sphere the charge acts as a point charge at the centre, so $r$ is the radius:
- It is directed towards the centre, because the charge is negative. Gravity gives $g = GM/R^2 = 9.83\ \text{N/kg}$, so $E/g \approx 11$.
- Say the direction as well as the size. For a negative source charge the field points inwards, and half the mark is there.
例题:带电的地球
- 把地球当作半径 $6.37 \times 10^6\ \text{m}$、带电 $-4.80 \times 10^5\ \text{C}$ 的导体球。求表面处的场强并与 $g$ 比较。
- 在球外电荷表现为位于球心的点电荷,所以 $r$ 就是半径:
- 它指向球心,因为电荷为负。引力给出 $g = GM/R^2 = 9.83\ \text{N/kg}$,所以 $E/g \approx 11$。
- 除了大小还要说方向。源电荷为负时场指向内,一半的分在那里。
A sphere of radius 6.37e6 m carries -4.80e5 C. What is the field strength at its surface, in V/m? (1/4 pi eps0 = 8.99e9) · 一个半径 6.37e6 m 的球带电 -4.80e5 C。它表面处的场强是多少 V/m?(1/4 pi eps0 = 8.99e9)
Outside a sphere the charge acts as a point charge at the centre, so r is the radius: E = 8.99e9 x 4.80e5 / (6.37e6)^2 = 106 V/m, directed towards the centre. · 球外电荷表现为球心处的点电荷,所以 r 是半径:E = 8.99e9 x 4.80e5 / (6.37e6)^2 = 106 V/m,指向球心。
Several charges: add as vectors
- The total field at a point is the vector sum 矢量和 of the fields of each charge separately.
- Add them as arrows, with size and direction, not as numbers. Two fields of $100\ \text{V/m}$ in opposite directions give zero, not two hundred.
- This is where the electric field differs from electric potential, which is a scalar and adds with sign alone. Keep the two operations apart.
多个电荷:按矢量相加
- 某点的总场是各个电荷单独产生的场的矢量和(vector sum)。
- 要按箭头相加——有大小也有方向——不是按数字相加。两个方向相反的 $100\ \text{V/m}$ 的场给出零,不是二百。
- 这正是电场与电势不同的地方:电势是标量,只按符号相加。要把这两种运算分清。
For several charges, the total field at a point is: · 对于几个电荷,某点的总场是:
Fields are vectors, so add them with both size and direction. · 场是矢量,所以要带着大小和方向相加。
Worked example: where is the field zero?
- Point charges of $+4.0\ \text{nC}$ and $+1.0\ \text{nC}$ are $30\ \text{cm}$ apart. Where on the line joining them is the resultant field zero?
- Between the charges the two fields point in opposite directions, so they can cancel there. Let $x$ be the distance from the $4.0\ \text{nC}$ charge:
- The zero lies twice as far from the larger charge, which is the sanity check: the bigger field needs more distance to weaken.
- For opposite charges the two fields between them point the same way and never cancel, so the zero lies outside, beyond the smaller charge.
例题:场在哪里为零?
- $+4.0\ \text{nC}$ 和 $+1.0\ \text{nC}$ 两个点电荷相距 $30\ \text{cm}$。在它们的连线上,合场强在哪里为零?
- 在两电荷之间,两个场方向相反,所以能在那里抵消。设 $x$ 为距 $4.0\ \text{nC}$ 电荷的距离:
- 零点距较大的电荷远一倍,这就是合理性检验:更大的场需要更远才能减弱下来。
- 对异号电荷,它们之间的两个场方向相同,永不抵消,所以零点在外侧,在较小电荷之外。
Charges of +4.0 nC and +1.0 nC are 0.30 m apart. How far from the 4.0 nC charge, in metres, is the resultant field zero? · +4.0 nC 和 +1.0 nC 的电荷相距 0.30 m。合场强为零的点距 4.0 nC 电荷多少米?
Set the magnitudes equal: 4/x^2 = 1/(0.30-x)^2, so (0.30-x)/x = 1/2 and x = 0.20 m, twice as far from the larger charge. · 令两者大小相等:4/x^2 = 1/(0.30-x)^2,得 (0.30-x)/x = 1/2,x = 0.20 m,距较大电荷远一倍。
For two opposite point charges, there is a point between them where the resultant field is zero. · 对两个异号点电荷,它们之间存在一点使合场强为零。
Between opposite charges the two fields point the same way and add, so they never cancel. The zero lies outside, beyond the smaller charge. · 在异号电荷之间,两个场方向相同、相加,所以永不抵消。零点在外侧,在较小电荷之外。
Reading a field-distance graph
- For a charged conducting sphere, $E$ is zero inside, jumps to its maximum at the surface, and then falls as $1/r^2$ outside.
- The radius is the distance at which the field jumps from zero, and the charge follows from any point on the curve using $Q = 4\pi\varepsilon_0 r^2 E$.
- Given the surface value $E_0$, the field is $E_0/4$ at $2R$ and $E_0/9$ at $3R$. Sketch through those points rather than guessing the curve.
读场强—距离图
- 对带电导体球,$E$ 在内部为零,在表面跳到最大值,然后在外部按 $1/r^2$ 衰减。
- 半径就是场从零跳起来的那个距离,而电荷可由曲线上任一点用 $Q = 4\pi\varepsilon_0 r^2 E$ 求出。
- 已知表面值 $E_0$ 时,在 $2R$ 处场为 $E_0/4$,在 $3R$ 处为 $E_0/9$。要经过这些点作图,而不是凭感觉画曲线。
On a graph of field strength against distance from the centre of a charged conducting sphere, which are true? Select all · 所有 that apply. · 在场强对距带电导体球球心距离的图上,哪些是对的?选出所有适用的。
The radius is where the field jumps from zero, and the charge follows from any point on the outside curve using Q = 4 pi eps0 r^2 E. · 半径就是场从零跳起来的地方,而电荷可由外部曲线上任一点用 Q = 4 pi eps0 r^2 E 求出。
Marks that slip away
- Field is $\dfrac{1}{r^2}$; potential is $\dfrac{1}{r}$. Confusing them is the single commonest error in this topic.
- Give the direction: outwards from positive, inwards towards negative.
- Fields add as vectors; potentials add as scalars. Two equal opposite fields cancel.
- For like charges the zero field lies between them; for opposite charges it lies outside, beyond the smaller one.
容易丢掉的分
- 场是 $\dfrac{1}{r^2}$;电势是 $\dfrac{1}{r}$。混淆这两个是本单元唯一最常见的错误。
- 要给出方向:由正电荷向外,朝负电荷向内。
- 场按矢量相加;电势按标量相加。两个大小相等方向相反的场会抵消。
- 同号电荷的零场点在它们之间;异号电荷的在外侧,在较小的那个之外。
You've got it
- dividing Coulomb's law by the test charge leaves the source charge alone: $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$, outwards from positive and inwards towards negative
- it is inverse-square, the same shape as $g = GM/r^2$, but charge has two signs so the field can repel
- several charges add as a vector sum, unlike potentials which are scalars
- the zero-field point lies between like charges, nearer the smaller one, and outside opposite charges, beyond the smaller one
你掌握了
- 用检验电荷去除库仑定律,剩下的只有源电荷:$E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$,由正向外、朝负向内
- 它是平方反比的,与 $g = GM/r^2$ 同形,但电荷有两种符号所以场能排斥
- 多个电荷按矢量和相加,而电势是标量,与此不同
- 零场点在同号电荷之间、靠近较小的那个,或在异号电荷的外侧、在较小的那个之外