Electric Fields of Charge Distributions · 电荷分布的电场
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| integral/ˈɪntɪɡrəl/ | 积分 | jī fēn |
| symmetry/ˈsɪmətri/ | 对称性 | duì chèn xìng |
What if the charge is smeared along a rod, not a dot?
- Real objects hold charge spread over a line, a surface, or a volume.
- A single $E = kQ/r^2$ won't do — there is no single distance $r$.
- The calculus trick: chop the charge into tiny pieces and add up their fields.
- This is the heart of AP Physics C: turning a sum into an integral 积分.
如果电荷沿一根棒铺开,而不是一个点呢?
- 真实物体上的电荷铺展在一条线、一个面或一个体积上。
- 单一的 $E = kQ/r^2$ 行不通——没有唯一的距离 $r$。
- 微积分的诀窍:把电荷切成许多小块,再把它们的场加起来。
- 这就是 AP 物理 C 的核心:把求和变成积分。
Charge densities
- Spread along a line: linear density $\lambda = \dfrac{dq}{d\ell}$ (C/m).
- Spread over a surface: surface density $\sigma = \dfrac{dq}{dA}$ (C/m²).
- Filling a volume: volume density $\rho = \dfrac{dq}{dV}$ (C/m³).
- Each lets you write a small charge as $dq = \lambda\,d\ell$, $\sigma\,dA$, or $\rho\,dV$.
电荷密度
- 沿线铺开:线密度 $\lambda = \dfrac{dq}{d\ell}$(C/m)。
- 铺在面上:面密度 $\sigma = \dfrac{dq}{dA}$(C/m²)。
- 充满体积:体密度 $\rho = \dfrac{dq}{dV}$(C/m³)。
- 每一个都让你把小电荷写成 $dq = \lambda\,d\ell$、$\sigma\,dA$ 或 $\rho\,dV$。
Linear charge density $\lambda$ has units of: · 线电荷密度$\lambda$的单位是：
$\lambda = dq/d\ell$ is charge per length, C/m. · $\lambda = dq/d\ell$是单位长度的电荷，单位为C/m。
Chop into elements dq
- Pick a tiny element $dq$ at distance $r$ from your point $P$.
- It makes a tiny field $dE = \dfrac{k\,dq}{r^2}$, pointing from $dq$ toward $P$.
- Every element points a slightly different way — $d\vec E$ is a vector.
- Sum them: $\vec E = \displaystyle\int d\vec E$.
切成微元 dq
- 在距你的点 $P$ 距离 $r$ 处取一个微小元 $dq$。
- 它产生一个微小场 $dE = \dfrac{k\,dq}{r^2}$,从 $dq$ 指向 $P$。
- 每个元的方向都略有不同——$d\vec E$ 是矢量。
- 把它们相加:$\vec E = \displaystyle\int d\vec E$。
Build up a field · 构建电场
See how a larger, denser charge builds a stronger field from many contributions. · 观察更大、更密集的电荷如何通过多个贡献建立更强的电场。
The field from a small element $dq$ at distance $r$ is $dE = k\,dq/$ ____. · 微元$dq$在距离$r$处产生的电场为$dE = k\,dq/$ ____。
Each element is a point charge: $dE = k\,dq/r^2$. · 每个微元是一个点电荷：$dE = k\,dq/r^2$。
Select all · 所有 correct steps for the field of a charge distribution. · 选择计算电荷分布电场的所有正确步骤。
Density → dE → integrate. A single kQ/r² fails because r differs for each element. · 密度→dE→积分。单个kQ/r²失效，因为每个微元的r不同。
Use symmetry to kill components
- Adding vectors is hard, so let symmetry 对称性 do the work.
- For a symmetric shape, sideways components cancel in pairs.
- Only the component along the symmetry axis survives.
- Integrate just that surviving component — far less algebra.
用对称消去分量
- 矢量相加很难,所以让对称来帮忙。
- 对于对称形状,侧向分量成对抵消。
- 只有沿对称轴的分量存活下来。
- 只对这个存活的分量积分——代数量大大减少。
For a symmetric charge, sideways field components often cancel in pairs. · 对于对称电荷分布，侧向电场分量通常成对抵消。
Symmetry cancels perpendicular components, leaving only the axial part. · 对称性抵消垂直分量，仅保留轴向分量。
A worked shape: the infinite line
- For an infinite line of charge, the integral gives $E = \dfrac{2k\lambda}{r}$.
- Notice: it falls off as $1/r$, not $1/r^2$ — the extra charge farther along helps.
- A charged plane gives an even flatter field, $E = \dfrac{\sigma}{2\varepsilon_0}$ (constant).
- The shape of the object sets how fast the field fades.
一个算好的形状:无限长线
- 对无限长带电线,积分给出 $E = \dfrac{2k\lambda}{r}$。
- 注意:它随 $1/r$ 减弱,不是 $1/r^2$——远处更多的电荷帮了忙。
- 带电平面给出更平的场,$E = \dfrac{\sigma}{2\varepsilon_0}$(恒定)。
- 物体的形状决定场减弱的快慢。
The field of an infinite line of charge falls off as: · 无限长直线的电场随距离衰减规律为：
An infinite line gives $E = 2k\lambda/r$ — a $1/r$ falloff. · 无限长线产生$E = 2k\lambda/r$——即$1/r$衰减。
A charged plane gives $E = \sigma/(2\varepsilon_0)$. If $\sigma$ doubles, $E$ becomes how many times bigger? · 带电平面产生$E = \sigma/(2\varepsilon_0)$。若$\sigma$加倍，$E$变为原来的多少倍？
$E \propto \sigma$, so doubling $\sigma$ doubles $E$. · $E \propto \sigma$，因此加倍$\sigma$会使$E$加倍。
A ring of radius $a$ carries charge $Q$. Find $E$ on its axis, distance $x$ from the centre.
- By symmetry, only the along-axis part survives.
- The integral gives $E = \dfrac{kQx}{(x^2 + a^2)^{3/2}}$, pointing along the axis.
半径 $a$ 的圆环带电荷 $Q$。求其轴线上距圆心 $x$ 处的 $E$。
- 由对称,只有沿轴分量存活。
- 积分给出 $E = \dfrac{kQx}{(x^2 + a^2)^{3/2}}$,方向沿轴。
You must add the little fields as vectors, not just their sizes. Skipping symmetry and summing magnitudes gives a wrong (too big) answer — always cancel the components that symmetry kills first.
你必须把小场按矢量相加,而不是只加它们的大小。跳过对称、直接加大小会给出错误(偏大)的答案——一定先抵消对称消去的那些分量。
For spread-out charge, write $dq$ with a density ($\lambda,\sigma,\rho$), find each element's $dE = k\,dq/r^2$, and integrate $\vec E = \int d\vec E$. Use symmetry to cancel components. An infinite line gives $E = 2k\lambda/r$; a plane gives a constant $E = \sigma/2\varepsilon_0$.
对铺开的电荷,用密度($\lambda,\sigma,\rho$)写出 $dq$,求每个元的 $dE = k\,dq/r^2$,再积分 $\vec E = \int d\vec E$。用对称抵消分量。无限长线给出 $E = 2k\lambda/r$;平面给出恒定的 $E = \sigma/2\varepsilon_0$。