Electric Fields of Charge Distributions
Introduced| English |
|---|
| integral/ˈɪntɪɡrəl/ |
| symmetry/ˈsɪmətri/ |
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.
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 $\rho\,dV$.
Linear charge density $\lambda$ has units of:
$\lambda = dq/d\ell$ is charge per length, 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: $\vec E = \displaystyle\int d\vec E$.
Build up a field
See how a larger, denser charge builds a stronger field from many contributions.
The field from a small element $dq$ at distance $r$ is $dE = k\,dq/$ ____.
Each element is a point charge: $dE = k\,dq/r^2$.
Select all correct steps for the field of a charge distribution.
Density → dE → integrate. A single kQ/r² fails because r differs for each element.
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.
For a symmetric charge, sideways field components often cancel in pairs.
Symmetry cancels perpendicular components, leaving only the axial part.
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.
The field of an infinite line of charge falls off as:
An infinite line gives $E = 2k\lambda/r$ — a $1/r$ falloff.
A charged plane gives $E = \sigma/(2\varepsilon_0)$. If $\sigma$ doubles, $E$ becomes how many times bigger?
$E \propto \sigma$, so doubling $\sigma$ doubles $E$.
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.
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.
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 $E = \sigma/2\varepsilon_0$.