电荷、电场与高斯定律

Electric charge 电荷 comes in positive and negative; like charges repel, opposites attract. Charge is quantized 量子化 – every charge is a whole-number multiple of the elementary charge 基本电荷 $e=1.6\times10^{-19}\ \text{C}$ – and obeys conservation of charge 电荷守恒: charge is never created or destroyed, only moved. The force between two point charges 点电荷 is Coulomb's law 库仑定律:

an inverse-square 平方反比 force along the line joining them, with $\dfrac{1}{4\pi\varepsilon_0}=k=9.0\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$. For several charges (AP uses four or fewer, unless symmetry helps), add the force vectors on the charge you care about – superposition 叠加.

Charge is a fundamental property 基本属性 of matter, and a point charge is a model that ignores the size of a charged object. The constant $\varepsilon_0$ is the permittivity of free space 真空介电常数 – a material property of the vacuum. The permittivity of matter differs from $\varepsilon_0$, set by how easily the material polarizes, which is why a dielectric between charges weakens the force. Note too that although the electric force is far stronger than gravity, gravity dominates at astronomical scales because large bodies are nearly neutral.

Worked example. Two $+2.0\ \mu\text{C}$ charges sit $0.30\ \text{m}$ apart: $F=\dfrac{kq_1q_2}{r^2}=\dfrac{9.0\times10^{9}(2.0\times10^{-6})^2}{(0.30)^2}=0.40\ \text{N}$, repulsive.

Worked example (vectors). Charges $+3.0\ \mu\text{C}$ and $-3.0\ \mu\text{C}$ sit at two corners of a right angle, each $0.30\ \text{m}$ from a $+1.0\ \mu\text{C}$ charge at the corner. Each exerts $F=\dfrac{(9.0\times10^{9})(3.0\times10^{-6})(1.0\times10^{-6})}{0.09}=0.30\ \text{N}$ – one a push, one a pull, at right angles to each other. The net force is $F_{\text{net}}=\sqrt{0.30^2+0.30^2}=0.42\ \text{N}$, pointing between them. Never add magnitudes blindly: components first.

A conductor 导体 lets charge move freely; an insulator 绝缘体 holds it in place. Objects gain net charge three ways:

• Charging by friction 摩擦起电: rubbing transfers electrons from one surface to the other.
• Charging by conduction 传导起电: touching a charged object shares charge of the same sign.
• Charging by induction 感应起电: a nearby charge pushes the conductor's charge apart; ground 接地 the far side, remove the ground wire first, and the conductor is left with the opposite sign – all without contact.

An insulator cannot pass charge, but it can develop polarization 极化: its molecules stretch or turn so one face is slightly positive and the other slightly negative. That is why a charged comb picks up neutral paper scraps. An electroscope 验电器 shows net charge by its leaves repelling; on any isolated conductor the excess charge sits entirely on the outer surface.

The electric field 电场 at a point is the force per unit charge that a small positive test charge 检验电荷 would feel there:

A charge placed in a field feels $\vec{F}=q\vec{E}$ – positive charges along the field, negative charges against it. Fields from several sources add as vectors (superposition again). Field lines 电场线 make the field visible: they point away from positive charge and toward negative, their density shows the strength, and they never cross.

In a uniform 均匀 field a charge feels a constant force, so it accelerates uniformly – launched sideways, it follows a parabola, just like a projectile in gravity.

For a continuous charge distribution 连续电荷分布, cut the object into pieces $dq$ and integrate their point-charge fields:

Write $dq$ using the right density: $\lambda\,dl$ with the linear charge density 线电荷密度 (a line or ring), $\sigma\,dA$ with the surface charge density 面电荷密度, or $\rho\,dV$ with the volume charge density 体电荷密度. Then use symmetry 对称性 to cancel components before integrating – that sentence is usually a scored step. AP's calculus cases: a finite line (collinear point or perpendicular bisector), an infinite wire or cylinder, a ring on its axis, and an arc at its centre.

Worked example (ring, on axis). A ring of radius $R$ carries charge $Q$. At a point $z$ along the axis, each element is a distance $\sqrt{R^2+z^2}$ away, and by symmetry the sideways components cancel, leaving only the axial part ($\cos\alpha=z/\sqrt{R^2+z^2}$):

Check the limits: $E=0$ at the centre ($z=0$), and far away ($z\gg R$) it becomes $kQ/z^2$ – a point charge, as it must.

Electric flux 电通量 measures how much field passes through a surface:

For a uniform field through a flat area, $\Phi_E=EA\cos\theta$, where the area vector 面积矢量 is perpendicular to the surface. Flux is positive where field lines exit a closed surface and negative where they enter – only the perpendicular component of $\vec{E}$ counts.

Gauss's law 高斯定律 – the first of Maxwell's equations – relates the flux through any closed surface to the charge inside it:

It is true for any closed surface, but it solves for $E$ only when symmetry lets you choose a Gaussian surface 高斯面 on which $E$ is constant and perpendicular (or parallel, contributing zero). The three symmetries AP tests: spherical symmetry 球对称 (concentric sphere), cylindrical symmetry 柱对称 (coaxial cylinder), and planar symmetry 平面对称 (a straddling "pillbox").

Worked example (sphere). Outside a sphere of total charge $Q$, a spherical Gaussian surface of radius $r$ gives $E(4\pi r^2)=Q/\varepsilon_0$, so $E=\dfrac{kQ}{r^2}$ – identical to a point charge at the centre. Inside a uniformly charged solid sphere of radius $R$, the surface encloses only $Q_{\text{enc}}=Q\,r^3/R^3$, so $E=\dfrac{kQr}{R^3}$: zero at the centre, growing linearly to the surface.

Worked example (wire). For an infinite wire with charge per length $\lambda$, take a coaxial cylinder of radius $r$ and length $L$. The ends contribute nothing ($\vec{E}\perp d\vec{A}$), so $E(2\pi rL)=\dfrac{\lambda L}{\varepsilon_0}$ and $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$. The same pillbox method gives an infinite sheet's field, $E=\dfrac{\sigma}{2\varepsilon_0}$, uniform on both sides.

Exam skill. A full-credit Gauss's-law answer has four parts: name the surface, state the symmetry argument (why $E$ is constant and perpendicular on it), count $Q_{\text{enc}}$, then solve. Skipping the symmetry sentence loses the reasoning point – and remember Gauss's law also explains why $E=0$ inside any conductor in equilibrium.

• Use Coulomb's law $F=\tfrac{1}{4\pi\varepsilon_0}\tfrac{q_1 q_2}{r^2}$ and superpose forces as vectors (components, not magnitudes).
• For a continuous charge, integrate $d\vec E$ over $dq=\lambda\,dl,\ \sigma\,dA,\ \rho\,dV$; use symmetry to cancel components.
• The field points away from positive and toward negative charge; draw field lines correctly.
• Distinguish the force on a charge ($\vec F=q\vec E$) from the field the charge creates.
• Keep the constant $k=\tfrac{1}{4\pi\varepsilon_0}$ straight and check units.

Common mistakes

• Adding field or force magnitudes from several charges as numbers. Resolve into components and add as vectors.
• Using Gauss's law to find $E$ without symmetry. The law is always true, but it gives $E$ only when $E$ is constant and perpendicular on the chosen surface.
• Counting charge outside the Gaussian surface. Only $Q_{\text{enc}}$ enters $\oint \vec E\cdot d\vec A = Q_{\text{enc}}/\varepsilon_0$.
• Putting the full charge $Q$ in for a point inside a uniformly charged sphere. Inside, $Q_{\text{enc}} = Q r^3/R^3$.