Electric Charges, Fields, and Gauss's Law

Peel a strip of sticky tape off a desk and it pulls hairs on your arm towards it from a few centimetres away. Nothing touches them. Somewhere a few trillion electrons have moved from one surface to the other, and that tiny imbalance is enough to lift a hair against the whole Earth's gravity. This unit builds the tools to describe that: the force between charges, the field around them, and Gauss's law, which finds the field in one line when the charge is arranged symmetrically.

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.

Worked example (electric versus gravity). In a hydrogen atom the electron and proton are $5.3\times10^{-11}\ \text{m}$ apart. The electric force is $F_E=\dfrac{ke^2}{r^2}=\dfrac{(9.0\times10^{9})(1.6\times10^{-19})^2}{(5.3\times10^{-11})^2}=8.2\times10^{-8}\ \text{N}$. The gravitational force is $F_G=\dfrac{Gm_em_p}{r^2}=\dfrac{(6.67\times10^{-11})(9.11\times10^{-31})(1.67\times10^{-27})}{(5.3\times10^{-11})^2}=3.6\times10^{-47}\ \text{N}$. The ratio is about $10^{39}$, and it does not depend on $r$, because both forces fall off as $1/r^2$. Electric forces can attract or repel; gravity only attracts. A planet carries almost equal amounts of $+$ and $-$ charge, so the electric forces cancel and gravity is left.

Exam skill (hanging spheres). A common lab hangs a small charged sphere of mass $m$ on a thread near a second charged sphere. At equilibrium the thread makes angle $\theta$ with the vertical, and the forces balance: $T\cos\theta=mg$ and $T\sin\theta=F_E$, so $F_E=mg\tan\theta$. Measure $\theta$ for several separations $d$ and plot $F_E$ against $1/d^2$: Coulomb's law predicts a straight line through the origin with slope $kq_1q_2$. If the spheres are conductors, the charges on them move. Like charges push each other to the far sides, so the centres of charge are further apart than $d$ and the force is smaller than $kq_1q_2/d^2$. Opposite charges pull each other to the near sides, so the force is larger.

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.

In every method, charge only moves; the total never changes. Charging almost always means moving electrons, because protons are locked inside nuclei. An object is positive because it has lost electrons, not because it gained protons. Grounding connects an object to the Earth, a body so large that it can give or take any amount of charge while staying neutral.

Worked example (sharing charge). Two identical metal spheres carry $+8.0\ \text{nC}$ and $-2.0\ \text{nC}$. They touch and are then separated. The total is $+6.0\ \text{nC}$, and identical spheres share it equally: $+3.0\ \text{nC}$ each. The second sphere gained $5.0\ \text{nC}$ of positive charge, which really means it lost $5.0\times10^{-9}/1.6\times10^{-19}=3.1\times10^{10}$ electrons to the first sphere.

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. "Small" matters: a test charge is too weak to push the source charges around, so it measures the field without changing it.

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.

Conductors and insulators in equilibrium. In a conductor at rest, the excess charge sits on the surface, the field inside is zero, and the field just outside is perpendicular to the surface; a sideways part would push charge along the surface. In an insulator the charge cannot move away, so it can stay inside the material as well as on the surface, and the field inside can be non-zero. A charged sphere of either kind, seen from outside, has the field of a point charge at its centre.

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.

Worked example (beyond the end of a rod). A thin rod of length $L$ carries charge $Q$ spread evenly, $\lambda=Q/L$. Find $E$ at a point on the rod's line, a distance $a$ beyond one end. Put the point at the origin, so the rod runs from $x=a$ to $x=a+L$. Every element pushes along the same line, so no components cancel and the magnitudes simply add:

Far away ($a\gg L$) this becomes $kQ/a^2$, a point charge again. With two rods, find each rod's field this way and add the two as vectors.

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

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

Worked example (flux through a flat card). A uniform field of $400\ \text{N/C}$ meets a $0.20\ \text{m}\times0.30\ \text{m}$ card. The angle between $\vec E$ and the area vector is $60^\circ$: $\Phi_E=EA\cos\theta=400(0.060)(0.50)=12\ \text{N}\cdot\text{m}^2/\text{C}$. Tilt the card until it lies along the field ($\theta=90^\circ$) and the flux is zero: no field lines pass through it. In a uniform field, a closed box has zero total flux: every line that enters on one face leaves through another.

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 flux through a closed surface does not depend on its size or shape, only on the charge inside: a bigger sphere around a point charge has a weaker field over a bigger area, and the two changes cancel exactly. 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 (charge at the centre of a cube). A $+4.0\ \text{nC}$ charge sits at the centre of a cube. The total flux is $\Phi_E=\dfrac{q}{\varepsilon_0}=\dfrac{4.0\times10^{-9}}{8.85\times10^{-12}}=450\ \text{N}\cdot\text{m}^2/\text{C}$. Gauss's law cannot give $E$ on a cube face, because $E$ is not constant there, but symmetry still splits the flux evenly: each of the six faces gets $75\ \text{N}\cdot\text{m}^2/\text{C}$.

Worked example (non-uniform charge density). A ball of radius $R$ has charge density $\rho=ar$, where $a$ is a constant, so it holds more charge near its surface. Integrate over thin shells of volume $4\pi r'^2\,dr'$: $Q_{\text{enc}}=\int_0^r a r'\,4\pi r'^2\,dr'=\pi a r^4$. Gauss's law gives $E(4\pi r^2)=\dfrac{\pi a r^4}{\varepsilon_0}$, so inside $E=\dfrac{ar^2}{4\varepsilon_0}$. Outside, the total charge is $\pi aR^4$, and $E=\dfrac{aR^4}{4\varepsilon_0 r^2}$. Check: the two answers agree at $r=R$.

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.

Worked example (inside a slab). A very large slab of thickness $2h$ has uniform charge density $\rho$. By symmetry the field points away from the middle plane and has the same size at equal distances on each side. Take a pillbox of face area $A$ with its faces at $+z$ and $-z$ ($z<h$). Both faces carry outward flux $EA$, and the pillbox encloses $\rho(2zA)$, so $2EA=\dfrac{2\rho zA}{\varepsilon_0}$ and $E=\dfrac{\rho z}{\varepsilon_0}$. The field is zero on the middle plane and grows linearly to $\rho h/\varepsilon_0$ at the surface, and it stays at that value outside.

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, $Q_{\text{enc}} = Q r^3/R^3$.
• In charging by induction, removing the charged rod before the ground wire. The induced charge then flows straight back and the conductor ends neutral.
• Saying a positive object "gained protons". It lost electrons; protons do not move.
• Getting the sign of flux wrong. The area vector of a closed surface points outward, so field lines going in give negative flux.
• Saying "the flux is zero, so the field is zero". Zero net flux means zero net charge inside; the field on the surface can still be large.