2 · Continuous Charge and Gauss's Law

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Main results（核心结论）

A continuous source is handled by adding the fields of infinitesimal charge elements. Gauss's law instead relates the total signed flux through a closed surface to enclosed charge. It always holds, but determines a field directly only when symmetry sufficiently constrains that field.

The classroom contrasts 化整为零 (divide a source into elements and integrate) with 化零为整 (use a global flux law). Neither method removes the need to reason about symmetry.

1. Charge densities and the Coulomb integral

Distribution	Charge element	Total charge
Line	\(dq=\lambda\,d\ell\)	\(Q=\int\lambda\,d\ell\)
Surface	\(dq=\sigma_s\,dA\)	\(Q=\int\sigma_s\,dA\)
Volume	\(dq=\rho_q\,d\mathcal V\)	\(Q=\int\rho_q\,d\mathcal V\)

Densities are signed and can depend on position. Let \(\mathbf r'\) locate a source element and \(\mathbf r\) the observation point. Then

\[ d\mathbf E=k\,dq\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3},\qquad \mathbf E(\mathbf r)=k\int\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}\,dq. \]

Do not mix the source coordinate \(\mathbf r'\) with the observation coordinate \(\mathbf r\). The separation vector changes as the source element moves through the integral.

A reliable sequence is: choose \(dq\) → write the separation vector → project into components → use symmetry → integrate the surviving components → check limits.

2. Uniformly charged ring: field on its axis

Consider a ring of radius \(R\), lying in the \(xy\) plane and carrying total charge \(Q\) uniformly. At \(P=(0,0,z)\), every element is at the same distance \(s=\sqrt{R^2+z^2}\). Opposite elements cancel their transverse contributions. Rotation about the axis also rules out any preferred transverse direction.

The axial component is

\[ dE_z=\frac{k\,dq}{s^2}\frac{z}{s} =k\frac{z\,dq}{(R^2+z^2)^{3/2}}. \]

Integrating over the ring gives the exact axial field

\[ \boxed{\mathbf E(0,0,z)=\frac{kQz}{(R^2+z^2)^{3/2}}\,\hat{\mathbf z}.} \]

For positive \(Q\), the sign of \(z\) determines which way the field points. At the center, \(\mathbf E=0\). Far away, \(\mathbf E\simeq kQ\,\mathrm{sgn}(z)\hat{\mathbf z}/z^2\), as expected for a localized source of total charge \(Q\).

A useful extremum check

For \(z>0\),

\[ \frac{dE_z}{dz}=kQ\frac{R^2-2z^2}{(R^2+z^2)^{5/2}}. \]

The maximum occurs at \(z=R/\sqrt2\), with \(E_{\max}=2kQ/(3\sqrt3R^2)\). The field starts at zero, increases, then decays; it cannot monotonically decrease from the ring center.

The two curves use different normalizations, as stated in the legend; the graph compares their shapes rather than equal-charge field magnitudes.

3. Uniformly charged disk: build it from rings

A disk of radius \(R\) has constant surface charge density \(\sigma_s\). A thin ring of radius \(s\) and width \(ds\) carries

\[ dq=\sigma_s\,2\pi s\,ds. \]

For \(z>0\), insert this element into the ring result:

\[ E_z=2\pi k\sigma_s z\int_0^R\frac{s\,ds}{(z^2+s^2)^{3/2}} =\frac{\sigma_s}{2\epsilon_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right). \]

For either side of the disk, away from its ideal zero-thickness surface,

\[ \mathbf E(0,0,z)=\frac{\sigma_s}{2\epsilon_0} \left[\mathrm{sgn}(z)-\frac{z}{\sqrt{z^2+R^2}}\right]\hat{\mathbf z},\qquad z\ne0. \]

Three essential limits are:

1. Far field, \(|z|\gg R\): expansion of the square root gives \(|E_z|\simeq k|Q|/z^2\), where \(Q=\pi R^2\sigma_s\).
2. Near the center, \(z\to0^\pm\): \(E_z\to\pm\sigma_s/(2\epsilon_0)\). These are one-sided limits. Do not assign a unique continuous field to the charged ideal surface itself.
3. Infinite-sheet limit, \(R\to\infty\) at fixed \(z\): the field becomes distance independent, with magnitude \(|\sigma_s|/(2\epsilon_0)\) on each side.

The field jump across a charged sheet is consistent with the general normal-boundary condition discussed in class 5.

4. Electric flux（电通量）

An oriented surface element has area vector \(d\mathbf A=\hat{\mathbf n}\,dA\). Electric flux is

\[ \Phi_E=\int_S\mathbf E\cdot d\mathbf A. \]

For a uniform field through a flat area \(A\), \(\Phi_E=EA\cos\theta\), where \(\theta\) is the angle between the field and the normal, not the plane. Flux is a scalar and has unit N·m²/C.

For a closed surface, choose the outward normal everywhere:

\[ \Phi_E=\oint_S\mathbf E\cdot d\mathbf A. \]

Outward contributions are positive and inward contributions negative. A field tangent to a patch contributes zero flux through that patch. A Gaussian surface is an imaginary mathematical surface, not necessarily a physical object.

中文提示：“穿过多少电场线”只能帮助理解；精确定义是带方向的点积面积分。通量既可正也可负，不能把每个面的通量绝对值相加。

5. Gauss's law and why it works

In vacuum,

\[ \boxed{\oint_S\mathbf E\cdot d\mathbf A=\frac{Q_{\mathrm{enc}}}{\epsilon_0}.} \]

The field on the left is due to all charges, inside and outside. Only the net enclosed charge appears on the right.

For a point charge at the center of a sphere, \(E=kq/r^2\) and \(\Phi_E=E\,4\pi r^2=q/\epsilon_0\). The shrinking field and growing area cancel. The classroom connects this cancellation to the inverse-square law in three dimensions.

For a general closed surface, a point charge contributes

\[ d\Phi_E=kq\frac{\hat{\mathbf R}\cdot\hat{\mathbf n}}{R^2}\,dA =kq\,d\Omega, \]

where \(d\Omega\) is the signed solid angle seen from the charge. The net solid angle is \(4\pi\) for a charge inside and zero for a charge outside. Superposition extends the result to many charges and continuous distributions. A charge exactly on the chosen surface requires special limiting treatment; choose a surface that avoids it.

What Gauss's law does not say

• \(Q_{\mathrm{enc}}=0\) implies zero net flux, not zero field. A uniform field crosses a charge-free box, entering one side and leaving another.
• External charges contribute zero net flux but can change the local field at every point on the surface.
• Knowing one number, the flux, does not determine an arbitrary vector field on a surface.
• Enclosing a dipole gives zero flux even though its surrounding field is nonzero.

6. How symmetry makes Gauss's law useful

To extract an unknown \(E\) from a flux integral, establish from the physical source and boundary conditions that \(E\) is constant on the relevant surface patches and has a known orientation. Drawing a symmetric Gaussian surface cannot create symmetry absent from the source.

Spherical symmetry

If the charge distribution is spherically symmetric, \(\mathbf E=E_r(r)\hat{\mathbf r}\). A concentric sphere gives

\[ E_r(r)=\frac{Q_{\mathrm{enc}}(r)}{4\pi\epsilon_0r^2}. \]

For a uniformly charged solid insulating ball of radius \(R\) and total charge \(Q\),

\[ \rho_q=\frac{3Q}{4\pi R^3},\qquad \mathbf E(r)= \begin{cases} \displaystyle\frac{\rho_qr}{3\epsilon_0}\hat{\mathbf r} =\frac{kQr}{R^3}\hat{\mathbf r},&r<R,\\[4pt] \displaystyle\frac{kQ}{r^2}\hat{\mathbf r},&r\ge R. \end{cases} \]

Inside, only \(Q(r/R)^3\) is enclosed. The field increases linearly and is continuous at \(R\). A uniform thin spherical shell instead has zero field inside and the same external point-charge field. A conductor can redistribute charge; it is not the uniformly volume-charged insulating ball.

For a nonuniform spherically symmetric density,

\[ Q_{\mathrm{enc}}(r)=4\pi\int_0^r\rho_q(s)s^2\,ds. \]

Planar symmetry: an infinite nonconducting sheet

Uniformity under translations and rotations within the plane implies a perpendicular field of constant magnitude on each side. Reflection symmetry gives equal magnitudes on the two sides. A pillbox of cap area \(A\) has flux \(2EA\), hence

\[ E=\frac{|\sigma_s|}{2\epsilon_0}. \]

For sheets at \(z=0\) and \(z=d\) carrying \(+\sigma_s\) and \(-\sigma_s\), respectively, superposition gives \(\mathbf E=(\sigma_s/\epsilon_0)\hat{\mathbf z}\) between them and zero outside, in the infinite-sheet approximation.

Distinguish the field on one side of a single prescribed sheet from the field immediately outside a conducting surface: the latter uses \(E_n=\sigma_s/\epsilon_0\) because the field inside the metal is zero.

Cylindrical symmetry: an infinite line of charge

For line density \(\lambda\), translation along the axis, rotation about it, and reflection perpendicular to it imply a radial field. A coaxial cylinder of radius \(r\) and length \(L\) has no flux through its ends:

\[ E_r(2\pi rL)=\frac{\lambda L}{\epsilon_0},\qquad \boxed{\mathbf E=\frac{\lambda}{2\pi\epsilon_0r}\hat{\mathbf r}.} \]

The \(1/r\) scaling differs from the \(1/r^2\) point-charge field because the source is extended indefinitely in one dimension. A finite rod approximates this result only near its central region when the perpendicular distance is much smaller than its length.

As an extension of the same enclosed-charge argument, a uniformly charged infinite solid cylinder of radius \(R\) has \(E_r=\rho_qr/(2\epsilon_0)\) inside and \(E_r=\rho_qR^2/(2\epsilon_0r)\) outside. This is a deduction from the taught method rather than a separate later topic.

7. Common errors and self-checks

• Ring: every element has the same distance to an axial point, but its field is not entirely axial; project before adding.
• Disk: the annulus area is \(2\pi s\,ds\), not \(\pi(ds)^2\).
• Sphere: use enclosed charge, not total charge, for an observation point inside a distributed source.
• Sheet: a two-sided pillbox has two flux-producing caps; this creates the factor of two.
• Line: a coaxial Gaussian cylinder has lateral area \(2\pi rL\), not \(\pi r^2\).
• Symmetry: an off-axis ring point or a finite sheet edge generally prevents a simple Gauss-law solution.

For a quick dimensional check, \(\lambda/(\epsilon_0r)\), \(\sigma_s/\epsilon_0\), and \(\rho_qr/\epsilon_0\) must all have units of electric field.

References

Lecture 2 PDF pp. 3–8 covers the ring and disk; pp. 10–24 covers flux and Gauss's law; pp. 25–30 covers spherical, planar, and cylindrical applications. Class 2, 17:12–95:16, adds the symmetry reasoning, solid-angle explanation, and integral/global comparison. Its mathematical appendices are summarized in the toolkit.