Gauss's law,

$$\oint_S\mathbf E\cdot d\mathbf A=\frac{Q_{\rm enc}}{\varepsilon_0},$$

becomes a practical field-calculation tool when symmetry fixes the field direction and makes its magnitude constant on useful parts of a closed Gaussian surface.

Three ideal symmetries are especially useful:

• spherical symmetry: $\mathbf E$ is radial and depends only on radius $r$;
• cylindrical symmetry: $\mathbf E$ is radial from an axis and depends only on perpendicular distance $r$;
• planar symmetry: $\mathbf E$ is normal to the plane and depends only on distance from it.

Uniformly charged solid sphere

Let a solid sphere of radius $R$ have uniform volume charge density $\rho_q$. For $r<R$, choose a concentric Gaussian sphere. Then

$$E(4\pi r^2)=\frac{\rho_q(4\pi r^3/3)}{\varepsilon_0},$$

so

$$\boxed{E(r)=\frac{\rho_q r}{3\varepsilon_0}\qquad(r<R)}.$$

Inside, the field grows linearly with radius.

The total charge is

$$Q=\rho_q\frac43\pi R^3.$$

For $r>R$,

$$E(4\pi r^2)=\frac{Q}{\varepsilon_0},$$

so

$$\boxed{E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\qquad(r>R)}.$$

Outside, the spherically symmetric distribution produces the same field as a point charge $Q$ at its center.

Infinite line charge

Let an ideal infinite line have linear charge density $\lambda$. Choose a cylindrical Gaussian surface of radius $r$ and length $L$. The field is normal to the curved surface and tangent to the end caps, so only the curved surface contributes flux:

$$E(2\pi rL)=\frac{\lambda L}{\varepsilon_0}.$$

Thus

$$\boxed{E(r)=\frac{\lambda}{2\pi\varepsilon_0r}}.$$

The ideal line-charge field falls as $1/r$, not $1/r^2$.

Infinite sheet of charge

Let an ideal infinite sheet have uniform surface charge density $\sigma$. Symmetry requires equal field magnitudes on the two sides, normal to the sheet.

Choose a pillbox crossing the sheet. The curved side contributes no flux, while each flat face contributes $EA$:

$$2EA=\frac{\sigma A}{\varepsilon_0}.$$

Therefore

$$\boxed{E=\frac{\sigma}{2\varepsilon_0}}.$$

For the ideal infinite sheet, the field magnitude is independent of distance.

A reliable procedure

To use Gauss's law to solve for $E$:

1. identify the symmetry of the entire charge distribution;
2. infer the allowed direction and spatial dependence of $\mathbf E$;
3. choose a closed surface on which $\mathbf E\cdot d\mathbf A$ is simple;
4. determine which parts of the surface contribute flux;
5. calculate $Q_{\rm enc}$;
6. apply Gauss's law.

A chosen surface cannot create symmetry that the source lacks. For a finite line, finite sheet, or irregular source, Gauss's law remains true but usually does not directly determine the field. The key skill is recognizing when symmetry turns the flux law into a field solution.