Applications of Gauss's Law

Gauss's law shines when the field has high symmetry. Four canonical cases occur again and again in JEE/NEET: infinite line of charge, infinite plane sheet, thin spherical shell, and uniformly charged solid sphere.

Concept

The recipe is always the same:

1. Identify the symmetry (cylindrical, planar, spherical).
2. Choose a Gaussian surface that respects the symmetry.
3. Argue that $\vec E$ is constant in magnitude and perpendicular (or parallel) to each patch.
4. Equate flux to $Q_\text{enc}/\varepsilon_0$.

Derivation

Infinite line of charge, $\lambda\,C/m$. Symmetry: cylindrical. Take a coaxial cylinder of radius $r$ and length $L$. Flux through the curved surface: $E\cdot 2\pi r L$. End caps contribute zero (field is radial). Enclosed charge: $\lambda L$. $E\cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0} \Longrightarrow E = \frac{\lambda}{2\pi\varepsilon_0 r}.$ Field falls as $1/r$, slower than a point charge.

Infinite plane sheet, $\sigma\,C/m^2$. Symmetry: planar. Take a cylinder ("pillbox") of cross-section $A$ piercing the sheet. Flux through each cap: $EA$. Side: zero. Enclosed: $\sigma A$. $2EA = \frac{\sigma A}{\varepsilon_0} \Longrightarrow E = \frac{\sigma}{2\varepsilon_0}.$ Field is uniform — independent of distance.

Thin spherical shell, charge $Q$, radius $R$. Spherical symmetry. For $r > R$: $E\cdot 4\pi r^2 = Q/\varepsilon_0 \Longrightarrow E = \frac{kQ}{r^2}.$ The shell looks like a point charge at its centre. For $r < R$: enclosed charge is zero, hence $E = 0$ inside.

Uniformly charged solid sphere, charge $Q$, radius $R$. Volume charge density $\rho = Q/(\tfrac{4}{3}\pi R^3)$.

• $r > R$: same as point charge: $E = kQ/r^2$.
• $r < R$: enclosed charge $= \rho\cdot \tfrac{4}{3}\pi r^3 = Q\,r^3/R^3$. Then $E\cdot 4\pi r^2 = \frac{Q r^3/R^3}{\varepsilon_0} \Longrightarrow E = \frac{kQr}{R^3}.$ Inside, $E$ grows linearly with $r$; outside, it falls as $1/r^2$. The maximum is at $r=R$.

Worked Example

A long wire has linear charge density $\lambda = 4\,nC/m$. Find the field at $r = 5\,cm$.

$E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k\lambda}{r} = \frac{2\times 9\times 10^9 \times 4\times 10^{-9}}{0.05} = 1440\,N/C.$

A sphere of radius $10\,cm$ carries $Q = 1\,\mu C$ uniformly. Field at $r = 5\,cm$ (inside): $E = \frac{kQr}{R^3} = \frac{9\times 10^9 \times 10^{-6} \times 0.05}{(0.10)^3} = \frac{0.45}{10^{-3}} = 4.5\times 10^{5}\,N/C.$

Common Confusions

• The infinite sheet has $E = \sigma/2\varepsilon_0$, but a conductor with surface charge $\sigma$ has $E = \sigma/\varepsilon_0$ just outside. The factor of 2 differs because in a conductor all flux exits on one side.
• Inside a shell, $E = 0$, but the potential is not zero. Don't confuse field and potential.
• Inside a uniform solid sphere, $E$ grows linearly with $r$. It is NOT zero.
• For an infinite line, $E \propto 1/r$ — not $1/r^2$.

Key Takeaways

• Wire: $E = \lambda/(2\pi\varepsilon_0 r)$.
• Sheet: $E = \sigma/(2\varepsilon_0)$, uniform.
• Shell: outside $kQ/r^2$, inside $0$.
• Solid sphere: outside $kQ/r^2$, inside $kQr/R^3$.
• Always use symmetry to pick a Gaussian surface where $\vec E$ is constant on each piece.