Gauss's law applications — sphere, cylinder, plane symmetry selection

Question

A long straight wire carries a uniform linear charge density $\lambda = 5 \times 10^{-6}$ C/m. Using Gauss’s law, find the electric field at a distance $r = 0.1$ m from the wire. Explain why we chose a cylindrical Gaussian surface and not a spherical one.

(JEE Main & CBSE 12 — high weightage)

Solution — Step by Step

The wire has cylindrical symmetry — it looks the same from any angle around the axis. The electric field must point radially outward (perpendicular to the wire) and depend only on $r$ , the distance from the wire.

We pick a cylindrical Gaussian surface of radius $r$ and length $L$ , coaxial with the wire. Why? Because:

On the curved surface, $\vec{E}$ is parallel to $d\vec{A}$ and constant in magnitude
On the flat end caps, $\vec{E}$ is perpendicular to $d\vec{A}$ , so $\vec{E} \cdot d\vec{A} = 0$

This makes the flux integral trivial.

\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}

Left side: $E \times 2\pi r L$ (curved surface only, caps contribute zero)

Right side: $\dfrac{\lambda L}{\varepsilon_0}$ (charge enclosed = $\lambda \times L$ )

E \times 2\pi r L = \frac{\lambda L}{\varepsilon_0}

E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k\lambda}{r}

E = \frac{2 \times 9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = \mathbf{9 \times 10^5 \text{ N/C}}

The field falls as $1/r$ — not $1/r^2$ like a point charge. This is characteristic of line charges.

Why This Works

Gauss’s law is always true, but it’s only useful for calculating $E$ when the symmetry lets us pull $E$ out of the integral. The trick is matching the Gaussian surface to the charge distribution’s symmetry.

graph TD
    A["Gauss's Law Problem"] --> B{"Identify symmetry"}
    B -->|"Point charge or<br/>spherical shell"| C["Spherical Gaussian surface<br/>E ∝ 1/r²"]
    B -->|"Infinite line charge<br/>or long wire"| D["Cylindrical Gaussian surface<br/>E ∝ 1/r"]
    B -->|"Infinite plane<br/>or large sheet"| E["Pillbox Gaussian surface<br/>E = constant"]
    C --> F["E × 4πr² = q/ε₀"]
    D --> G["E × 2πrL = λL/ε₀"]
    E --> H["E × 2A = σA/ε₀"]

If you chose a sphere around a wire, $E$ would not be constant on the surface (it varies with angle), and you couldn’t simplify the integral. The Gaussian surface must match the symmetry — that’s the whole point.

Alternative Method — Coulomb’s Law Integration

Without Gauss’s law, you’d integrate Coulomb’s law over the entire wire:

E = \int_{-\infty}^{\infty} \frac{k\lambda\,dz}{(r^2 + z^2)} \cdot \frac{r}{\sqrt{r^2 + z^2}}

This integral gives the same $\lambda/(2\pi\varepsilon_0 r)$ but takes much longer. Gauss’s law gives the answer in three lines when symmetry permits.

For JEE: memorise the three results — sphere gives $1/r^2$ , cylinder gives $1/r$ , infinite plane gives constant $E = \sigma/(2\varepsilon_0)$ . The power of $r$ in the denominator drops by one for each dimension of symmetry you add.

Common Mistake

Students apply Gauss’s law to find the field of a finite wire or finite plane. Gauss’s law simplification works only for infinite (or very long/large) distributions where the symmetry argument holds. For a finite wire, the field at the ends is not purely radial — you must use integration. If the problem says “long wire” or “large sheet,” Gauss’s law is fine. If it gives a specific length, think twice.

Question

Solution — Step by Step

Why This Works

Alternative Method — Coulomb’s Law Integration

Common Mistake

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