Chapter 1 / 1.14 Applications of Gauss’s Law

1.14 Applications of Gauss’s Law ⚡️

Gauss’s law is like a shortcut for finding electric fields when the charge layout has plenty of symmetry. Pick the right “imaginary surface” (a Gaussian surface) and the math becomes a breeze. Let’s look at three classic cases and one neat example, all powered by the same idea. 😀

1. Infinitely Long Straight Wire 🧵

The wire carries a uniform linear charge density $ \lambda $. Symmetry says the field must shoot straight outward (or inward if $ \lambda < 0 $) and depend only on the distance $ r $. :contentReference[oaicite:0]{index=0}
Choose a cylindrical Gaussian surface (radius $ r $, length $ l $). Flux flows only through the curved side: $ \Phi = E \, 2\pi r l $. :contentReference[oaicite:1]{index=1}
Enclosed charge is $ q_{\text{in}} = \lambda l $. Gauss’s law $ \Phi = q_{\text{in}}/\varepsilon_0 $ gives $$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$ :contentReference[oaicite:2]{index=2}
Vector form: $$\mathbf{E} = \frac{\lambda}{2\pi\varepsilon_0 r}\,\hat{n}$$ with $ \hat{n} $ pointing radially. :contentReference[oaicite:3]{index=3}
This result is spot-on near the middle of very long wires (end effects vanish). 👌 :contentReference[oaicite:4]{index=4}

2. Uniformly Charged Infinite Plane Sheet 📄

Surface charge density is $ \sigma $. Because the sheet looks the same everywhere, the field must be perpendicular to it and constant. :contentReference[oaicite:5]{index=5}
Use a “pillbox” Gaussian surface (cross-section $ A $). Flux through both faces: $ \Phi = 2EA $. Enclosed charge: $ q_{\text{in}} = \sigma A $. :contentReference[oaicite:6]{index=6}
Gauss’s law → $$E = \frac{\sigma}{2\varepsilon_0}$$ Vector form: $$\mathbf{E} = \frac{\sigma}{2\varepsilon_0}\,\hat{n}$$ (points away if $ \sigma > 0 $, toward if $ \sigma < 0 $). :contentReference[oaicite:7]{index=7}
Notice how the field stays the same no matter how far you go from the sheet — a very exam-friendly fact! 😎

3. Uniformly Charged Thin Spherical Shell ⚽️

Surface charge density is $ \sigma $, shell radius $ R $. Spherical symmetry means the field is radial and depends only on $ r $. :contentReference[oaicite:8]{index=8}
Outside ($ r > R $)
Enclosed charge $ q = 4\pi R^{2}\sigma $. $$E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^{2}},\qquad \mathbf{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^{2}}\hat{r}$$ :contentReference[oaicite:9]{index=9}
Inside ($ r < R $)
The Gaussian sphere encloses no charge, so $$E = 0$$ everywhere inside. 🛑 :contentReference[oaicite:10]{index=10}
Cool takeaway: outside, the shell acts just like a point charge at its centre. Inside, you feel nothing at all!

4. Example 🌟 — Early “Plum-Pudding” Atom

A point nucleus with charge $ +Ze $ sits at the centre of a sphere of negative charge (density $ \rho $, radius $ R $). The whole atom is neutral.

Charge density is $ \rho = -\dfrac{3Ze}{4\pi R^{3}} $. :contentReference[oaicite:11]{index=11}
Inside ($ r < R $) the field grows linearly: $$E(r) = \frac{Ze}{4\pi\varepsilon_0}\,\frac{r}{R^{3}}$$ pointing outward. :contentReference[oaicite:12]{index=12}
Outside ($ r \ge R $) the field drops to zero because the positive and negative charges cancel perfectly. :contentReference[oaicite:13]{index=13}

High-Yield NEET Pointers 🎯

$ E \propto \dfrac{1}{r} $ near a long charged wire — great for quick calculations and MCQs.
An infinite plane sheet creates a distance-independent field: $ E = \sigma / (2\varepsilon_0) $.
For a charged shell, remember: inside $ E = 0 $; outside it behaves like a point charge.
Smart choice of Gaussian surface (cylinder, pillbox, sphere) is the secret sauce for easy field calculations.
Uniform charge distributions inside spheres often give linear fields (here, $ E \propto r $) — a pattern worth spotting in complex problems.

Keep practising, and Gauss’s law will feel as natural as breathing. 🚀