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. 🚀
