⚡ Electrostatics – Electric Charges & Fields

Page 7: Applications of Gauss’s Law (JEE Advanced Level)

---

1. Why Applications of Gauss’s Law are Important?

In JEE Advanced, Gauss’s law is rarely asked directly. Instead, it is tested through applications involving symmetric charge distributions and tricky concepts.

• Reduces complex integrations
• Gives quick results with symmetry
• Conceptual + numerical questions

---

2. Electric Field of Infinite Line Charge (Revisited)

Choose a cylindrical Gaussian surface of radius r and length L.

E(2πrL) = λL / ϵ₀

E = λ / (2πϵ₀r)

Used in long wire, coaxial cable problems.

---

3. Electric Field of Charged Conducting Shell

• Inside shell (r < R): E = 0
• On surface: E = kQ / R²
• Outside shell: E = kQ / r²

Entire charge resides on outer surface.

⚠ Common trap: Field inside conductor is ZERO even if hollow.

---

4. Electric Field of Uniformly Charged Solid Sphere

For a non-conducting sphere of radius R:

• Inside (r < R): E ∝ r
• Outside (r ≥ R): E ∝ 1/r²

E_inside = (1 / 4πϵ₀) · (Qr / R³)

Linear variation inside → Straight line graph.

---

5. Electric Field of Infinite Plane Sheet

Using pillbox Gaussian surface:

E = σ / (2ϵ₀)

• Independent of distance
• Field is uniform

Important for capacitor and slab problems.

---

6. Two Parallel Charged Sheets

Field due to each sheet adds vectorially.

• Same sign → fields cancel inside
• Opposite sign → fields add inside

Resulting field between plates = σ / ϵ₀

---

7. Gauss’s Law in Dielectric Medium

Replace ϵ₀ with ϵ = Kϵ₀

∮ E · dA = q_free / ϵ

• K = dielectric constant
• Field reduces by factor K

---

8. JEE Advanced Numerical Example

Question:

A solid sphere of radius R has uniform charge density. Find electric field at r = R/2.

Solution:

E = (1 / 4πϵ₀) · (Qr / R³)

Put r = R/2:

E = (1 / 8πϵ₀) · (Q / R²)

---

9. Graph-Based Questions

• Conducting sphere → sudden drop to zero inside
• Non-conducting sphere → straight line inside
• Infinite sheet → constant line

---

10. High-Yield JEE Tips

• Always identify symmetry first
• Choose correct Gaussian surface
• Check enclosed charge carefully
• Field inside conductor = ZERO (always)

---

🎯 Page 7 Summary

• Gauss’s law simplifies field calculation
• Used only with symmetry
• Key tool for JEE Advanced numericals