Electromagnetism & Propagation of Waves (8hrs) · Engineering Physics

Electric fields

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Electric fields

1. What an Electric Field Is

Electric field EE at point PP = force per unit positive test charge placed at PP due to source charges, vector:

E=F/q0E = F/q_0

SI unit V/m (or N/C). It exists even without test charge; F=qEF = q E for charge qq in field.

Visualized as field lines (tangent gives EE direction, density ∝ ∣E∣|E|).

2. Coulomb's Law — Point Charge

For point charge QQ at origin:

E(r)=(1/4πε0)Q/r2r^=kQ/r2r^E(r) = (1/4π ε_0) Q/r² \hat{r} = k Q/r² \hat{r}

with k=9×109k=9×10^9 N·m²/C², ε0=8.85×10−12ε_0=8.85×10^{-12} F/m. For many charges, superposition:

E=Σ(1/4πε0)Qi/ri2r^iE = Σ (1/4π ε_0) Q_i/r_i² \hat{r}_i

For continuous distribution:

E=(1/4πε0)∫ρ(r′)(r−r′)/∣r−r′∣3dV′E = (1/4π ε_0) ∫ ρ(r') (r - r')/|r - r'|³ dV'

3. Key Geometries

  • Infinite line charge λλ (C/m):
E=λ/(2πε0r)r^E = λ/(2π ε_0 r) \hat{r}

(radial, ∝1/r)

  • Infinite sheet σσ (C/m²):
E=σ/(2ε0)n^E = σ/(2 ε_0) \hat{n}

(uniform, independent of distance)

  • Parallel plate capacitor ±σ±σ:
E=σ/ε0=V/dE = σ/ε_0 = V/d

uniform between plates.

4. Relation to Potential

EE is conservative for electrostatics:

E=−∇VE = -∇V

where V(r)=(1/4πε0)Q/rV(r)= (1/4π ε_0) Q/r for point charge. Equipotentials ⊥ field lines.

5. Worked Examples

(a) E at 1 m from 1 nC: E=9e9·1e-9/1=9 V/m radially outward ✓

(b) Two +Q at (±a,0): E at origin cancels to 0 by symmetry; at (0,y): E_y =2·(kQ/(a²+y²))· y/√(a²+y²) ✓

(c) Field between plates V=10 V, d=1 mm: E=10/1e-3=10⁴ V/m ✓

(d) Uniformly charged ring axis: E= kQ z/(z²+R²)^{3/2} \hat{z}

6. Exam Points

  • Define E=F/q0E=F/q_0 — 3 marks, frequent.
  • Coulomb's law and superposition — 5 marks, very frequent.
  • Sketch field lines for point, dipole, line — 3-5 marks, frequent.
  • E=−∇VE=-∇V — 3 marks.
  • Compute EE for line/sheet/ring — 5-7 marks, frequent.
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