Electrostatics

Electrostatics governs static electric fields, capacitance and energy — the foundation of the ~11-mark Electromagnetics section. GATE EC tests Gauss's law, potential and boundary conditions (1–2 marks).

Key results

• Coulomb: F = Q₁Q₂/(4πε r²). Field of a point charge E = Q/(4πε r²).
• Gauss's law: ∮ D·dS = Q_enclosed; ∇·D = ρ_v; D = εE.
• Potential: E = −∇V; V = Q/(4πε r) for a point charge. Work W = QV.
• Energy density = ½εE² (= ½ D·E); capacitor energy = ½CV².
• Boundary conditions: tangential E continuous (E_t1 = E_t2); normal D differs by surface charge (D_n1 − D_n2 = ρ_s). Conductor surface: E is normal, interior E = 0.
• Poisson ∇²V = −ρ/ε; Laplace ∇²V = 0 (charge-free).

Quantity	Expression
Point-charge field	Q/(4πεr²)
Gauss's law	∮D·dS = Q_enc
Energy density	½εE²
Parallel-plate C	εA/d
E–V relation	E = −∇V

Exam Tricks & Tips

• 🎯 Use Gauss's law with symmetry (spherical/cylindrical/planar) to get E without integrating Coulomb's law.
• 🎯 Tangential E is continuous across any boundary; normal D jumps by the free surface charge ρ_s.
• 🎯 Inside a conductor E = 0 and it is an equipotential; just outside, E is purely normal = ρ_s/ε.
• 🎯 Energy density ½εE² — integrate over volume for total stored energy (= ½CV²).
• 🎯 Field of an infinite sheet = ρ_s/2ε (independent of distance); infinite line = ρ_L/(2πεr).
• ❌ Common mistake: making normal E continuous — it's normal D that's governed by surface charge, not E.

Expected exam pattern

1–2 marks: compute E/V/flux via Gauss's law, capacitance, stored energy, or apply boundary conditions at a dielectric/conductor interface.

Quick recap

Gauss: ∮D·dS=Q_enc, D=εE. E=−∇V. Energy density ½εE². Boundary: E_t continuous, D_n jumps by ρ_s. Conductor: E=0 inside, normal outside. Use symmetry for E. Poisson/Laplace for V.

Cover the answer, recall, then check. 12 cards on electrostatics for GATE EC.

Q1. State Gauss's law in integral and differential form.
A1. ∮ D·dS = Q_enclosed; ∇·D = ρ_v (with D = εE).

Q2. Electric field of a point charge?
A2. E = Q/(4πε r²), directed radially.

Q3. Relation between E and potential V?
A3. E = −∇V.

Q4. Electrostatic energy density?
A4. w = ½εE² = ½ D·E (J/m³).

Q5. Boundary condition on tangential E?
A5. It is continuous across an interface: E_t1 = E_t2.

Q6. Boundary condition on normal D?
A6. D_n1 − D_n2 = ρ_s (the free surface charge density).

Q7. Field and potential inside a conductor in electrostatics?
A7. E = 0 inside; the conductor is an equipotential; just outside, E is normal = ρ_s/ε.

Q8. Field of an infinite line charge?
A8. E = ρ_L/(2πε r), radial.

Q9. Field of an infinite charged sheet?
A9. E = ρ_s/(2ε), independent of distance.

Q10. Capacitance of a parallel-plate capacitor?
A10. C = εA/d.

Q11. State Poisson's and Laplace's equations.
A11. ∇²V = −ρ/ε (Poisson); ∇²V = 0 in charge-free regions (Laplace).

Q12. Energy stored in a capacitor?
A12. W = ½CV² = ½QV = Q²/(2C).

Electrostatics is the study of charges at rest and the fields they create — the foundation of capacitance, dielectrics, and the boundary conditions that all of electromagnetics builds on. GATE reliably tests Gauss's law for symmetric charge distributions, potential/energy relations, and dielectric boundary behaviour.

Core concept: Static charges produce an electric field E (and potential V) governed by Coulomb's and Gauss's laws; energy is stored in the field, and boundary conditions link fields across material interfaces.

The theory

Beginner — Coulomb, field, and potential

• Coulomb's law: force F = Q1Q2/(4πε r²) between point charges.
• Electric field E = F/q; for a point charge E = Q/(4πε r²) r̂.
• Potential V = −∫E·dl; E = −∇V. For a point charge V = Q/(4πε r).
  Permittivity ε = ε0·εr, with ε0 = 8.854×10⁻¹² F/m.

Intermediate — Gauss's law and capacitance

Gauss's law: ∮ D·dS = Q_enclosed, where D = εE is the electric flux density. For symmetric charge distributions it gives E instantly:

• Point/sphere: E = Q/(4πε r²).
• Infinite line charge ρL: E = ρL/(2πε r).
• Infinite sheet σ: E = σ/(2ε) (independent of distance).
  Capacitance C = Q/V; parallel-plate C = εA/d; energy stored W = ½CV² = ½∫εE² dV (energy density ½εE²).

Advanced — dielectrics and boundary conditions

• A dielectric polarises (P) in a field, so D = ε0E + P = εE. Bound charges reduce the field inside.
• Boundary conditions at an interface between media 1 and 2:
  • Tangential E is continuous: Et1 = Et2.
  • Normal D differs by free surface charge: Dn1 − Dn2 = ρs (continuous if ρs = 0).
• Poisson/Laplace: ∇²V = −ρ/ε (Poisson), ∇²V = 0 in charge-free regions (Laplace) — solved with boundary conditions for field problems.
• Method of images handles charges near conductors; a conductor forces E tangential = 0 and is an equipotential.

Worked example

A parallel-plate capacitor has plate area A = 100 cm², separation d = 1 mm, filled with a dielectric εr = 4. Find C and the energy stored at 100 V.
C = ε0εr·A/d = (8.854×10⁻¹²)(4)(100×10⁻⁴)/(10⁻³) = 8.854e−12 × 4 × 10⁻²/10⁻³ = 8.854e−12 × 40 = 354 pF.
Energy W = ½CV² = ½(354×10⁻¹²)(100²) = ½ × 354e−12 × 10⁴ = 1.77 µJ.

GATE relevance

1–2 mark NAT/MCQ: field from a symmetric charge via Gauss's law, potential/potential-energy, capacitance and stored energy, dielectric boundary conditions, and Laplace/Poisson setups. Boundary-condition and Gauss-symmetry problems are common.

Exam tricks and shortcuts

• Gauss's law for symmetry: sphere ∝ 1/r², line ∝ 1/r, sheet = constant.
• E = −∇V; energy density = ½εE²; parallel-plate C = εA/d.
• Boundary: Et continuous, Dn jumps by free surface charge ρs.
• Conductor surface: E is purely normal, interior E = 0 (equipotential).
• Mnemonic: "Tangential E and normal D are the well-behaved ones."

Key formulas

• Coulomb / field: E = Q/(4πε₀r²); F = qE.
• Gauss's law: ∮D·dS = Q_enc; D = εE.
• Potential: V = Q/(4πε₀r); E = −∇V.
• Energy density: w_E = ½εE².
• Boundary conditions: E_tan continuous; D_normal difference = ρ_s.
• Poisson/Laplace: ∇²V = −ρ/ε (Poisson), ∇²V = 0 (Laplace).

Gauss's law gives fields for symmetric charges; boundary conditions relate fields across dielectric interfaces.

Worked Example

Problem: Two point charges Q₁ = +2 nC and Q₂ = −3 nC are separated by 10 cm in free space. Find (a) the electrostatic force between them and (b) the electric field magnitude at the midpoint of the line joining them. (Use k = 1/4πε₀ = 9 × 10⁹ N·m²/C².)

Solution:
(a) Coulomb's law:
F = k·|Q₁Q₂|/r² = 9 × 10⁹ × (2 × 10⁻⁹)(3 × 10⁻⁹)/(0.1)²
= 9 × 10⁹ × 6 × 10⁻¹⁸ / 0.01
= 54 × 10⁻⁹ / 0.01 = 5.4 × 10⁻⁶ N = 5.4 µN.
The charges have opposite signs, so the force is attractive.

(b) The midpoint is r = 0.05 m from each charge.
Field from Q₁ (points away from the positive charge, i.e. toward Q₂):
E₁ = kQ₁/r² = 9 × 10⁹ × 2 × 10⁻⁹ / (0.05)² = 18/0.0025 = 7200 V/m.
Field from Q₂ (points toward the negative charge, i.e. toward Q₂):
E₂ = 9 × 10⁹ × 3 × 10⁻⁹ / 0.0025 = 27/0.0025 = 10800 V/m.
Both fields point the same way (from Q₁ toward Q₂), so they add:
E = E₁ + E₂ = 7200 + 10800 = 18000 V/m = 18 kV/m.

Answer: (a) F = 5.4 µN (attractive); (b) E = 18 kV/m directed toward the negative charge.