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Qwestrum Engineering360 · Electrical & Electronics · Electromagnetic Theory

Electrostatic Fields

Electrostatics describes the field of stationary charges through Coulomb’s law and Gauss’s law; symmetry lets Gauss’s law give the field of line, sheet and spherical charge distributions almost by inspection.
Exam tip: keep SI units consistent end-to-end, write the governing relation symbolically before substituting, and sanity-check magnitude and sign.

Key formulas & points

Skim these first — then read the full notes below.

  • PotentialV=−∫E⋅dl;equipotential⊥fieldlinesPotential V = -\int E \cdot dl; equipotential ⊥ field lines
  • Boundaryconditions:tangentialEcontinuous;normalDjump=σBoundary conditions: tangential E continuous; normal D jump = \sigma
  • CapacitanceC=QV;energyW=12CV2Capacitance C = \frac{Q}{V}; energy W = \frac{1}{2} C V^{2}

Topic details

Introduction

Gauss’s law ∮D·dA = Q_enc is the exam workhorse whenever the charge has spherical, cylindrical or planar symmetry, because you can pull the field out of the integral. Coulomb’s law handles discrete point charges.

Scope in B.Tech and GATE syllabus

The potential V is the line integral of E, and E = −∇V lets you recover the field from a known potential. Capacitance follows from C = Q/V after finding the field and integrating to get the voltage.

Key relations & formulas

Formulas (Indian textbook notation)

  • Coulomb′slaw:F=q1q2(4πε0r2)Coulomb's law: F = q_{1} \frac{q_{2}}{(4\pi \varepsilon_{0} r^{2})}

Formulas (Indian textbook notation)

  • E=−∇V;D=εE=ε0E+PE = -∇V; D = \varepsilon E = \varepsilon_{0} E + P

Formulas (Indian textbook notation)

  • Gausslaw:∮D⋅dA=QenclosedGauss law: ∮ D \cdot dA = Q_{enclosed}

Notation and sign conventions

Relation 1 —
Coulomb′slaw:F=q1q2/Coulomb's law: F = q_{1} q_{2}/

Formulas (Indian textbook notation)

  • Coulomb′slaw:F=q1q2(4πε0r2)Coulomb's law: F = q_{1} \frac{q_{2}}{(4\pi \varepsilon_{0} r^{2})}
Write this relation with symbols exactly as in Elements of Electromagnetics — Matthew Sadiku before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
E=−∇V;D=εE=ε0E+PE = -∇V; D = \varepsilon E = \varepsilon_{0} E + P

Formulas (Indian textbook notation)

  • E=−∇V;D=εE=ε0E+PE = -∇V; D = \varepsilon E = \varepsilon_{0} E + P
Write this relation with symbols exactly as in Elements of Electromagnetics — Matthew Sadiku before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
Gausslaw:∮D⋅dA=QenclosedGauss law: ∮ D \cdot dA = Q_{enclosed}

Formulas (Indian textbook notation)

  • Gausslaw:∮D⋅dA=QenclosedGauss law: ∮ D \cdot dA = Q_{enclosed}
Write this relation with symbols exactly as in Elements of Electromagnetics — Matthew Sadiku before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

Choose a Gaussian surface matching the symmetry: a sphere for point/spherical charge (E = Q/4πε₀r²), a coaxial cylinder for a line charge (E = λ/2πε₀r), and a pillbox for an infinite sheet (E = σ/2ε₀, independent of distance).

Governing relations in practice

At a dielectric boundary the tangential component of E is continuous and the normal component of D jumps by the free surface charge σ. Getting these boundary conditions right is essential for layered-dielectric capacitor problems.

Design and analysis considerations

Stored energy is W = ½CV² = ½QV = Q²/2C; energy density is ½εE². Use whichever form matches the given data.

Assumptions and validity limits

State assumptions explicitly before using any relation for electrostatic fields — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Electromagnetic Theory viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Electromagnetic Theory papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to electrostatic fields.
4. Use equation 1:
Coulomb′slaw:F=q1q2/Coulomb's law: F = q_{1} q_{2}/
.
5. Use equation 2:
E=−∇V;D=εE=ε0E+PE = -∇V; D = \varepsilon E = \varepsilon_{0} E + P
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Electrostatic Fields appears in RF, power apparatus, and communications. In Indian electrical curricula this topic is tested because it connects theory to fields, Maxwell equations, and waves.
GATE and semester exams often combine electrostatic fields with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use electrostatic fields?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Forgetting the factor 2 difference between an infinite sheet (σ/2ε₀) and a conductor surface (σ/ε₀)
• Using ε₀ alone inside a dielectric instead of ε = ε_rε₀
• Mixing tangential-E and normal-D boundary conditions
• Dropping the 1/r vs 1/r² distinction (line charge vs point charge field)

Quick revision checklist

Before attempting electrostatic fields problems, confirm you can:
1.
PotentialV=−∫E⋅dl;equipotential⊥fieldlinesPotential V = -\int E \cdot dl; equipotential ⊥ field lines

2.
Boundaryconditions:tangentialEcontinuous;normalDjump=σBoundary conditions: tangential E continuous; normal D jump = \sigma

3.
CapacitanceC=QV;energyW=12CV2Capacitance C = \frac{Q}{V}; energy W = \frac{1}{2} C V^{2}
Revise the solved examples in Elements of Electromagnetics — Matthew Sadiku and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Field and force from a point charge

Problem

A point charge of 20 nC sits in air. Find the electric field magnitude at 3 cm and the force on a 5 nC charge placed there. Use 1/4πε₀ = 9×10⁹.

Solution

E = Q/(4πε₀ r²) = 9×10⁹ × 20×10⁻⁹ / (0.03)².
Numerator = 9×10⁹ × 20×10⁻⁹ = 180; r² = 9×10⁻⁴.
E = 180 / 9×10⁻⁴ = 2×10⁵ V/m.
F = qE = 5×10⁻⁹ × 2×10⁵ = 1×10⁻³ N = 1 mN.

Conceptual check — Electrostatic Fields

Problem

In a Electromagnetic Theory semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of electrostatic fields." What should a complete answer include?

Exams & GATE

Sadiku — field of line charge, sheet charge, spherical capacitor.

📖 Standard books (India)

  • Elements of Electromagnetics — Matthew Sadiku

    Read: Syllabus unit

    Fields, Maxwell equations, and waves

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