Electrostatic energy and Poisson's and Laplace's equations

Why it matters

The energy stored in an electric field sets how much a capacitor can deliver, how strongly electrodes attract (the principle of electrostatic actuators and some voltmeters), and how much damage a breakdown can do. Poisson's and Laplace's equations are the tools for finding the potential when the geometry is too awkward for Gauss's law — the basis of field plotting, finite-difference solvers and the analysis of semiconductor junctions.

Key ideas

Energy of a charge system. Assembling charges from infinity needs work; that work is stored as electrostatic potential energy. For point charges, W = ½ Σ Qk Vk, where Vk is the potential at Qk due to all the other charges (the ½ corrects for counting each pair twice). For two charges it reduces to W = Q1Q2/(4πεR12): positive (work needed) for like charges, negative for unlike charges.

Energy stored in the field. The same energy can be written as an integral over the field: W = ½ ∫ D·E dv. So every region of space where E ≠ 0 carries an energy density w = ½εE² (J/m³). This view tells you where the energy is: in a capacitor it sits in the dielectric; around a charged sphere it extends to infinity.

Capacitor energy. W = ½CV² = Q²/(2C) = ½QV. With V fixed (battery connected), inserting a dielectric raises C and the energy rises. With Q fixed (isolated), raising C lowers the energy — the field pulls the dielectric in.

Force from energy. At constant charge, F = −∂W/∂x. For parallel plates this gives the attraction F = Q²/(2εA) = ½εE²A — the principle of the electrostatic (attracted-disc) voltmeter and MEMS actuators.

Poisson's equation. Combining ∇·D = ρv with D = εE and E = −∇V gives ∇²V = −ρv/ε in a homogeneous medium. It links the curvature of V to the local charge.

Laplace's equation. In a charge-free region, ∇²V = 0. Most of the space between electrodes is charge-free, so the potential there obeys Laplace's equation with the electrode potentials as boundary conditions.

Uniqueness theorem. A solution of Laplace's (or Poisson's) equation that satisfies all the boundary conditions — potential given on the boundaries (Dirichlet) or its normal derivative given (Neumann) — is the only solution. So any method that gives a function meeting the equation and the boundary values (guessing, symmetry, image charges, numerical iteration) gives the answer.

Properties of Laplace solutions. A function satisfying ∇²V = 0 has no maximum or minimum inside the region (extremes are on the boundary), and V at a point equals the average of V on a small sphere around it — the idea behind the finite-difference relaxation method (each node = mean of its four neighbours in 2-D).

Formulas

• W = ½ Σ Qk Vk — energy of point charges (J).
• W = Q1Q2 / (4πεR) — two point charges.
• W = ½ ∫ ρv V dv = ½ ∫ D·E dv = ½ ∫ εE² dv — continuous distribution / field form.
• w = ½εE² = ½D·E = D²/(2ε) — energy density (J/m³).
• W = ½CV² = Q²/(2C) = ½QV — capacitor.
• F = Q² / (2εA) = ½εE²A — force between parallel plates (N).
• W = 4πρv²a⁵ / (15ε) = 3Q² / (20πεa) — uniformly charged sphere of radius a.
• ∇²V = −ρv / ε (Poisson) and ∇²V = 0 (Laplace); in Cartesian ∇²V = ∂²V/∂x² + ∂²V/∂y² + ∂²V/∂z².
• ∇²V = (1/ρ)∂(ρ ∂V/∂ρ)/∂ρ + (1/ρ²)∂²V/∂φ² + ∂²V/∂z² (cylindrical); ∇²V = (1/r²)∂(r² ∂V/∂r)/∂r + … (spherical, radial part).

Worked examples

Example 1 (standard). The 442.7 pF capacitor (A = 0.02 m², d = 1 mm, εr = 2.5) is charged to 100 V. Find the stored energy two ways.

1. W = ½CV² = 0.5 × 442.7 × 10⁻¹² × 100² = 2.21 μJ.
2. Field: E = V/d = 10⁵ V/m; w = ½ε₀εr E² = 0.5 × 2.5 × 8.854 × 10⁻¹² × 10¹⁰ = 0.1107 J/m³.
3. Volume = A × d = 0.02 × 0.001 = 2 × 10⁻⁵ m³, so W = 0.1107 × 2 × 10⁻⁵ = 2.21 μJ ✓.

Example 2 (GATE level, Poisson). Two large grounded plates at x = 0 and x = d = 0.1 m enclose a uniform charge ρv = 1 μC/m³ in free space. Find V(x), the maximum potential and the field at the plates.

1. Poisson in 1-D: d²V/dx² = −ρv/ε₀.
2. Integrate twice: V = −(ρv/2ε₀)x² + Ax + B. V(0) = 0 gives B = 0; V(d) = 0 gives A = ρv d/(2ε₀).
3. So V(x) = (ρv / 2ε₀) x (d − x), a parabola, maximum at x = d/2.
4. V_max = ρv d²/(8ε₀) = 10⁻⁶ × 0.01 / (8 × 8.854 × 10⁻¹²) = 141 V.
5. E = −dV/dx = (ρv/ε₀)(x − d/2); at x = 0, |E| = ρv d/(2ε₀) = 10⁻⁶ × 0.1 / (2 × 8.854 × 10⁻¹²) = 5.65 kV/m, pointing towards the plate (−x), as expected for positive charge between grounded plates.

Example 3 (energy of a charged sphere). A sphere of radius a = 0.1 m holds uniform ρv = 1 μC/m³ in free space. Find the total field energy.

1. Inside E = ρv r/(3ε₀); outside E = ρv a³/(3ε₀r²).
2. Integrating ½ε₀E² × 4πr² dr over both regions gives W = 4πρv²a⁵ / (15ε₀).
3. W = 4π × (10⁻⁶)² × (0.1)⁵ / (15 × 8.854 × 10⁻¹²) = 9.46 × 10⁻⁷ J (about 0.95 μJ); five-sixths of it lies outside the sphere.

Common mistakes

• Forgetting the ½ in W = ½ΣQV or W = ½CV².
• Claiming a dielectric always increases stored energy: true at constant V, false at constant Q.
• Sign of Poisson's equation: ∇²V = −ρv/ε (minus sign), because E = −∇V and ∇·E = ρv/ε.
• Using Laplace's equation in a region that contains charge.
• Assuming V = 0 at infinity inside a sphere formula: inside a uniform sphere V = ρv(3a² − r²)/(6ε₀), not ρv(a² − r²)/(6ε₀).
• Testing a potential for Laplace by checking only one second derivative — all three must sum to zero.

For GATE IN

Typical items: energy stored in a capacitor or in a field region; energy density; force between plates; checking whether a given V satisfies Laplace's equation, or finding ρv from V using Poisson's equation; 1-D solutions between plates or coaxial cylinders with given boundary potentials. Practise the Laplacian in all three coordinate systems.

Quick check

1. Does V = x² − y² satisfy Laplace's equation?
2. What is the energy in a 10 μF capacitor at 5 V?
3. If V = 3x² in free space, what is ρv?
4. At constant charge, what happens to stored energy when a dielectric is inserted?
5. Where can a solution of Laplace's equation have its maximum?

Answers: 1. Yes (2 − 2 = 0). 2. 125 μJ. 3. −6ε₀ ≈ −53.1 pC/m³. 4. It decreases. 5. Only on the boundary of the region.