Poisson and Laplace Equations

In many practical electrostatic problems, the charge distribution is not explicitly known, but the potential on certain boundaries (conductors or surfaces) is specified. In such situations, the electric potential must be found by solving a partial differential equation derived from combining Gauss's law with the relationship E = negative gradient of V. The resulting equations, known as Poisson's equation and its special case Laplace's equation, form the mathematical foundation for solving boundary value problems in electrostatics.

Core Concept: Derivation and Physical Meaning

Starting from Gauss's law in differential form, div(D) = rho_v, and substituting D = epsilon * E and E = negative grad(V), we get: div(epsilon * grad(V)) = negative rho_v. For a uniform (homogeneous) medium where epsilon is constant, this simplifies to del^2 V = negative rho_v / epsilon. This is Poisson's equation. The operator del^2 is the Laplacian, which is the divergence of the gradient.

In regions where there is no free charge, rho_v = 0, and Poisson's equation reduces to Laplace's equation: del^2 V = 0. Most practical electrostatic problems involve charge-free regions between conductors (where potential is specified on boundaries). In such cases, Laplace's equation must be solved subject to the given boundary conditions on the conductor surfaces. The field in most of the space between capacitor plates, around transmission lines, or inside waveguides is governed by Laplace's equation.

The physical meaning of del^2 V = 0 is that the potential has no local maxima or minima inside the charge-free region. The potential at any interior point equals the average of its values over any surrounding sphere. This is called the mean value theorem for harmonic functions. It implies that charges cannot be in stable equilibrium in a free-space region (Earnshaw's theorem).

Mathematical Expression

In Cartesian coordinates, the Laplacian is del^2 V = d^2V/dx^2 + d^2V/dy^2 + d^2V/dz^2. In cylindrical coordinates (r, phi, z), it is del^2 V = (1/r)*d/dr(r*dV/dr) + (1/r^2)*d^2V/dphi^2 + d^2V/dz^2. In spherical coordinates (r, theta, phi), the Laplacian takes its most complex form. For problems with high symmetry, many terms vanish. For example, in a coaxial capacitor with V depending only on r (cylindrical), Laplace's equation reduces to (1/r)*d/dr(r*dV/dr) = 0, whose general solution is V = A*ln(r) + B, where A and B are determined from boundary conditions.

The uniqueness theorem states that if a solution to Laplace's or Poisson's equation exists that satisfies all the boundary conditions of a problem, then that solution is the unique correct answer. This theorem justifies using any method to guess or construct a solution: if it satisfies del^2V = 0 or del^2V = -rho/epsilon and all boundary conditions, it is the one and only correct solution. Methods of images, separation of variables, and conformal mapping all rely on this theorem.

Practical Understanding

Laplace's equation is solved in practice to find capacitance, field distribution, and conductor surface charge density. The general procedure is: (1) Identify the coordinate system matching the geometry. (2) Write Laplace's equation in that system and reduce using symmetry. (3) Solve the resulting ordinary differential equation. (4) Apply boundary conditions to find constants. (5) Compute E = negative grad V and then find D, rho_s, or capacitance as needed.

For a parallel plate capacitor with plates at x = 0 (V = 0) and x = d (V = V0), the 1D Laplace equation is d^2V/dx^2 = 0, giving V = V0 * x/d. Then E = negative dV/dx = negative V0/d (pointing in negative x, from high to low potential). Surface charge density on the plates is rho_s = epsilon * En = epsilon * V0 / d, and capacitance C = rho_s * A / V0 = epsilon * A / d. This is the systematic derivation of the parallel plate formula using Laplace's equation.

Given:
Coaxial capacitor: inner conductor at r=a=2mm, V=100V
Outer conductor at r=b=8mm, V=0
Medium: epsilon_r=3, epsilon_0=8.85e-12

Why this formula applies:
Cylindrical symmetry, V depends only on r.
Laplace eq reduces to: (1/r)*d/dr(r*dV/dr)=0
General solution: V = A*ln(r) + B

Applying BCs:
At r=b=0.008: 0 = A*ln(0.008) + B
At r=a=0.002: 100 = A*ln(0.002) + B

Subtraction:
100 = A*(ln(0.002) - ln(0.008)) = A*ln(0.002/0.008) = A*ln(0.25)
A = 100 / ln(0.25) = 100 / (-1.386) = -72.13
B = -A*ln(0.008) = 72.13 * (-4.828) = 348.1

Potential distribution:
V(r) = -72.13*ln(r) + 348.1

Electric field (radial):
Er = -dV/dr = 72.13/r V/m

At r = a = 0.002 m:
Er = 72.13/0.002 = 36065 V/m

Final Answer:
V(r) = -72.13*ln(r) + 348.1 V
Er(r) = 72.13/r V/m
Er at inner surface = 36.1 kV/m

Exam Tip: For GATE, Laplace's equation problems always involve symmetry reduction. In 1D planar: d^2V/dx^2=0, solution V=Ax+B. In 1D cylindrical (r only): V=A*ln(r)+B. In 1D spherical (r only): V=A/r+B. Memorize these three solutions; they cover most GATE boundary value problems. Always determine A and B from the two given boundary potentials.

• Poisson: del^2 V = -rho_v/epsilon applies in regions with volume charge density.
• Laplace: del^2 V = 0 applies in charge-free regions (most practical problems).
• Planar solution: V = Ax + B (linear); cylindrical: V = A*ln(r) + B; spherical: V = A/r + B.
• Uniqueness theorem ensures the solution satisfying BCs is the only correct solution.
• After finding V, compute E = -grad V, then D, rho_s, and capacitance.

Quick Revision

• Poisson: del^2 V = -rho_v/epsilon (charge present).
• Laplace: del^2 V = 0 (charge-free, most exam problems).
• Planar 1D: V = Ax + B. Cylindrical 1D: V = A*ln(r)+B. Spherical 1D: V = A/r+B.
• Uniqueness: one solution satisfying all BCs is THE solution.
• Steps: write Laplace, reduce by symmetry, solve ODE, apply BCs, get V, compute E.
• GATE trap: Laplace applies between conductors (no charge there), not inside conductors.
• E = -grad V: always negative gradient. Normal E at conductor gives rho_s = epsilon*En.