Derivation of Laplace Equation from Gauss’s Law
Step 1: Gauss's Law (Integral Form)
Gauss’s law states that the total electric flux through a closed surface is proportional to the enclosed charge:
∮S E · dA = Qenclosed / ε₀
Where:
- E = electric field vector
- dA = infinitesimal area vector
- Qenclosed = ∫V ρ dV
Step 2: Convert to Differential Form
Using the divergence theorem:
∮S E · dA = ∫V (∇ · E) dV
So Gauss’s law becomes:
∫V (∇ · E) dV = (1/ε₀) ∫V ρ dV
Since this is true for any volume V, the integrands must be equal:
∇ · E = ρ / ε₀
Step 3: Express E in Terms of Potential
Electric field is the negative gradient of potential:
E = -∇V
Substitute into Gauss’s law:
∇ · (-∇V) = ρ / ε₀
Simplify:
-∇²V = ρ / ε₀ → ∇²V = -ρ / ε₀
Note: The negative sign comes from the definition E = -∇V, since the electric field points from high to low potential.
Step 4: Laplace Equation in Charge-Free Region
For regions where there is no charge (ρ = 0):
∇²V = 0
This is the Laplace equation.
Step 5: Summary
- Start with Gauss’s law:
∇ · E = ρ / ε₀ - Express E in terms of potential:
E = -∇V - Substitute to get:
∇²V = -ρ / ε₀ - For ρ = 0 → Laplace equation:
∇²V = 0 - The second derivative arises naturally from
∇ · ∇V(divergence of gradient)
Key Idea: The Laplace equation sets the curvature of potential = 0 in a charge-free region, allowing the slope (electric field) to exist but ensuring no net sources or sinks.