Poisson and Laplace’s equation

The motivation for using the Laplace’s equation in electrostatics is to solve for the electric potential \(V(\boldsymbol r)\). Recall that from Gauss’ law one may derive the Poisson equation,

\[ \nabla^2 V = -\frac{\rho}{\epsilon_0} \]

The Laplace’s equation applies to electrostatics is the cases that one attempts to determine the electric potential for a region of space with no electric charges (thus, \(\rho = 0\)). With the RHS being zero we arrive at the Laplace’s equation,

\[ \nabla^2 V = 0 \tag{Region of Zero Charge} \]

Separation of Variables

The direct approach to solve the Laplace’s equation given that we are dealing with partial differential equation is to use the separation of variables.

Uniqueness Theorem

A tool for solving the Laplace’s equation is the uniqueness theorem.

Theorem 1 : The solution to Laplace’s equation in some volume \(\mathcal V\) is uniquely determined if \(V\) is specified in the boundary surface \(S\).

In other words if the solution to the Laplace's equation $V=V(\boldsymbol r)$ satisfies the boundary surface $\Big(V(\boldsymbol r' \in S) = V(\boldsymbol r)\Big)$ then this solution will uniquely satisfies the volume $\Big(V(\boldsymbol r' \in \mathcal V) = V(\boldsymbol r)\Big)$.

**Usage: ** To solve a problem you may simplify the problem to one that also satisfies the boundary conditions at the boundary surface. You must also make sure that the equivalent problem stil satisfies that $\rho=0$ for $V$ within the region wanted.The solution $V$ to this simplified problem is also the solution to the original problem.

Theorem 2 : For a region of volume \(\mathcal V\) surrounded by conductors of charge \(Q_i\) and surrounded by a charge density \(\rho\) where there are no conductors, the electric field is uniquely determined if all \(Q_i\) and \(\rho\) is known.