Electrostatic fields can often be found more naturally by solving for the electric potential throughout a region than by summing the field of every charge directly.

The electric field is related to potential by

$$\boxed{\mathbf E=-\nabla V}.$$

Gauss's law for a continuous charge density can be converted from its integral form to the local relation

$$\boxed{\nabla\cdot\mathbf E=\frac{\rho}{\varepsilon_0}}.$$

Substituting $\mathbf E=-\nabla V$ gives

$$-\nabla\cdot\nabla V=\frac{\rho}{\varepsilon_0},$$

so the electrostatic potential obeys

$$\boxed{\nabla^2V=-\frac{\rho}{\varepsilon_0}}.$$

This is Poisson's equation with charge density as the source.

In a region containing no charge,

$$\rho=0,$$

and therefore

$$\boxed{\nabla^2V=0}.$$

Thus electrostatic potential is harmonic in every charge-free region.

Electrostatic boundary data

The field equation must be combined with boundary conditions.

For an ideal conductor in electrostatic equilibrium, the conductor is an equipotential. A conductor held at a specified voltage therefore provides a Dirichlet boundary condition

$$V=V_b.$$

The normal electric field is

$$E_n=-\frac{\partial V}{\partial n}.$$

Thus specifying normal field or surface-charge information can provide a Neumann-type boundary condition.

The general uniqueness properties of Poisson and Laplace boundary-value problems are especially useful here: once a candidate potential satisfies the electrostatic field equation and the complete physical boundary conditions, it is not necessary that the candidate have been obtained by direct charge summation.

Method of images

The method of images exploits uniqueness. Fictitious charges are placed outside the physical solution region so that their combined potential satisfies the actual conductor boundary conditions.

The image charges are a mathematical construction, not physical charges in the original problem.

Point charge above a grounded conducting plane

Place a real point charge $q$ at

$$z=a$$

above the infinite grounded plane

$$z=0.$$

For the region $z>0$, introduce a fictitious image charge

$$-q$$

at

$$z=-a.$$

The candidate potential is

$$\boxed{ V(x,y,z)=\frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{\sqrt{x^2+y^2+(z-a)^2}}

\frac{1}{\sqrt{x^2+y^2+(z+a)^2}} \right] }$$

for $z>0$ away from the real charge.

On the grounded plane $z=0$, the two distances are equal, so the two terms cancel:

$$\boxed{V(x,y,0)=0}.$$

The candidate therefore satisfies the conductor boundary condition.

Inside the physical region $z>0$, the image charge lies outside the domain. The potential obeys the correct Poisson equation at the real charge and Laplace's equation elsewhere in the charge-free region. By uniqueness, this candidate is the physical electrostatic potential above the grounded plane.

After the potential is known, the electric field follows from

$$\boxed{\mathbf E=-\nabla V}.$$

Why boundary-value reasoning matters

Conductors, cavities, electrodes, and material interfaces constrain electric fields globally. The important modeling sequence is therefore

1. identify the charge distribution in the region;
2. write Poisson's or Laplace's equation;
3. translate the physical boundaries into conditions on $V$ or its normal derivative;
4. find any solution satisfying both the differential equation and the boundary data;
5. use uniqueness to justify that solution.

Electrostatics is one application of the broader Poisson-Laplace boundary-value framework, with charge density supplying the source and electric potential supplying the scalar field.