# Method of images and Laplace Equation

In the method of images Section in the Griffiths' Electrodynamics A classic image problem is used to illustrate the significance of the of the uniqueness theorem.the whole point is about if a solution that satisfies the boundary condition, then it is the only solution. it doesn't matter how got to the solution. but the problem is about a charge free region, so we are solving Laplace equation.

the solution we obtain is V(x,y,z) = kq ($$\frac{1}{\sqrt(x^2+y^2+(z-d)^2)}- \frac{1}{\sqrt(x^2+y^2+(z+d)^2)}$$)
in this problem the boundary conditions are V=0 when z=0 and V=0 when $$x^2+y^2+z^2$$ approaches to infinity. ( V is potential at the point (x,y,z))

so, at both extremes the potential is zero. by the uniqueness theorem the extreme values should occur at boundaries since the Laplace equation does not tolerates local maxima and minima inside the region of concern. but the solution we have obtained above does have a maximum within the boundaries since the function have the value zero both in the origin and in infinity and taking the partial derivatives showed the function does have a maximum in fact an absolute maximum, but this contradicts the theory of uniqueness how is it possible? how can there be a maximum potential inside the boundaries when it should occur only at the boundaries?

There's a charge inside the domain, so the governing equation is not Laplace equation for the potential, but the Poisson equation, $$-\nabla^2 v = \frac{\rho}{\varepsilon} \ .$$ with a point source in $$\mathbf{r}_0 = d \, \mathbf{\hat{z}}$$ that can be modeled as a Dirac's delta.

Remarks.

• you're referring to the extreme conditions for a Laplace equation, namely the homogeneous version of a Poisson equation, not for Poisson equation: thus, these conditions are not true for the equations describing the potential field in a region of space containing the electric charge

• if you cut out the charge from the domain (prescribing the proper boundary conditions on the boundaries of the cut), there's no charge in the domain and thus the potential field is governed by a Laplace equation, supplied with the proper boundary conditions. The proper boundary conditions may be a Dirichlet condition prescribing the value of the potential on the boundary: to get an equivalent problem, you need to prescribe the right value, i.e. the value you get solving the problem governed by the Poisson equation, without any cut-out.

For this kind of problem, governed by a Laplace equation, all the extreme conditions for the Laplace equation hold, i.e. minimum and maximum values of the unknown field are on the boundaries.

In a stationary condition, the electric field is irrotational, $$\mathbf{\nabla} \times \mathbf{e} = \mathbf{0}$$, so that it can be written as a gradient of a potential $$\mathbf{e} = - \nabla v$$. Gauss' law for the electric field in vacuum reads

$$\frac{\rho}{\varepsilon_0} = \nabla \cdot \mathbf{e} = - \nabla^2 v \ .$$

Thus, the governing equation is not a Lagrange equation (for which the extreme values occur at the boundary) but it's the Poisson equation

$$- \nabla^2 v = \frac{\rho}{\varepsilon_0} \ ,$$

being $$\rho(\mathbf{r})$$ the charge density. A point charge in $$\mathbf{r}_0$$ with electric charge $$q$$ can be represented by the density (generalized) function $$\rho(\mathbf{r}) = q \delta(\mathbf{r}- \mathbf{r}_0) \ .$$

• Thankyou. Yes the charge is inside the boundaries so it's poisson equation we are solving. Commented May 23 at 15:43