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A pair of any two opposing charges $+Q=\rho/d$ and $-Q=-\rho/d$ creates an electric field. The limit of the field when $d\to 0$ is known as the dipole. For any $x\in \mathbb R^3$, the dipole potential at the point $p\in S$ has the formwhere $$\Phi(x)=-\frac{1}{4\pi\sqrt{x_1^2+x_2^2+x_3^2}},\quad \Phi(x)=-(4\pi|x|)^{-1},\quad x\in \mathbb R^3,$$dp,\quad x\in\mathbb R^3,$$is the double-layerpotential of the whole distributionat x\in\mathbb R^3.$$u_{-}(x_0)=u(x_0)-2\pi \rho(x_0),\quad u_{+}(x_0)=u(x_0)-2\pi u_{+}(x_0)=u(x_0)+2\pi \rho(x_0),\quad x_0\in S,\qquad\qquad S,\qquad\qquad\qquad\qquad (2)$$Relations (2) are easy to show when the density \rho is constant. In this case u(x)is constant inside, outside and on the surface and the integral has a direct geometric interpretation. It is the solid angle subtended by the surface when seen from a point x\in\mathbb R^3. The computation in general case can be reduced to the case of constant density. Now, relations (2) can be viewed as integral equations w.r.t. the unknown density (assuming that the potential on the surface is known). The kernels of the equations have an integrable singularity and the equations can be solved via using the Fredholm approach. The function u(x) then gives a solution to the Dirichlet problem. Edit. See a nice little textbook by Arnold where he shows how to make the physical intuition rigorous in this problem. 1 The electrostatic intuition does lead to a correct mathematical formulation of the Dirichlet problem although it doesn't solve the latter. (Well, except for some special domains such as a ball or a halfspace when the Dirichlet problem admits a closed form solution.) Let's consider an electric charge distribution of two thin layers (one layer is positive and the other is negative) located along a closed surface S\subset\mathbb R^3. Assume that d>0 is the distance between charges along the normal n_p to the surface at point p. Let \rho\in C(S,\mathbb R) denote the distribution's density. A pair of any two opposing charges +Q=\rho/d and -Q=-\rho/d creates an electric field. The limit of the field when d\to 0 is known as the dipole. For any x\in \mathbb R^3, the dipole potential at the point p\in S has the form$$\frac{\rho}{d}\Phi(x-(p+n_pd/2))-\frac{\rho}{d}\Phi(x-(p-n_pd/2))=\rho\frac{\partial \Phi(x-p)}{\partial n_p} +o(1)\quad{\rm as\ \ } d\to0,\qquad(1)$$where$$\Phi(x)=-\frac{1}{4\pi\sqrt{x_1^2+x_2^2+x_3^2}},\quad x\in \mathbb R^3,$$is the fundamental solution of Laplace's equation. (1) gives the dipole potential of a single dipole at the point p\in S and the integral$$u(x)=\int_{S}\rho(p) \frac{\partial \Phi(x-p)}{\partial n_p} dp$$is the double-layer potential of the whole distribution at x\in\mathbb R^3. Now, a simple computation shows that u(x) is a harmonic function when x is not on the surface. It has a jump when x passes through S:$$u_{-}(x_0)=u(x_0)-2\pi \rho(x_0),\quad u_{+}(x_0)=u(x_0)-2\pi \rho(x_0),\quad x_0\in S,\qquad\qquad (2) where $u_{-}(x_0)$ ($u_{+}(x_0)$) is the limit from the interior (exterior) of the surface. Relations (2) are easy to show when the density $\rho$ is constant. In this case $u(x)$ is constant inside, outside and on the surface and the integral has a direct geometric interpretation. It is the solid angle subtended by the surface when seen from a point $x\in\mathbb R^3$. The computation in general case can be reduced to the case of constant density.