Consider the integral form of Poisson's equation
$$\iiint_V(- \Phi(\vec x')\,4 \pi \delta(\vec x' – \vec x)\,+ \frac{1}{R} 4 \,\pi\, \rho(\vec x'))\,dv'=\iint_S(\,\Phi(\vec x')\,\frac{\partial(\frac{1}{R})}{\partial n'}\,-\frac{1}{R}\frac{\partial(\Phi(\vec x'))}{\partial n'})\,da'$$
Where surface integral is over the surface bounding V $,$
$\,\vec x'$ is the integration variable , $\,R=\frac{1}{|(\vec x-\vec x')|}$, $\, \Phi $ is the solution to Poisson equation,$\,n'$ is unit normal to S
- if $\vec x'$ lies inside V then the equation becomes:
$$\Phi(\vec x)=\iiint_V(\frac{\rho(\vec x')\,dv'}{R})+\,\,(\frac{1}{4\,\pi}\iint_S(\frac{1}{R}\frac{\partial(\Phi(\vec x')}{\partial n'}-\,\Phi(\vec x')\frac{\partial(\frac{1}{R}}{\partial n'})da')\,\,\,\,\,\,\,\, \rightarrow 1$$
Here first term on $RHS$ is the potential due to continuous charge distribution inside the volume V, next term is the surface term which can be interpreted as the sum of potential due to surface Charges at boundary and due to surface dipole moment density,which is determined by the charges outside V(boundary conditions).
Q1)Does this surface term also includes surface Charges induced by charge inside V?
Q2)When $\vec x$ lies outside V then $LHS$ becomes zero because delta function gives zero,thus we have lost information about $\Phi$ but since RHS of equation 1 equals zero implies that sum of potential due to charges inside V and surface should be zero,thus if there is point charge otside V then potential at any point outside V should be the potential due to this point charge only, also if there is no charges outside V then potential outside V should be zero. Is this correct?
Best Answer
In order to have a real surface charge density, you need a discontinuity in the normal electric field, that is, a discontinuity in the normal derivative of the potential. Similarly, a surface dipole layer requires a discontinuity in the potential. Nowhere in the mathematics you have written is such a discontinuity required.
Answering your question 2 first:
This means that interpreting the two terms as a surface charge and dipole layer means that you have enforced the condition that $\Phi(\vec x)$ is zero outside your volume. Since you enforce the condition that $\Phi = 0$ outside, $\Phi$ will be zero outside.
For your first question:
If you don't enforce the condition that $\Phi$ is zero outside, the equation is still correct. The coulomb integral will give the correct contribution for the potential of the charge inside, while the surface integrals will give the correct contribution for the charges outside. In that case, there is no surface charge or dipole layer on $S$ unless there is a physical surface charge or physical dipole. The surface terms can equally well contribute from other charges outside the interior volume.
For the case where there really is a surface charge and/or dipole layer, then it contains all the terms needed to satisfy Poisson's equation, including the induced charge. Otherwise it would not satisfy Poisson's equation.
I suggest looking at some simple cases where you know the complete potential. For example 1) a mathematical spherical surface and two point charges in vacuum, one inside and one outside, and 2) a conductor with a spherical cavity with a point charge inside.