If we place a charge $q$ over an infinite grounded plane we can find the potential in all points of space using the method of images and find the induced surface charge on the plane.

However, how does this case differ when instead of a point charge we have a dipole $p=qa$?

My intuition tells me that, through the superposition principle and assuming an ideal dipole (separation between charges is small), we can calculate the potential as approximately the double of the previous cases. Thus we could theoretically place an image dipole at the other side such that the potential is zero on the surface of where the plane used to be.

However, I'm not entirely sure if this is the best approach to this situation, or if it would be better to find a general solution using spherical harmonics through Poisson equation.


You can indeed use an image dipole to solve for the potential above the conductive plane. It doesn't even have to be an ideal dipole for it to work.

Or even more generally, for any arbitrary charge distribution $\rho(x,y,z)$ above the ground plane, you can solve for the potential $V(\mathbf x)$ using an image charge distribution mirrored with respect to the plane with the opposite sign, i.e. an image charge distribution $\rho_i(\mathbf x) \equiv -\rho(x,y,-z)$ (I've assumed that the conductor coincides with the $xy$ plane).

Why? well the image method is nothing but a clever way to exploit the uniqueness theorem for the solution of the Poisson equation. We simply need to show that for the actual problem (charge + conductor), and the fictitious problem (charge + image charge without the conductor):

1- The PDE governing the potential at $z\geq0$ is the same.
2- The boundary conditions for the potential in this region is also the same.

If both conditions are true, the potential function of the two problems has to be identical by the uniqueness theorem. The first point is fairly obvious. The potential obeys the same Poisson equation in both problems $\nabla^2V(x,y,z)=-\rho(x,y,z)/{\epsilon_0}$ at $z\geq0$.

The second point is also true. In the actual problem (charge+conductor) the boundary conditions are $V(x,y,0)=0$ and the potential vanishing at infinity. In the fictitious problem with the image charge distribution, the potential on the location of the conductor is: $$V(x,y,0)=\frac{1}{4\pi\epsilon_0}\int dV' \frac{\rho(\mathbf x')}{|\mathbf x' - \mathbf x|}$$ $$=\frac{1}{4\pi\epsilon_0}\int dV' \frac{\rho(x',y',z')+\rho_i(x',y',z')}{\Big[(x'-x)^2+(y'-y)^2+z'^2 \Big]^{1/2}}$$ $$=\frac{1}{4\pi\epsilon_0}\int dV' \frac{\rho(x',y',z')-\rho(x',y',-z')}{\Big[(x'-x)^2+(y'-y)^2+z'^2 \Big]^{1/2}}$$ $$=\frac{1}{4\pi\epsilon_0}\int_{z'\geq 0} dV' \frac{\rho(x',y',z')}{\Big[(x'-x)^2+(y'-y)^2+z'^2 \Big]^{1/2}}$$ $$\qquad -\frac{1}{4\pi\epsilon_0}\int_{z'\leq 0} dV' \frac{\rho(x',y',-z')}{\Big[(x'-x)^2+(y'-y)^2+z'^2 \Big]^{1/2}}$$ Now let's just change variables in the second integral as $z'\to -z'$. You can easily see that it simply turns into the first integral, resulting in: $$V(x,y,0)=0$$ In other words, the boundary conditions of the potential are also the same for the two problems, meaning that you can calculate the potential using the easier (fictitious) problem, and simply use its result for the actual problem (at $z\geq 0 $).

  • $\begingroup$ Thank you, your answer was very complete and it really cleared by doubts about using it for arbitrary charge distributions. I'll work some cases to practice. $\endgroup$ – Charlie Feb 7 '19 at 5:17

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