Electrostatics - Potential near conductor using Kelvins method of image charge

Question

As the figure shows, I have a 2D image of a conductor seen from the front. It is located a height, h1, above a grounded plane of infinite area. The conductor has a voltage, Vc, as shown. Around the conductor is air. The conductor has a radius, r, but is so narrow that it does not effect the forming of the electrical field. The ground plane however will (as shown in the figure) effect how the electric field forms.

Objective: Find the potential, Vp, at some location outside the conductor. The location is marked in the figure as a red cross and is located as indicated by d and h2.

I have seen others solve this with the method of image charge, but my problem is that only the potential/voltage of the conductor is known, and not the charge. I don't see how one can go from voltage to charge, and especially since the charge will depend on the length on the conductor, witch is not relevant in this 2D scenario. Thank you! (:

Figure 1:

Figure 2:

Answer 1

You are given the potentials at two surfaces. These are boundary conditions. They are sufficient to solve the problem without knowing the charge.

The question seems to suggest that $r \ll h$, so the charge distribution on the conductor, and the potential outside it, could be assumed to be spherically symmetric. The potential at any point is then the sum of the potentials due to the object charge and its image.

Answer 2

According to Gauss's law in the free space

$$ \mathbf{\nabla} \cdot \mathbf{E} = \mathbf{0}$$

since there there are no charges. However,

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

Finally, we got Laplace's equation:

$$\Delta V = 0$$

However, you could notice that for every $G$ verifying $\Delta G = 0$, then $V+G$ will be a solution of Laplace's equation. Thus, you must specify boundary conditions, for example Dirichlet's condition, which consists in specifying the value of $V$ along the boundary of the domain. Here, the domain is the half-plan $\{z>0\}$, without the conductor. Then, it happens that if you find a solution $V$ of Poisson's law which satisfies the boundary conditions, then $-\nabla V$ will be the real electric field. Thus, you can try to find such a solution.

Let's forget the grounded plane, and add a similar conductor with opposite charge as you did in Figure 2. Using the symmetry of the situation, you can see that the potential for $z=0$ is null. Plus, $V$ is still solution of Poisson's law. Since $V$ satisfies the same boundary conditions as with the grounded plane, and is solution of Poisson's law, you can affirm that the electric field, and the potential, for $z>0$ is the same in the two situations.

Now, using the potential created by a single conductor, you can find what the potential $V_p$ is.