Electrostatics - Potential near conductor using Kelvins method of image charge

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.

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.