If potential at centre of a uniformly volume charged cube (side $a$) is $V⁰$, what is potential at its corner? If potential at centre of a uniformly volume charged cube is V⁰, what is potential at its corner?
Through internet, I came to knew that answer is V⁰/2, but everywhere it has been answered logically by putting 7 more imaginary similar cubes, then the corner becomes centre of the (2a) cube;
Now, they do: potential is proportional to charge/dimension,hence potential at the imaginary 2a cube's center is 4V⁰, and thus due to 1 cube it becomes V⁰/2.
The problem I have here is that I think potential is not directly related to dimension of the body, it is related to distance of the charges from the point where potential is calculated. Here can it be proved that as the side is doubled, distance of every small charge element of cube  from the centre is two times as that was before.
I mean the Charge/ side, thing to calculate potential by 2a, was just an analogy or logic, I doubt that it would be proved by rigrous mathematics. Please help..
 A: I have seen many others using the scaling argument in the question so I expect my solution has missed an obvious point! Hopefully, someone can point it out?
Poisson's equation for this case is:
$$\nabla^2\phi=\cases{\frac{\rho}{\epsilon\epsilon_0}, & \text{in cube}\\0, & \text{otherwise}}$$
To solve the problem we can solve this equation in both regions and match the boundary conditions. However, as the provided boundary region is in the cube and the point of interest is on the surface of the cube, we need only consider:
$$\nabla^2\phi=\frac{\rho}{\epsilon\epsilon_0}$$
In cartesian coordinates the Laplacian $\nabla^2\equiv\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$ so by inspection:
$$\phi=\underbrace{a_0xyz+a_1xy+a_2yz+a_3xz+a_4x+a_5y+a_6z+V^0}_\text{complementary solution}+\underbrace{\frac{\rho}{6\epsilon\epsilon_0}\left(x^2+y^2+z^2\right)}_\text{particular integral}$$
where the constant term in the complementary solution $V^0$ is the potential at the centre of the cube. If we assume no external fields are applied then the potential must have the same rotational symmetries as a cube and so $a_n=0$ for all $n$. Thus, the solution within the cube is:
$$\phi=V^0+\frac{\rho}{6\epsilon\epsilon_0}\left(x^2+y^2+z^2\right)$$
At the corner of the cube of side lengths $2a$ the potential is then:
$$\phi=V^0+\frac{\rho a^2}{2\epsilon\epsilon_0}$$
