Find the general solution to Laplace's equation (∇²V = 0) for V(r) in spherical coordinates.

Laplace's equation in spherical coordinates $(r, \theta, \phi)$ is:

$$\nabla^2 V = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial V}{\partial r} \right) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta} \left( \sin\theta \frac{\partial V}{\partial \theta} \right) + \frac{1}{r^2 \sin^2\theta} \frac{\partial^2 V}{\partial \phi^2} = 0$$

Given that $V$ depends only on $r$, the partial derivatives with respect to $\theta$ and $\phi$ are zero ($\frac{\partial V}{\partial \theta} = 0$ and $\frac{\partial V}{\partial \phi} = 0$). The equation simplifies to:

$$\frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial V}{\partial r} \right) = 0$$

Multiply by $r^2$ and integrate with respect to $r$:

$$\frac{\partial}{\partial r} \left( r^2 \frac{\partial V}{\partial r} \right) = 0$$ $$r^2 \frac{\partial V}{\partial r} = A$$

where $A$ is an integration constant.

$$\frac{\partial V}{\partial r} = \frac{A}{r^2}$$ $$V(r) = \int \frac{A}{r^2} dr = A \int r^{-2} dr = A \left( -\frac{1}{r} \right) + B$$

Let $C_1 = -A$ and $C_2 = B$.
The general solution in spherical coordinates for $V(r)$ is:

$$\boxed{\mathbf{V(r) = \frac{C_1}{r} + C_2}}$$

Case 2: Cylindrical Coordinates

Laplace's equation in cylindrical coordinates $(r, \phi, z)$ is:

$$\nabla^2 V = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2} = 0$$

Given that $V$ depends only on $r$, the partial derivatives with respect to $\phi$ and $z$ are zero ($\frac{\partial V}{\partial \phi} = 0$ and $\frac{\partial V}{\partial z} = 0$). The equation simplifies to:

$$\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) = 0$$

Multiply by $r$ and integrate with respect to $r$:

$$\frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) = 0$$ $$r \frac{\partial V}{\partial r} = A$$

where $A$ is an integration constant.

$$\frac{\partial V}{\partial r} = \frac{A}{r}$$ $$V(r) = \int \frac{A}{r} dr = A \ln(r) + B$$

Let $C_1 = A$ and $C_2 = B$.
The general solution in cylindrical coordinates for $V(r)$ is:

$$\boxed{\mathbf{V(r) = C_1 \ln(r) + C_2}}$$