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Ive been unfortunately trying to apply simple boundary conditions on your equation with no luck, because it's wrong. so First of all, let me correct you:

Laplaces equation for r dependance is: $\frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial \phi}{\partial r}) = 0$

$(r^2\frac{\partial \phi}{\partial r}) = C_{0}$

$(\frac{\partial \phi}{\partial r}) = \frac{C_{0}}{r^2}$

$\phi= \frac{C_{1}}{r} + \phi_{0}$

Poissons equation for electrostatics.

$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$

There are 2 ways you can think about setting $\rho =0$ here.

Line of reasoning 1:

Lets say I have the equation $$z=yx$$ If I say $$y=1$$ This equation reduces to $$z=x$$

By substituting y=1, we are finding what z is, in regions where y=1.

That is not to say, that y=1 always. We are just finding the value of z that satisfy the condition that y=1

The same can be said for poissons equation.

$$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$$

Reduces to laplaces equation:

$$\nabla^2 \phi = 0$$

Using the former as a guide, what does this equation actually mean? By substituting $\rho = 0$ into poissons equation. We are saying "what equation does the potential $\phi$ follow, in regions where the charge density is zero?"

That is not to say that $\rho = 0$ everywhere, but where $\rho$ IS zero, we know the potential must follow this equation. This is true for all distributions. Take the laplacian of any potential in a region where there is no charge density, and the result is zero!

If poissons equation did automatically yield zero, then that would be even more confusing that if it didn't! As we KNOW the potential of e.g a point charge isn't zero, and most of the functions domain $\rho=0$.

Here the solution we find from laplaces equation ONLY works in regions where $\rho=0$, you may hear the phrase "finding the potential on a specific domain"

To find the constants we need to enter boundary conditions. Like the field is continuous about the boundary between the charged sphere/outside of it, or that the potential vanishes at infinity.

Line of reasoning 2:

We have discussed setting $\rho$ to zero to find the potential in regions where the charge density is zero. But what about if the charge density is actually zero everywhere?

The same equation should equally apply!

$$\phi= \frac{C_{1}}{r} + \phi_{0}$$

Given $$\phi(\infty) = 0$$ We then have $$\phi_{0} = 0$$

Given we know $\rho=0$ everywhere, we have no domain restriction, this fact will allow us to solve for $C_{1}$

We now need to say that $$\phi(0) ≠ \infty$$ Which is pleasing given we are saying there is no charge.

Doing so gives:

$$C_{1} = 0$$

Meaning $$\phi = 0$$

Non uniqueness of maxwells equations:

Maxwells equations are not uniquely determined by the curl and divergence of the fields.

Infact, the equation $\nabla^2 \phi = 0$ where $\rho$ is infact zero everywhere, does not need to be zero. The only reason that it is zero, is due to our boundary conditions, and assumption about the symmetry of our potential

In general, Any harmonic function can be added to the potential that does not need to be a constant, and will equally satisfy maxwells equations. This is the homogenous solution to maxwells equations in electrostatics. The helmholtz theorem states that if the fields vanish at infinity, then there is a unique solution. This is the normal boundary condition that we impose on maxwells equations in order to find a unique solution, much like the field of a point charge.

Without it, there exists multiple fields for the same charge distribution! For example the homogenous wave equation that can be added to any distribution, that is not uniquely determined by the charges, but by the boundary conditions of the EM field! [I wonder what the true boundary conditions are!!,pretty exciting that there could be a background EM field existing independantly of charges]

Ive been unfortunately trying to apply simple boundary conditions on your equation with no luck, because it's wrong. so First of all, let me correct you:

Laplaces equation for $r-$dependance is: $$\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial \phi}{\partial r}\right) = 0\implies r^2\frac{\partial \phi}{\partial r} = C_0 \implies \frac{\partial \phi}{\partial r} = \frac{C_{0}}{r^2}\implies \phi= \frac{C_{1}}{r} + \phi_{0}$$

Poissons equation for electrostatics.

$$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$$

There are 2 ways you can think about setting $\rho =0$ here.

Line of reasoning 1:

Lets say I have the equation $$z=yx$$ If I say $$y=1$$ This equation reduces to $$z=x$$

By substituting y=1, we are finding what z is, in regions where y=1.

That is not to say, that y=1 always. We are just finding the value of z that satisfy the condition that y=1

The same can be said for poissons equation.

$$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$$

Reduces to laplaces equation:

$$\nabla^2 \phi = 0$$

Using the former as a guide, what does this equation actually mean? By substituting $\rho = 0$ into poissons equation. We are saying "what equation does the potential $\phi$ follow, in regions where the charge density is zero?"

That is not to say that $\rho = 0$ everywhere, but where $\rho$ IS zero, we know the potential must follow this equation. This is true for all distributions. Take the laplacian of any potential in a region where there is no charge density, and the result is zero!

If poissons equation did automatically yield zero, then that would be even more confusing that if it didn't! As we KNOW the potential of e.g a point charge isn't zero, and most of the functions domain $\rho=0$.

Here the solution we find from laplaces equation ONLY works in regions where $\rho=0$, you may hear the phrase "finding the potential on a specific domain"

To find the constants we need to enter boundary conditions. Like the field is continuous about the boundary between the charged sphere/outside of it, or that the potential vanishes at infinity.

Line of reasoning 2:

We have discussed setting $\rho$ to zero to find the potential in regions where the charge density is zero. But what about if the charge density is actually zero everywhere?

The same equation should equally apply!

$$\phi= \frac{C_{1}}{r} + \phi_{0}$$

Given $$\phi(\infty) = 0$$ We then have $$\phi_{0} = 0$$

Given we know $\rho=0$ everywhere, we have no domain restriction, this fact will allow us to solve for $C_{1}$

We now need to say that $$\phi(0) ≠ \infty$$ Which is pleasing given we are saying there is no charge.

Doing so gives:

$$C_{1} = 0$$

Meaning $$\phi = 0$$

Non uniqueness of maxwells equations:

Maxwells equations are not uniquely determined by the curl and divergence of the fields.

Infact, the equation $\nabla^2 \phi = 0$ where $\rho$ is infact zero everywhere, does not need to be zero. The only reason that it is zero, is due to our boundary conditions, and assumption about the symmetry of our potential

In general, Any harmonic function can be added to the potential that does not need to be a constant, and will equally satisfy maxwells equations. This is the homogenous solution to maxwells equations in electrostatics. The helmholtz theorem states that if the fields vanish at infinity, then there is a unique solution. This is the normal boundary condition that we impose on maxwells equations in order to find a unique solution, much like the field of a point charge.

Without it, there exists multiple fields for the same charge distribution! For example the homogenous wave equation that can be added to any distribution, that is not uniquely determined by the charges, but by the boundary conditions of the EM field! [I wonder what the true boundary conditions are!!,pretty exciting that there could be a background EM field existing independantly of charges]

Ive been unfortunately trying to apply simple boundary conditions on your equation with no luck, because it's wrong. so First of all, let me correct you:

Laplaces equation for r dependance is: $\frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial \phi}{\partial r}) = 0$

$(r^2\frac{\partial \phi}{\partial r}) = C_{0}$

$(\frac{\partial \phi}{\partial r}) = \frac{C_{0}}{r^2}$

$\phi= \frac{C_{1}}{r} + \phi_{0}$

Poissons equation for electrostatics.

$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$

There are 2 ways you can think about setting $\rho =0$ here.

Line of reasoning 1:

Lets say I have the equation $$z=yx$$ If I say $$y=1$$ This equation reduces to $$z=x$$

By substituting y=1, we are finding what z is, in regions where y=1.

That is not to say, that y=1 always. We are just finding the value of z that satisfy the condition that y=1

The same can be said for poissons equation.

$$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$$

Reduces to laplaces equation:

$$\nabla^2 \phi = 0$$

Using the former as a guide, what does this equation actually mean? By substituting $\rho = 0$ into poissons equation. We are saying "what equation does the potential $\phi$ follow, in regions where the charge density is zero?"

That is not to say that $\rho = 0$ everywhere, but where $\rho$ IS zero, we know the potential must follow this equation. This is true for all distributions. Take the laplacian of any potential in a region where there is no charge density, and the result is zero!

If poissons equation did automatically yield zero, then that would be even more confusing that if it didn't! As we KNOW the potential of e.g a point charge isn't zero, and most of the functions domain $\rho=0$.

Here the solution we find from laplaces equation ONLY works in regions where $\rho=0$, you may hear the phrase "finding the potential on a specific domain"

To find the constants we need to enter boundary conditions. Like the field is continuous about the boundary between the charged sphere/outside of it, or that the potential vanishes at infinity.

Line of reasoning 2:

We have discussed setting $\rho$ to zero to find the potential in regions where the charge density is zero. But what about if the charge density is actually zero everywhere?

The same equation should equally apply!

$$\phi= \frac{C_{1}}{r} + \phi_{0}$$

Given $$\phi(\infty) = 0$$ We then have $$\phi_{0} = 0$$

Given we know $\rho=0$ everywhere, we have no domain restriction, this fact will allow us to solve for $C_{1}$

We now need to say that $$\phi(0) ≠ \infty$$ Which is pleasing given we are saying there is no charge.

Doing so gives:

$$C_{1} = 0$$

Meaning $$\phi = 0$$

Non uniqueness of maxwells equations:

Maxwells equations are not uniquely determined by the curl and divergence of the fields.

Infact, the equation $\nabla^2 \phi = 0$ where $\rho$ is infact zero everywhere, does not need to be zero. The only reason that it is zero, is due to our boundary conditions, and assumption about the symmetry of our potential

In general, Any harmonic function can be added to the potential that does not need to be a constant, and will equally satisfy maxwells equations. This is the homogenous solution to maxwells equations in electrostatics. The helmholtz theorem states that if the fields vanish at infinity, then there is a unique solution. This is the normal boundary condition that we impose on maxwells equations in order to find a unique solution, much like the field of a point charge.

Without it, there exists multiple fields for the same charge distribution! For example the homogenous wave equation that can be added to any distribution, that is not uniquely determined by the charges, but by the boundary conditions of the EM field! [I wonder what the true boundary conditions are!!]

Ive been unfortunately trying to apply simple boundary conditions on your equation with no luck, because it's wrong. so First of all, let me correct you:

Laplaces equation for r dependance is: $\frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial \phi}{\partial r}) = 0$

$(r^2\frac{\partial \phi}{\partial r}) = C_{0}$

$(\frac{\partial \phi}{\partial r}) = \frac{C_{0}}{r^2}$

$\phi= \frac{C_{1}}{r} + \phi_{0}$

Poissons equation for electrostatics.

$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$

There are 2 ways you can think about setting $\rho =0$ here.

Line of reasoning 1:

Lets say I have the equation $$z=yx$$ If I say $$y=1$$ This equation reduces to $$z=x$$

By substituting y=1, we are finding what z is, in regions where y=1.

That is not to say, that y=1 always. We are just finding the value of z that satisfy the condition that y=1

The same can be said for poissons equation.

$$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$$

Reduces to laplaces equation:

$$\nabla^2 \phi = 0$$

Using the former as a guide, what does this equation actually mean? By substituting $\rho = 0$ into poissons equation. We are saying "what equation does the potential $\phi$ follow, in regions where the charge density is zero?"

That is not to say that $\rho = 0$ everywhere, but where $\rho$ IS zero, we know the potential must follow this equation. This is true for all distributions. Take the laplacian of any potential in a region where there is no charge density, and the result is zero!

If poissons equation did automatically yield zero, then that would be even more confusing that if it didn't! As we KNOW the potential of e.g a point charge isn't zero, and most of the functions domain $\rho=0$.

Here the solution we find from laplaces equation ONLY works in regions where $\rho=0$, you may hear the phrase "finding the potential on a specific domain"

To find the constants we need to enter boundary conditions. Like the field is continuous about the boundary between the charged sphere/outside of it, or that the potential vanishes at infinity.

Line of reasoning 2:

We have discussed setting $\rho$ to zero to find the potential in regions where the charge density is zero. But what about if the charge density is actually zero everywhere?

The same equation should equally apply!

$$\phi= \frac{C_{1}}{r} + \phi_{0}$$

Given $$\phi(\infty) = 0$$ We then have $$\phi_{0} = 0$$

Given we know $\rho=0$ everywhere, we have no domain restriction, this fact will allow us to solve for $C_{1}$

We now need to say that $$\phi(0) ≠ \infty$$ Which is pleasing given we are saying there is no charge.

Doing so gives:

$$C_{1} = 0$$

Meaning $$\phi = 0$$

Non uniqueness of maxwells equations:

Maxwells equations are not uniquely determined by the curl and divergence of the fields.

Infact, the equation $\nabla^2 \phi = 0$ where $\rho$ is infact zero everywhere, does not need to be zero. The only reason that it is zero, is due to our boundary conditions, and assumption about the symmetry of our potential

In general, Any harmonic function can be added to the potential that does not need to be a constant, and will equally satisfy maxwells equations. This is the homogenous solution to maxwells equations in electrostatics. The helmholtz theorem states that if the fields vanish at infinity, then there is a unique solution. This is the normal boundary condition that we impose on maxwells equations in order to find a unique solution, much like the field of a point charge.

Without it, there exists multiple fields for the same charge distribution! For example the homogenous wave equation that can be added to any distribution, that is not uniquely determined by the charges, but by the boundary conditions of the EM field! [I wonder what the true boundary conditions are!!,pretty exciting that there could be a background EM field existing independantly of charges]

Using the former as a guide, what does this equation actually mean? By substituting $\rho$ into poissons equation. We are saying "what equation does the potential $\phi$ follow, in regions where the charge density is zero?"

Infact, the equation $\nabla^2 \phi = 0$ where $\rho$ is infact zero everywhere, does not need to be zero. Your method forced it to be zero by assuming spherical symmetry[or maybe my r=0 boundary condition], but that isn't always the case.

Any harmonic function can be added to the potential that does not need to be a constant, and will equally satisfy maxwells equations. This is the homogenous solution to maxwells equations in electrostatics. The helmholtz theorem states that of the fields vanish at infinity, then there is a unique solution. This is the normal boundary condition that we impose on maxwells equations in order to find a unique solution, much like the field of a point charge.

Using the former as a guide, what does this equation actually mean? By substituting $\rho = 0$ into poissons equation. We are saying "what equation does the potential $\phi$ follow, in regions where the charge density is zero?"

Infact, the equation $\nabla^2 \phi = 0$ where $\rho$ is infact zero everywhere, does not need to be zero. The only reason that it is zero, is due to our boundary conditions, and assumption about the symmetry of our potential

In general, Any harmonic function can be added to the potential that does not need to be a constant, and will equally satisfy maxwells equations. This is the homogenous solution to maxwells equations in electrostatics. The helmholtz theorem states that if the fields vanish at infinity, then there is a unique solution. This is the normal boundary condition that we impose on maxwells equations in order to find a unique solution, much like the field of a point charge.

Without it, there exists multiple fields for the same charge distribution! For example the homogenous wave equation that can be added to any distribution, that is not uniquely determined by the charges, but by the boundary conditions of the EM field! [I wonder what the true boundary conditions are!!]