# Conceptual Laplace's/Poisson's Equation Solution Questions

• Dobuskis
In summary, by redefining the radial variable r as R2/r, a solution to Laplace's equation inside of a cylindrical region with radius R can be obtained for outside of R. This solution can also be used to construct a solution to Poisson's equation for a line charge inside the cylinder by using a linear combination of the solutions in the previous part. This is also related to the method of images, which is based on the uniqueness theorem. By applying the method of images, a solution for outside of the cylinder can be obtained by subtracting the R2/r solution from the r solution, and a solution for inside can be obtained by subtracting the r solution from the R2/r solution.
Dobuskis

## Homework Statement

-You are given a solution to Laplace's equation inside of a cylindrical region radius R.
-Show that by redefining the radial variable r as R2/r you get a solution for outside of R.
-Grounded conducting cylinder at r=R. Using a linear combination of the solutions in the previous part, construct a solution to Poisson's equation for a line charge inside the cylinder. Do the same for outside.
-What does this have to do with the method of images.

## Homework Equations

General Laplace's and Poisson's equations
ΔF = 0, ΔF = f

## The Attempt at a Solution

I spent a while trying to check the new solution by plugging it into Laplace's equation, got a bunch of chain rule terms that made something I didn't recognize, then gave up for a while. Later I realized I might be overthinking/moving in the wrong direction.
-range of R2/r with r>R is (0,R), thus the radial inputs have the same range (0,R), thus if one is a solution to Laplace's equation in its region then the same goes for the other.
-Since solutions to Laplace's equation are also solutions to Poisson's equation then and linear combination of the two previous equations is a solution. By the method of images for a conducting surface and a line charge, the solution is the object solution minus the image solution. Thus outside the cylinder is the R2/r solution minus the r solution, for inside the r solution minus the R2/r solution.
-This is the method of images for a cylindrical surface.

Dobuskis said:
range of R2/r with r>R is (0,R), thus the radial inputs have the same range (0,R), thus if one is a solution to Laplace's equation in its region then the same goes for the other
That cannot be a sufficient argument. If f is any function satisfying f(0)=f(R)=1 then rf(r) also has range (0,R), but that does not make rf(r) a solution in the range (0,R).

You know that V(r) = 0 at r = R. What is V(R2/r) at r = R? Think "Uniqueness Theorem". The method of images is one of its applications.

## 1. What is Laplace's/Poisson's equation?

Laplace's/Poisson's equation is a partial differential equation that describes the relationship between the potential of a scalar field and its sources. It is used in many areas of physics, including electrostatics, fluid dynamics, and heat transfer.

## 2. What is the difference between Laplace's and Poisson's equation?

Both equations are similar, but the main difference is that Laplace's equation assumes a zero source term, while Poisson's equation includes a non-zero source term. This means that Poisson's equation can account for the influence of external sources on the potential of the field.

## 3. How is Laplace's/Poisson's equation solved?

Laplace's/Poisson's equation can be solved analytically or numerically. Analytical solutions involve using mathematical techniques to find an exact solution, while numerical solutions involve using algorithms to approximate a solution. Boundary conditions must also be specified in order to solve the equation.

## 4. What are some applications of Laplace's/Poisson's equation?

Laplace's/Poisson's equation has many applications in physics, engineering, and mathematics. Some examples include calculating electric potential in electronic circuits, modeling heat transfer in materials, and analyzing fluid flow in pipes or channels.

## 5. What are some limitations of Laplace's/Poisson's equation?

Laplace's/Poisson's equation assumes that the field is steady-state, meaning that it does not change over time. This limits its applicability in dynamic systems. Additionally, it assumes that the field is continuous, which may not be true in certain situations. It also does not account for quantum effects, which may be important in some systems.

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