# Finding potential for a sphere, when to use laplace equation

• cshanny
In summary, the conversation discusses a problem involving a spherical shell with a specified charge density over its surface. The goal is to find the resulting potential inside and outside the sphere. The suggested method is to use the Laplace equation and separation of variables. The reason for this is that the charge is on the surface of the sphere, making the electric field inside equal to zero. The conversation also mentions the possibility of using Gauss' law, but it is noted that the assumption of a purely radial and constant electric field is incorrect. Alternative methods, such as using Green's function, are also mentioned.
cshanny

## Homework Statement

This is example 3.9 in Griffiths Electrodynamics.
"A specified charge density σ(θ) (inclination angle) is glued over the surface of a spherical shell of radius R. Find the resulting potential inside and outside the sphere." The problem suggests that although it is possible to compute this by using the integral for potential V=(1/4πε)int[σ(θ)dA], it is much easier to do so by separation of variables.[/B]

## Homework Equations

The solution to this example uses separation of variables, using the general solution to Laplace's equation in spherical coordinates.

## The Attempt at a Solution

At the suggestion of the book, I correctly found the potential inside and outside of the sphere using the Laplace equation. My question is what in the problem signifies that this is most easily solved with Laplace equation and not, say, Gauss' law. I could certainly say that because the charge is on the surface of the sphere that the electric field inside is 0 and therefore the potential is constant. I could also find the electric field outside by integrating the charge density to find the total enclosed charge outside of the sphere (which yields zero...). I'm obviously missing something here.

The enclosed charge for a gaussian surface outside the sphere is not zero, it is the total charge of the sphere. This way you would have to argue by symmetry that the electric field must be radial and integrate to find the potential.

Sorry I left something out in my post. They actually give the charge distribution as kcosθ...so I calculated Qenclosed by integrating the charge density over the surface of the sphere. The fact that there's a sinθcosθ in the integral made it go to zero.

Also, I don't quite understand why Gauss' law is not applicable.

Gauss law is applicable. Your assumption that the field is purely radial and constant is not.

Got it. How else can I approach this problem?

What was wrong with solving the Laplace equation?

Posting again to make sure you see it rather than miss my edit.

Of course, there are different methods available to solve the Laplace equation. One of them would be to integrate to sum up the contributions from each part of the surface. (Aka, Green's function methods)

Nothing it worked just fine and I got the correct answer. I was just curious if there was any other obvious way of doing this.

## 1. What is the Laplace equation and when is it used in finding potential for a sphere?

The Laplace equation is a partial differential equation used to describe the potential (or voltage) in a region of space where there are no charges present. It is commonly used in electrostatics to determine the potential of a conductive sphere, as the sphere has no net charge and thus the Laplace equation applies.

## 2. How is the potential of a sphere calculated using the Laplace equation?

The potential of a sphere can be calculated by solving the Laplace equation with appropriate boundary conditions. For a sphere, the boundary conditions would be that the potential is constant on the surface of the sphere and approaches zero at infinity.

## 3. Can the Laplace equation be used for non-spherical objects?

Yes, the Laplace equation can be used for any object or region of space where there are no charges present. However, the boundary conditions and method of solving the equation may differ depending on the shape and characteristics of the object.

## 4. Are there any limitations to using the Laplace equation in finding potential for a sphere?

One limitation is that the Laplace equation assumes a steady-state system, meaning that the potential is not changing over time. Additionally, the Laplace equation does not take into account the effects of non-uniform charge distributions or any external fields present in the system.

## 5. Can the Laplace equation be used to find the potential inside a hollow sphere?

No, the Laplace equation can only be used to find the potential outside of a hollow sphere. Inside the hollow sphere, there will be a constant potential due to the conducting material of the sphere, but the Laplace equation does not apply in this case.

Replies
1
Views
803
Replies
4
Views
2K
Replies
11
Views
2K
Replies
7
Views
2K
Replies
7
Views
983
Replies
1
Views
2K
Replies
4
Views
2K
Replies
2
Views
308