# Potential Inside and Outside of a Charged Spherical Shell

• jkthejetplane
In summary, the poster is seeking help on a question involving Legendre polynomials and surface charge distribution. They initially thought they understood the problem but became confused after looking at an example in the book. They are unsure how to cancel out any Legendre polynomials without knowing the surface charge distribution. The suggested solution is to consider two simple cases for a better understanding. However, without knowing the surface charge distribution and potential, one cannot find a solution.
jkthejetplane
Homework Statement
I need help on this question that i thought i understood but i really dont. A semi similar example in the book started to spark some brain juice but i ended up more confused than when i started.
Relevant Equations
Legendre polynomials

So here was my first go around at it:

At first it made sense in my head but don't think my process is correct. Then i noticed the example in the book:

I guess the reasoning isn't 100% there in my head and if i don't have an actual σ, how will i cancel out any legendre polynomials due to orthogonality?

Last edited by a moderator:
jkthejetplane said:
Homework Statement:: I need help on this question that i thought i understood but i really dont. A semi similar example in the book started to spark some brain juice but i ended up more confused than when i started.
Relevant Equations:: Legendre polynomials

View attachment 280846
So here was my first go around at it:
View attachment 280849
At first it made sense in my head but don't think my process is correct. Then i noticed the example in the book:
View attachment 280850
View attachment 280851
View attachment 280852
I guess the reasoning isn't 100% there in my head and if i don't have an actual σ, how will i cancel out any legendre polynomials due to orthogonality?
You cannot cancel any Legendre polynomials without knowing σ(θ). Take two simple cases that you can do in your head.
1. Uniform density over the sphere implies a 1/r potential and only P0 is non-zero.
2. σ(θ)=σ0cosθ implies a 1/r2 potential and only P1 is non-zero.

In short, if you don't know the surface charge distribution and you don't know the potential, you don't know nuttin'.

But you know the surface-charge distribution,
$$\sigma(\vartheta)=\sigma_0 \Theta(0 \leq \vartheta \leq \pi/2).$$
Now you can just use the solution for the coefficients posted in #1.

vela and Orodruin

## 1. What is the potential inside a charged spherical shell?

The potential inside a charged spherical shell is constant and does not depend on the distance from the center of the shell. This is because the electric field inside a conductor is zero, and therefore there is no work done in moving a charge from one point to another inside the shell.

## 2. How is the potential outside a charged spherical shell calculated?

The potential outside a charged spherical shell is calculated using the equation V = kQ/r, where k is the Coulomb constant, Q is the charge on the shell, and r is the distance from the center of the shell. This equation applies for points outside the shell, but still within the electric field of the shell.

## 3. What happens to the potential at the surface of a charged spherical shell?

The potential at the surface of a charged spherical shell is equal to the potential of the shell itself. This is because the surface of a conductor is an equipotential surface, meaning that all points on the surface have the same potential.

## 4. How does the potential inside a charged spherical shell change if the charge on the shell is doubled?

The potential inside a charged spherical shell is independent of the charge on the shell. Therefore, doubling the charge on the shell will not change the potential inside the shell.

## 5. Can the potential inside a charged spherical shell ever be negative?

No, the potential inside a charged spherical shell will always be positive or zero. This is because the potential is defined as the amount of work done in moving a unit positive charge from infinity to a point, and it is not possible to have a negative amount of work done.

Replies
2
Views
2K
Replies
1
Views
7K
Replies
3
Views
1K
Replies
7
Views
2K
Replies
6
Views
1K
Replies
4
Views
3K
• Classical Physics
Replies
3
Views
645
Replies
14
Views
10K
• Electromagnetism
Replies
14
Views
1K