Expressing a potential inside a spherical shell as

In summary, The potential inside a spherical shell is given by V_{-}(x,y,z)= \frac{V_0}{R^2}(6z^2-3x^2-3y^2)P_n(\cos(\theta)) where \theta is the polar angle. The potential on the surface carries a surface charge density \sigma and is rotationally symmetric around the z-axis both inside and outside the shell. To express the potential inside the spherical shell using a Legendre Polynomial, first express V in terms of spherical coordinates and write all the trig functions in terms of cos θ. Then compare the resulting expression to the Legendre polynomials to obtain the simplified solution of \frac{3V_0r^
  • #1
dikmikkel
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Homework Statement


The potential inside a spherical shell is given by:
[itex]V_{-}(x,y,z)= \frac{V_0}{R^2}(6z^2-3x^2-3y^2)[/itex]
[itex]P_n(\cos(\theta ))[/itex] where [itex] \theta [/itex] is the polar angle.

The potential on the surface carries a surface charge density [itex]\sigma[/itex]. Besides this, ther's no other charges and no outher field. The potential is rotational symmetric around the z-axis inside and outside, and goes to 0 far away from the sphere.

b) express the potential inside the spherical shell using a LegendrePolynomial

Homework Equations


In spherical coordinates i have:
[itex]V(r,\theta ) = \sum\limits_{l=0}^{\infty}(A_lr^lP_l(\cos(\theta)) = V_0(\theta)[/itex]

The Attempt at a Solution


This is how far i made it. Now i suppose i could multiply it with [itex]P_{l'}(\cos( \theta ))\sin(\theta)[/itex] and integrate, but i can't figure out how to simplify it and extract the solution.
I'm aware that the functions' are orthogonal, but the integral of a sum is something I've never done before.
 
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  • #2
You're making the problem more complicated than you need to. First, express the given V in terms of spherical coordinates and write all the trig functions in terms of cos θ. Then compare what you get to the Legendre polynomials.
 
  • #3
Ha ha, yeah now i get it :) It's because normally we want to find the potential. But now i get it as the Legendre from 0 to 2, which makes sense by the order of the cartesian polynomial. So i am getting
[itex]\frac{3V_0r^2}{R^2}(3\cos(\theta)^2-1) [/itex]
 

1. What is a spherical shell?

A spherical shell is a three-dimensional shape that is formed by rotating a circle about its diameter. It is a hollow, spherical object with a defined inner and outer radius.

2. How can a potential be expressed inside a spherical shell?

A potential inside a spherical shell can be expressed using the formula V(r) = (Q / 4πε0) * (1/r), where V is the potential, Q is the charge of the shell, ε0 is the permittivity of free space, and r is the distance from the center of the shell.

3. What is the significance of expressing potential inside a spherical shell?

Expressing potential inside a spherical shell allows us to understand the electric field and potential inside the shell. It also helps us to calculate the work done in moving a charge from one point to another inside the shell.

4. Can the potential inside a spherical shell ever be zero?

Yes, the potential inside a spherical shell can be zero if the charge of the shell is zero or if the distance from the center of the shell is infinite.

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

Yes, the potential inside a spherical shell can be negative if the charge of the shell is negative. This indicates that the electric field inside the shell is directed towards the center of the shell.

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