Solving for electric potential using separation of variables

In summary, the conversation discusses how to find the surface charge density of a point charge between two concentric spherical shells using the delta function and the electric potential between the shells using the separation of variables method. The relevant equations for this problem are the jump discontinuity in the orthogonal part of the field and the general form of the solution of the Laplace equations in spherical coordinates. The individual's attempts at solving the problem are shown in images, with some corrections and suggestions given for improvement.
  • #1
nashed
58
5

Homework Statement


Given two two grounded concentric spherical shells with radii a,b (a<b) and a point charge q between them at a<r=R<b find:

1.The surface charge density of the point charge using the delta function, assume the charge is on the z axis

2.By using the separation of variables method find the electric potential between the shells.

Homework Equations


I have no idea how to insert mathematical equation in the forum but the most relevant equations are the jump discontinuity in the orthogonal part of the field and the general form of the solution of the Laplace equations in spherical coordinates

The Attempt at a Solution



In my first try I tried to solve for the whole volume at once which produced contradictions, in my second attempt I divided the volume and solved for each volume seperately but I got stuck on the orthogonality condition for the Legendre polynomials, the precise attempt is shown in the images

HW solution attempt_1.jpg
HW solution attempt_2.jpg
HW solution attempt_3.jpg
HW solution attempt_4.jpg
HW solution attempt_5.jpg
HW solution attempt_6.jpg
HW solution attempt_7.jpg
 
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  • #2
Your second approach should work. However, I noticed a few errors:

Page 1: Your expression for σ is incorrect. You want to write a surface charge density for the surface r = R. Thus, ∫σdA = q for this surface. There is no integration over r here. Think about the expression for dA for this surface.

Page 4: Your powers of r in ##\Phi_I## and ##\Phi_{II}## don't appear to be correct. For example, see eq. 14 here http://www.luc.edu/faculty/dslavsk/courses/phys301/classnotes/laplacesequation.pdf

Page 4: Check the signs in the equation at the bottom of the page for the boundary condition involving R+ and R-. Recall that ##\vec{E} = -\nabla \Phi##. ( I don't understand the equation in the middle of the page, just before "in Volume II".)

Page 7: You forgot the ##\delta_{l,l'}## that occurs when doing the integral on the left at the top of the page.

There is another boundary condition that you can use. What can you say about the continuity of ##\Phi## at r = R when ##\theta \neq 0##?
 

1. What is the concept of separation of variables in solving for electric potential?

The separation of variables method involves breaking down a complex equation into smaller, simpler equations that can be solved individually. In the context of solving for electric potential, this means separating the variables of distance and charge into separate equations to make the problem more manageable.

2. How is separation of variables used to solve for electric potential in a system?

To solve for electric potential using separation of variables, the Laplace equation is used, which relates the potential to the charge distribution in the system. By separating the variables of distance and charge, the Laplace equation can be solved for each variable separately and then combined to find the overall electric potential.

3. What are the advantages of using separation of variables in solving for electric potential?

Separation of variables allows for the breaking down of a complex problem into smaller, more manageable parts. This can make the problem easier to solve and also allows for a better understanding of the relationship between different variables in the system.

4. What are some limitations of using separation of variables in solving for electric potential?

One limitation of using separation of variables is that it may not be applicable to all types of problems. In some cases, the equations may not be separable or the separation process may result in a more complex solution. Additionally, separation of variables may not work for systems with non-uniform charge distributions.

5. Can separation of variables be used for three-dimensional systems?

Yes, the separation of variables method can be extended to three-dimensional systems. However, the equations become more complex and may not have an exact analytical solution. In these cases, numerical methods may be used to approximate the solution.

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