What Is the Energy of an Electrostatic Field Between Concentric Spheres?

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Homework Help Overview

The problem involves calculating the energy of an electrostatic field between two concentric spheres with a given charge density. The charge density is defined as \(\rho=\frac{\alpha}{r^2}\), and the spheres have radii \(R_1\) and \(R_2\). The original poster presents their calculations and compares them to a solution provided in a textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use Gauss's law to derive the electric field in different regions and subsequently calculate the electrostatic energy. Some participants question the correctness of the textbook's solution and suggest that there may be a typo. Others propose exploring the problem using potentials as an alternative method.

Discussion Status

The discussion is ongoing, with participants examining the calculations provided by the original poster and debating the validity of the textbook's solution. There is a recognition of potential errors in the textbook, and some participants express agreement with the original poster's approach while others suggest verifying the results through different methods.

Contextual Notes

Participants note that the calculation of electrostatic energy typically requires integration over all space, which raises questions about the treatment of the electric field in region 3. There is also mention of boundary conditions that must be satisfied in the context of using Poisson's and Laplace's equations.

Petar Mali
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Homework Statement


The space between the two concentric spheres is charged by spatial density of charge [tex]\rho=\frac{\alpha}{r^2}[/tex]. The radius of spheres are [tex]R_1,R_2[/tex]. Integral charge is [tex]Q[/tex]. Find energy of electrostatic field.


Homework Equations


Gauss law

[tex]\oint_S\vec{E} \cdot d{\vec{S}}=\frac{q}{\epsilon_0}[/tex]

[tex]W_E=\frac{1}{2}\epsilon_0\int_VE^24\pi r^2dr[/tex]



The Attempt at a Solution



Using Gauss law I get

[tex]E^{(1)}=0[/tex], for [tex]r<R_1[/tex]

[tex]E^{(2)}(r)=\frac{1}{\epsilon_0}\cdot \frac{Q}{4\pi(R_2-R_1)}\frac{r-R_1}{r^2}[/tex]
for [tex]R_1\leq r \leq R_2[/tex]

[tex]E^{(3)}(r)=\frac{Q}{4\pi\epsilon_0r^2}[/tex], for [tex]r>R_2[/tex]


And get [tex]W_E^{(1)}=0[/tex]

[tex]W_E^{(2)}=\frac{Q^2}{8\pi\epsilon_0(R_2-R_1)}(1+\frac{2R_1}{R_2-R_1}ln\frac{R_2}{R_1}+\frac{R_1}{R_2})[/tex]

[tex]W_E^{(3)}=\frac{Q^2}{8\pi\epsilon_0R_2}[/tex]

This is my solution.

Final solution from book is

[tex]W_E^{(1)}=W_E^{(3)}=0[/tex]

[tex]W_E^{(2)}=\frac{Q^2}{4\pi\epsilon_0(R_2-R_1)}(1+\frac{2R_1}{R_2-R_1}ln\frac{R_2}{R_1})[/tex]

Where I make a mistake?
 
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I think you have no mistake with your calculation and there is perhaps typo in the book.
 


Is there some other idea, other way, to solve this and check the result? Maybe with potentials?
 


I don't understand. When we calculate the electrostatic energy, we must integrate in all space (from zero to infinity) and the formula depends only in electric field and doesn't mention the material. In region (3), there is a field here. Therefore, I think W_3 must not be zero.
 


ApexOfDE said:
I don't understand. When we calculate the electrostatic energy, we must integrate in all space (from zero to infinity) and the formula depends only in electric field and doesn't mention the material. In region (3), there is a field here. Therefore, I think W_3 must not be zero.

Sorry, my mistake. I had misread the question. They do ask for the energy of the electrostatic field, and you are right that it is then not zero in region 3. I removed my post.
 


Petar Mali said:

Homework Statement


The space between the two concentric spheres is charged by spatial density of charge [tex]\rho=\frac{\alpha}{r^2}[/tex]. The radius of spheres are [tex]R_1,R_2[/tex]. Integral charge is [tex]Q[/tex]. Find energy of electrostatic field.


Homework Equations


Gauss law

[tex]\oint_S\vec{E} \cdot d{\vec{S}}=\frac{q}{\epsilon_0}[/tex]

[tex]W_E=\frac{1}{2}\epsilon_0\int_VE^24\pi r^2dr[/tex]


The Attempt at a Solution



Using Gauss law I get

[tex]E^{(1)}=0[/tex], for [tex]r<R_1[/tex]

[tex]E^{(2)}(r)=\frac{1}{\epsilon_0}\cdot \frac{Q}{4\pi(R_2-R_1)}\frac{r-R_1}{r^2}[/tex]
for [tex]R_1\leq r \leq R_2[/tex]

[tex]E^{(3)}(r)=\frac{Q}{4\pi\epsilon_0r^2}[/tex], for [tex]r>R_2[/tex]


And get [tex]W_E^{(1)}=0[/tex]

[tex]W_E^{(2)}=\frac{Q^2}{8\pi\epsilon_0(R_2-R_1)}(1+\frac{2R_1}{R_2-R_1}ln\frac{R_2}{R_1}+\frac{R_1}{R_2})[/tex]

[tex]W_E^{(3)}=\frac{Q^2}{8\pi\epsilon_0R_2}[/tex]

This is my solution.

Final solution from book is

[tex]W_E^{(1)}=W_E^{(3)}=0[/tex]

[tex]W_E^{(2)}=\frac{Q^2}{4\pi\epsilon_0(R_2-R_1)}(1+\frac{2R_1}{R_2-R_1}ln\frac{R_2}{R_1})[/tex]

Where I make a mistake?
Al your steps look right (except for your sign of the ln term). So the book is incorrect, indeed, as ApexofDE pointed out
 


nrqed said:
Al your steps look right (except for your sign of the ln term). So the book is incorrect, indeed, as ApexofDE pointed out

In case the consensus is not enough, I'll add that I'm in agreement that the book is wrong, and the OPs approach is correct, and the OP's math is either correct or has only a slight error. I did not check the energy in the central region, but all other calculations look correct.
 


nrqed said:
Al your steps look right (except for your sign of the ln term). So the book is incorrect, indeed, as ApexofDE pointed out

Yes, this is mistake in writing :(

[tex] W_E^{(2)}=\frac{Q^2}{8\pi\epsilon_0(R_2-R_1)}(1+\frac{2R_1}{R_2-R_1}ln\frac{R_1}{R_2}+\frac{R_1}{R_2})[/tex]

Now is OK! Do you know how to solve this using vector and scalar potential [tex]A(\vec{r},t)[/tex], [tex]\varphi(\vec{r},t)[/tex]?
 


Petar Mali said:
Do you know how to solve this using vector and scalar potential [tex]A(\vec{r},t)[/tex], [tex]\varphi(\vec{r},t)[/tex]?

This is an electrostatics problem, so you can ignore the vector potential and just use the scalar potential. This problem is basically solvable in terms of voltage potential using Poisson's equation. Actually, Poisson's equation applies in the region of charge, while Laplace's equation applies both inside and outside the charged region. Boundary conditions at zero, infinity, R1 and R2 must be met.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/laplace.html

I haven't tried to work out this particular problem, but my gut feeling is that the way you have already done it is relatively easy and probably easier than using potentials. It would be good practice to verify your answer however.
 
Last edited:
  • #10


stevenb said:
This is an electrostatics problem, so you can ignore the vector potential and just use the scalar potential. This problem is basically solvable in terms of voltage potential using Poisson's equation. Actually, Poisson's equation applies in the region of charge, while Laplace's equation applies both inside and outside the charged region. Boundary conditions at zero, infinity, R1 and R2 must be met.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/laplace.html

I haven't tried to work out this particular problem, but my gut feeling is that the way you have already done it is relatively easy and probably easier than using potentials. It would be good practice to verify your answer however.

Ok! I get that! :) Thanks!
 

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