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

In summary, the conversation was about solving a problem involving the energy of an electrostatic field between two concentric spheres, with a spatial density of charge and a given integral charge. The Gauss law and the formula for the energy of an electrostatic field were used to calculate the electric field in each region and the resulting energy in each region. There was a discrepancy between the OP's solution and the final solution in a book, which was determined to be a typographical error. The question of using vector and scalar potentials to solve the problem was also raised, but it was concluded that the OP's method was sufficient.
  • #1
Petar Mali
290
0

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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  • #2


I think you have no mistake with your calculation and there is perhaps typo in the book.
 
  • #3


Is there some other idea, other way, to solve this and check the result? Maybe with potentials?
 
  • #4


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.
 
  • #5


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.
 
  • #6


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
 
  • #7


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.
 
  • #8


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]?
 
  • #9


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.
 
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  • #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!
 

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

What is electrostatic energy?

Electrostatic energy is the potential energy stored in an object or system due to the presence of electric charges. It is the energy that results from the interaction between these charges, and is typically measured in joules (J).

How do you calculate the energy of an electrostatic field?

The energy of an electrostatic field can be calculated using the equation U = kQ1Q2/r, where U is the energy in joules, k is the Coulomb's constant (8.99x10^9 Nm^2/C^2), Q1 and Q2 are the magnitudes of the charges in Coulombs, and r is the distance between the charges in meters.

What factors affect the energy of an electrostatic field?

The energy of an electrostatic field is affected by the magnitude of the charges involved, the distance between the charges, and the type of medium in which the charges are located. Additionally, the presence of other charges in the vicinity can also influence the energy of the field.

Why is the energy of an electrostatic field important?

The energy of an electrostatic field is important because it is the basis for many everyday phenomena, such as static electricity, electric discharge, and the behavior of electrically charged particles. It is also crucial in understanding and predicting the behavior of electric fields in various applications, such as in electronic devices and power systems.

Is the energy of an electrostatic field conserved?

Yes, the energy of an electrostatic field is conserved. This means that it cannot be created or destroyed, but can only be transferred or converted from one form to another. This is known as the law of conservation of energy, and it applies to all forms of energy, including electrostatic energy.

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