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

[Petar Mali]

Homework Statement

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

Homework Equations

Gauss law

\oint_S\vec{E} \cdot d{\vec{S}}=\frac{q}{\epsilon_0}

W_E=\frac{1}{2}\epsilon_0\int_VE^24\pi r^2dr

The Attempt at a Solution

Using Gauss law I get

E^{(1)}=0, for r<R_1

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

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


And get W_E^{(1)}=0

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})

W_E^{(3)}=\frac{Q^2}{8\pi\epsilon_0R_2}

This is my solution.

Final solution from book is

W_E^{(1)}=W_E^{(3)}=0

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})

Where I make a mistake?

 

[ApexOfDE]

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

 

[Petar Mali]

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

 

[ApexOfDE]

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.

 

[nrqed]

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.

 

[nrqed]

Homework Statement

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

Homework Equations

Gauss law

\oint_S\vec{E} \cdot d{\vec{S}}=\frac{q}{\epsilon_0}

W_E=\frac{1}{2}\epsilon_0\int_VE^24\pi r^2dr

The Attempt at a Solution

Using Gauss law I get

E^{(1)}=0, for r<R_1

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

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


And get W_E^{(1)}=0

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})

W_E^{(3)}=\frac{Q^2}{8\pi\epsilon_0R_2}

This is my solution.

Final solution from book is

W_E^{(1)}=W_E^{(3)}=0

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})

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

 

[stevenb]

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.

 

[Petar Mali]

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 :(

<br /> 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})<br />

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

 

[stevenb]

Do you know how to solve this using vector and scalar potential A(\vec{r},t), \varphi(\vec{r},t)?

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.

 

[Petar Mali]

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!