Find energy of electrostatic field

  • Thread starter Petar Mali
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  • #1
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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?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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I think you have no mistake with your calculation and there is perhaps typo in the book.
 
  • #3
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Is there some other idea, other way, to solve this and check the result? Maybe with potentials?
 
  • #4
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I dont 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 doesnt mention the material. In region (3), there is a field here. Therefore, I think W_3 must not be zero.
 
  • #5
nrqed
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I dont 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 doesnt 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
nrqed
Science Advisor
Homework Helper
Gold Member
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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?

Homework Statement





Homework Equations





The Attempt at a Solution

Al your steps look right (except for your sign of the ln term). So the book is incorrect, indeed, as ApexofDE pointed out
 
  • #7
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6


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
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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
699
6


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
290
0


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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