Electrostatic potential energy - dielectric between concentric spheres

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Homework Statement


A conducting sphere of radius ##a## carries an initial charge ##q_o##. It is surrounded by another concentric sphere of radius ##b##. The space between the two spheres is filled with a dielectric of permittivity ##\epsilon## and conductivity ##\sigma##. Find the electrostatic potential energy stored in the system at time ##t##.


Homework Equations





The Attempt at a Solution


The electric field at a distance ##r## from the centre of smaller sphere is ##\frac{q_0}{4\pi \epsilon r^2}##. The electrostatic potential energy stored can be calculated by the following integral:
[tex]U=\int \frac{1}{2}\epsilon E^2dV[/tex]
Replacing dV with ##4\pi r^2dr## and solving, I don't get the right answer. Also, I don't see how can I express U in terms of t and ##\sigma##. :confused:

Any help is appreciated. Thanks!
 

Answers and Replies

  • #2
ehild
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That conductivity allows charge to flow from one plate to the other. It is the same as if a resistor was connected across the capacitor.

ehild
 
  • #3
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That conductivity allows charge to flow from one plate to the other. It is the same as if a resistor was connected across the capacitor.

ehild
What if there was no conductivity given in the question? Would my approach be correct then?

And I still don't know how to begin with this question. :(
 
  • #4
AGNuke
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You have HCV Part II? There's a formula given for finding out Capacitance of Spherical Capacitor. That's ought to help.
 
  • #5
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You have HCV Part II? There's a formula given for finding out Capacitance of Spherical Capacitor. That's ought to help.
I don't think that formula would help. A current flows here so I think a good equation to start with would be ##J=\sigma E## (that's the only equation I remember which involves conductivity), where J is current density and E is the electric field but I am not sure if it would work though.
 
  • #6
AGNuke
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The charge will be conducted from Inner Sphere to outer Sphere. The instantaneous charge can be found out by the help of current. Then we can find the energy of the system. Let's see if that helps.

BTW, is your D-Day this year or next? (You know what I mean)
 
  • #7
ehild
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I don't think that formula would help. A current flows here so I think a good equation to start with would be ##J=\sigma E## (that's the only equation I remember which involves conductivity), where J is current density and E is the electric field but I am not sure if it would work though.
You are on the right track. J=σE and JE is the power dissipated in unit volume in unit time.

You also can find both the capacitance and the resistance of the spherical shell, and treat the problem as discharging a capacitor.

ehild
 
  • #8
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You also can find both the capacitance and the resistance of the spherical shell, and treat the problem as discharging a capacitor.
How can I calculate the resistance here? Can I have a few hints to start with? :confused:
 
  • #9
AGNuke
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Use conductivity as inverse of resistivity to find resistance. Take an elemental spherical shell of thickness dx and radius x. Now use integral calculus to find the overall resistance.
 
  • #10
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Use conductivity as inverse of resistivity to find resistance. Take an elemental spherical shell of thickness dx and radius x. Now use integral calculus to find the overall resistance.
That's the first thought that came to my mind but what should I use as cross-section area?
We have
[tex]R=\frac{1}{\sigma}\cdot \frac{\ell}{A}[/tex]
What should be A for the differential element?
 
  • #11
AGNuke
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$$dR=\frac{1}{\sigma}\times \frac{dx}{4\pi x^2}$$
 
  • #12
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$$dR=\frac{1}{\sigma}\times \frac{dx}{4\pi x^2}$$
That gives me,
[tex]R=\frac{1}{4\pi \sigma}\cdot \left(\frac{1}{a}-\frac{1}{b}\right)[/tex]

The capacitance is
[tex]4\pi \epsilon\left(\frac{ab}{b-a}\right)[/tex]

From ehild's suggestion,
[tex]I=I_oe^{-t/RC}[/tex]
where ##RC=\epsilon/\sigma## and how can I find ##I_o##? :confused:
 
  • #13
AGNuke
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This is what I am coming to[tex]E=\frac{Q_0^2}{8C}\times (1 - e^{-t/RC})^2[/tex]

It really helps if you have some solution to reach to...:frown:
 
  • #14
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This is what I am coming to[tex]E=\frac{Q_0^2}{8C}\times (1 - e^{-t/RC})^2[/tex]

It really helps if you have some solution to reach to...:frown:
That's wrong and I am still clueless about what should be ##I_o##. :(
 
  • #15
AGNuke
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How about finding the potential at the inner sphere and the outer sphere due to the charge on inner sphere, find their difference and divide by resistance?
 
  • #16
ehild
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From ehild's suggestion,
[tex]I=I_oe^{-t/RC}[/tex]
where ##RC=\epsilon/\sigma## and how can I find ##I_o##? :confused:
Use Q(t)=Q0e-t/RC instead.

ehild
 
  • #17
ehild
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This is what I am coming to[tex]E=\frac{Q_0^2}{8C}\times (1 - e^{-t/RC})^2[/tex]

It really helps if you have some solution to reach to...:frown:
Conduction means dissipation, loss of electric energy. The energy of the electric field has to decrease.

ehild
 
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  • #18
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How about finding the potential at the inner sphere and the outer sphere due to the charge on inner sphere, find their difference and divide by resistance?
Use Q(t)=Q0e-t/RC instead.

ehild
Thank you both, I have reached close to the answer but it is still not correct.
[tex]U=\frac{Q_o^2}{2C}e^{-\frac{2t}{RC}}[/tex]
Substituting RC and C,
[tex]U=\frac{Q_o^2(b-a)}{8\pi \epsilon ab}e^{-\frac{2\sigma t}{\epsilon}}[/tex]
But the given answer is:
[tex]U=\frac{q_o^2(b-a)e^{-2\sigma \epsilon t}}{4\pi \epsilon ab}[/tex]
I am sure that my exponential term is correct. ##2\sigma \epsilon t## is not dimensionless so the given answer is wrong but what about the factor of 1/2?

Thanks!
 
Last edited:
  • #19
ehild
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Thank you both, I have reached close to the answer but it is still not correct.
[tex]U=\frac{Q_o^2}{2C}e^{-\frac{t}{RC}}[/tex]
Substituting RC and C,
[tex]U=\frac{Q_o^2(b-a)}{8\pi \epsilon ab}e^{-\frac{2\sigma t}{\epsilon}}[/tex]
But the given answer is:
[tex]U=\frac{q_o^2(b-a)e^{-2\sigma \epsilon t}}{4\pi \epsilon ab}[/tex]
I am sure that my exponential term is correct. ##2\sigma \epsilon t## is not dimensionless so the given answer is wrong but what about the factor of 1/2?

Thanks!
The given answer is wrong. Yours is correct (at least the same, I got) :)

ehild
 
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  • #20
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The given answer is wrong. Yours is correct (at least the same, I got) :)

ehild
Thanks a lot ehild! :smile:
 
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  • #21
ehild
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  • #22
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Thanks for the thanks:biggrin:

ehild
One thanks isn't enough. You deserve a lot more of them. :smile:
 
  • #23
AGNuke
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Asking ehild whether or not he's willing to help tomorrow too in real time!? :p
 
  • #24
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Asking ehild whether or not he's willing to help tomorrow too in real time!? :p
Don't you know they don't allow other gadgets in the examination hall? :biggrin:
 
  • #25
AGNuke
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This IS India my son! There's always a way or two around!
 

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