Inserting a dielectric material between two plates of a capacitor

  • #1
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Homework Statement:

Is the external mechanical work done by inserting a dielectric material between two plates of a capacitor (plates capacitor) when the power source remains connected always negative?
and if so, does it mean that the process can proceed spontaneously after being initiated?

Relevant Equations:

The original capacity is C0, the voltage is V0, and the dielectric constant of the material is K >1.
1566120347488.png

1. new capacity: ##C = K\cdot C_{0}##.
2. initial potential energy: ##U_{i} = \frac 1 2 \cdot C_{0}V_{0}^2##.
3. final potential energy: ##U_{f} = \frac 1 2 \cdot KC_{0}V_{0}^2##.
4. total work done: ##W_{tot} =\Delta U = U_{f} - U_{i} = \frac 1 2 \cdot (K-1)C_{0}V_{0}^2 > 0##.
5. also total work done: ##W_{tot} = W_{p}~\text{(power source work for streaming more charge)} + W_{ext}~\text{(external work for inserting the material).}##
6. power source work: ##W_{p} = \Delta q \cdot V_{0} = (q_{f} - q_{i})V_{0} = (C \cdot V_{0} - C_{0} \cdot V_{0}) \cdot V_{0} = (K-1) \cdot C_{0}V_{0}^2.##
7. external mechanical work: ##W_{ext} = W_{tot} - W_{p} = \frac 1 2 \cdot (K-1) \cdot C_{0}V_{0}^2 - (K-1) \cdot C_{0}V_{0}^2 =
\\= -\frac 1 2 \cdot (K-1)C_{0}V_{0}^2 < 0.##
 
Last edited:

Answers and Replies

  • #2
BvU
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Hello Idor, ##\qquad## :welcome: ##\qquad## !

Well, you work out your own question immacculately ! For K > 1 you get a definite W < 0.
So: yes !
And it indeed means the capacitor 'pulls in' the dielectric. If the force is enough to overcome friction it will actually do it seemingly spontaneously.

So where does the energy come from ?
 
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  • #3
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Hi,
Thank you very much! your confirmation really helps.

It just seemed odd to me at first that the capacitor would 'pull in' the dielectric material even when the power source remains connected, after looking at a similar example where it was disconnected first, and the total amount of work done was negative because the capacitor lost potential energy.

However, in the second case (my question) the total amount of work done (positive from the power source + negative from the insertion of the material) is positive (as the capacitor gains potential energy), and that is why I wonder what makes this process favorable.

I assume that the energy comes from the power source, but what does the system earn from this?
Is it because the dielectric material lowers the electric field between the plates of the capacitor, and therefore somehow makes the system more stable?
 
  • #4
rude man
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Energy density (volumtric) is 1/2 DE. D = ##\epsilon E ##. So the final energy in the capacitor's E field is incremented by ## \epsilon_r = \epsilon/\epsilon_0. ##
 
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