I need help with the following problem: Plate capacitor, with area of electrodes S, and distance between them d has perfect dielectric, of relative permitivitty Ɛr. Capacitor is connected to constant voltage. Then, it is disconnected and dielectric is removed. What is the energy of a capacitor after dielectric is removed?

In the first case (with dielectric), energy of a capacitor is $$W_e^{(1)}=\frac{\epsilon_0\epsilon_rSU^2}{2d}$$

In my book's solution, energy in the second case (removed dielectric) is: $$W_e^{(2)}=\frac{\epsilon_r^2\epsilon_0SU^2}{2d}$$

which doesn't change anything because dielectric is perfect (Ɛr=1). So, in both cases, energy is the same.

Is this correct?

  • \$\begingroup\$ Is there work done (by or on) the dielectric to remove it? \$\endgroup\$ – Chu Aug 21 '15 at 14:51
  • \$\begingroup\$ Do you mean the physical work? I think it is irrelevant in the problem. \$\endgroup\$ – user300045 Aug 21 '15 at 14:54
  • \$\begingroup\$ But work = energy which must transfer to/from somewhere. Compare with inserting/removing a solenoid core. I don't know the answer, just asking the question! \$\endgroup\$ – Chu Aug 21 '15 at 14:55
  • \$\begingroup\$ Also, does 'perfect dielectric' mean \$\epsilon_r =1\$ or that there is no leakage? \$\endgroup\$ – Chu Aug 21 '15 at 15:03
  • \$\begingroup\$ I don't know. Perfect dielectric has zero conductivity. \$\endgroup\$ – user300045 Aug 21 '15 at 15:08

A 'perfect dielectric' in this case simply means that we can treat it as an ideal capacitor. We can show using Gauss' law that \$C=\frac{\epsilon_0\epsilon_rS}{d}\$. Recall the relationship \$\epsilon_0\epsilon_r = \epsilon\$ , and in particular, the fact that \$\epsilon_r=\frac{C_d}{C_0} \$, where \$C_d \$ is the capacitance with a vacuum between the plates. This implies that the relative permittivity \$\epsilon_r\$ is the factor by which capacitance changes when a dielectric is added.

Note that knowing only that \$W_e=\frac{CV^2}{2}\$, in combination with the definition of \$\epsilon_r\$, we can arrive at the solution given since the general form for energy stored in a capacitor doesn't change. By a simple discrete mathematical argument, we can safely say that the energy after removing the dielectric is larger. You probably would have arrived at this conclusion had you not misunderstood what a perfect dielectric was. While quick and dirty, this solution doesn't do much for the intuition, so I'll proceed another way.

Consider \$C=\frac{Q}{V}\implies\frac{Q}{C}=V\$; since the cap is essentially an open circuit, the charge cannot change when we remove the dielectric, the assumption here being that the dielectric has no net charge. From this simple observation, we have determined that the change in capacitance brought about by removing the dielectric must cause a change in voltage in this case. Therefore, voltage and electric field are changing. If we decrease the capacitance, the voltage must rise in this case (no battery). If there was a battery, then we could follow similar arguments, except in this case the voltage \$V\$, and hence \$\vec{E}\$ does not change.

Let's put this another way considering the electric field: If we take the electric field \$\vec{E}\$ as \$E=\frac{V}{d}\$, or alternatively that \$E=\frac{\sigma}{\epsilon_0\epsilon_r}\$ , where \$\sigma=\frac{Q}{A}\$, we see again that \$\vec{E}\$ must change because the charge will remain the same while the permittivity \$\epsilon_0\epsilon_r = \epsilon\$ does not. (Again, this is not the case if the battery is still connected.) Ergo, we see an increase in voltage and electric field by removing the dielectric with no battery attached.

So, if we go through the plug and chug knowing the permittivity between the plates affects capacitance, and that the electric field and voltage must change given a constant initial charge, we get that the final energy is actually higher since we've done work on the capacitor.

In short, the energy is higher after the dielectric is removed. Your belief that a perfect dielectric means the \$\epsilon_r=1\$ is not correct I think; rather, 'perfection' in this case simply means that there is no electrical conductivity whatsoever.

  • \$\begingroup\$ How can I make my math not look like crap \$\endgroup\$ – RYS Aug 21 '15 at 16:13
  • \$\begingroup\$ So, why are you not using the built in math markup. Actually, I have never posted any math in this SE but MathJAX is the way to go as described: meta.math.stackexchange.com/questions/5020/… \$\endgroup\$ – K7PEH Aug 21 '15 at 18:16
  • \$\begingroup\$ We don't normally do homework for OPs \$\endgroup\$ – Chu Aug 21 '15 at 18:49
  • \$\begingroup\$ @Chu He's already got the answer, might as well provide some intuition :) \$\endgroup\$ – RYS Aug 21 '15 at 21:26
  • \$\begingroup\$ @K7PEH thanks for the link, forgot to escape my dollar signs \$\endgroup\$ – RYS Aug 22 '15 at 0:20

This is a homework question so I'm not going to give you the answer directly. The work done in removing the dielectric is significant, but you don't need to consider it directly.

Instead, consider that the charge on the plates will remain constant (there's no connection so it has nowhere to go). From that you should be able to calculate the voltage and/or energy.


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