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?
Best Answer
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.