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A parallel-plate capacitor with plate area $A$, separation $d$ and dielectric with $\epsilon = 2\epsilon_0$ is charged up to a voltage $V_0$. Therefore:

  • the capacitance is $C_0=2\epsilon_0\frac{A}{d}$
  • the electric field is $E_0=V_0/d$ (ignoring fringing fields)

Now the battery is disconnected, so the charge $Q$ on the plates is reserved such that: $$Q=C_0V_0$$ is constant. The stored energy in the capacitor is: $$W_0=\frac{1}{2}C_0V_0^2$$

Now, assume we remove the dielectric between the plates. No change happens to the stored charge $Q$, unlike the capacitance,the electric field and the voltage between the plates.

  • the new capacitance is $C=\frac{1}{2}C_0$
  • the new electric field is $E=2E_0$
  • the new voltage is $V=2V_0$

Now: what about the stored energy in the capacitor? $$W=\frac{1}{2}CV^2=\frac{1}{2}(\frac{1}{2}C_0)(4V_0^2)=C_0V_0^2=2W_0$$

This means that removing the dielectric will double the energy stored in the capacitor, although the battery is disconnected.

My question: where did this extra energy come from? Or did I get it wrong?

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2 Answers 2

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The energy come from the work you must do to pull the dielectric out of the capacitor. The dielectric is atracted to the charges because it is polarizable -- jut like a pice of paper is attracted to a charged object or a rubbed balloon is attracted to a wall,

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You've had to do work removing the dielectric. Remember that while between the charged capacitor plates the molecules of the dielectric were polarised by the electric field, resulting in the face of the dielectric adjacent to the positive plate having a negative charge and the face of the dielectric adjacent to the negative plate having a positive charge. Work is done separating these adjacent charges.

[It is, of course, the charges on the faces of the dielectric that reduced the electric field and the voltage for a given charge on the plates, and therefore increased the capacitance while the dielectric was in place.]

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