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?