# Where did this extra energy in the capacitor come from?

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?