Charging a capacitor

Below is an excerpt from a physics textbook:

One common way to charge a capacitor is to connect these two wires to opposite terminals of a battery. Once the charges $Q$ and $-Q$ are established on the conductors, the battery is disconnected. This gives a fixed potential difference $V_{ab}$ between the conductors (that is, the potential of the positively charged conductor $a$ with respect to the negatively charged conductor $b$) that is just equal to the voltage of the battery.

Why does the potential difference between the conductors equal the voltage of the battery when the battery is disconnected? Could someone please provide a detailed explanation? Thanks in advance.

Once the battery is disconnected, the charge on the capacitor plates is stuck where it is and has no path to go anywhere else. Since the charge remains on the plates, there is an electric field between the plates. And because there's a electric field between the plates there must be a voltage difference between them.

We know the voltage was equal to the battery voltage when the battery was connected. And since it doesn't change when the battery is disconnected, it must still be equal to the battery voltage afterwards.

• 'We know the voltage was equal to the battery voltage when the battery was connected', why is this so? sorry for keep asking stupid question... Commented Mar 22, 2014 at 6:00
• We assume the voltage drop across the connecting wires is 0 (or negligible). This is a very good assumption in this case because after the capacitor charges up, there's no current going through them. Then, since the positive terminal of the battery is connected to the first terminal of the capacitor, their potentials are equal. And since the negative terminal of the battery is connected to the second terminal of the capacitor, their potentials are equal. Therefore the potential difference across the capacitor is the same as the potential difference across the battery. Commented Mar 22, 2014 at 6:05