The voltage between the plates of a capacitor

Question

I have read in one book that:

the potential difference between the plates of a capacitor when it's connected to a DC source (a battery) is in opposite direction to the potential difference of that battery.

I don't really get that point, and I don't know how could this help in the break of the DC after the capacitor charges. I used to think that when connecting a battery to a capacitor the plate connected to the positive terminal gets a positive charge and the plate connected to the negative terminal gets a negative charge. So, how will the PD be in the opposite direction to the PD of the battery? Could someone clarify it?

Answer 1

That's a somewhat confusing way of putting it. It's related to the direction you define as "the opposite direction"; here's a diagram.

^{simulate this circuit – Schematic created using CircuitLab}

The arrows here always point from negative to positive. The left schematic shows how you might think of it as "the same direction"; the right one shows how you might think of it as "the opposite direction". Note that these two schematics are identical.

Answer 2

The capacitor charges up to the voltage of the battery and, as a result, opposes the battery's voltage sufficiently to stop any further current.

If you connect the capacitor to the battery and wait long enough (not long, really) the capacitor will be charged to a voltage very similar to the battery so that when you pull it away and remove it the polarity of the capacitor will match the battery's polarity. Here, one might see that it is the same and not opposite. But it did oppose the battery's voltage while connected, too.

It's just a matter of what you want to convey, I suppose.