Constant Voltage:

Constant Charge:

Playing with a capacitor

Take a parallel-plate capacitor and connect it to a power supply. The power supply sets the potential difference between the plates of the capacitor.

The distance between the capacitor plates can be changed. While the capacitor is still connected to the power supply, the distance between the plates is increased. When this occurs, what happens to Q, C, and DV?

The potential difference across the capacitor:

  1. increases
  2. decreases
  3. stays the same

The capacitance:

  1. increases
  2. decreases
  3. stays the same

The charge on the capacitor:

  1. increases
  2. decreases
  3. stays the same

Because the capacitor is still connected to the power supply the potential difference can't change. Moving the plates further apart decreases the capacitance, also reducing the charge stored by the capacitor.

Now the capacitor is charged by the power supply and then the connections to the power supply are removed. When the distance between the plates is increased now, what happens to Q, C, and DV?

The potential difference across the capacitor:

  1. increases
  2. decreases
  3. stays the same

The capacitance:

  1. increases
  2. decreases
  3. stays the same

The charge on the capacitor:

  1. increases
  2. decreases
  3. stays the same

In this case the charge stays constant, because there's nowhere for it to go. The capacitance is still decreased, and the potential difference increases because
DV = Q/C

Another argument for why the potential difference increases is that the potential difference is the field multiplied by the distance between the plates. If the charge is constant the field is constant, so increasing the plate separation increases DV.