If the capacitor is connected to a constant voltage source (Voltage =V),

The object will oscillate if it was not fit right into the capacitor.

Click play to start the simulation.

While in paused mode, you can adjust k,d => so capacitance C is changed,too.

You can watch how charge Q changed as C is changed in constant voltage mode.

If you turn into constant Q mode (switch off the charging circuit by selecting constant Q radio button),

the change Q will become constant,

and voltage V will vary if you move the dielectric object with slider (change y).

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The energy stored in the capacitor C can be expressed as when the capacitor are connected to a constant voltage source.

For two parallel plate capacitor with area A and separation d:

From Gauss law:

The electric field where is the charge density.

Since ,

so .

If an dielectric material with dielectric constant k is insert between the parallel plate,

electric field will introduce dipole field (in opposite direction to electric field) inside the dielectric material.

So that the net elctric field become (from defition of dielectric constant).

The voltage between capacitor plate will reduced to if the capacitor is not connected to voltage source.

(The charge Q remain the same --- constant Q mode)

However, if the capacitor is connected to constant voltage source, then voltage source will charge the capacitor to keep up the same voltage V (from V' to V).

So more charge (Q'=kQ )will be stored into the capacitor, so the capacitance .

If the dielectric meterial is moved side way, it will be attracted back to it's original position.

assume the side way displacement is y, and area .

The capacitoance become

The total energy , so the force can be calculated as

(it is a constant acceleration, but in the direction opposite to displacement).