00:01
Hi there, so for this problem we are told that a parallel plate capacitor of capacitance c -0 has plates of area a and with a separation d between them.
00:20
Now when it is connected to a battery of voltage b -0, it has charge of magnitude q -0 on its plate.
00:35
So it is then disconnected from the battery, and the space between the plates is filled with a material of dielectric constant that is equal to 3.
00:48
So we will have a dilectric constant that is equal to 3.
00:53
And after the dilectric is added, the magnitudes of the charge on the plates and the potential difference between them are the following.
01:02
So we need to find the charge.
01:06
And the potential after we added the dilectric.
01:16
So in this case, we know that the charge does not change.
01:25
Once we introduce the dilectric, this does not change.
01:30
But the potential, we know that initially is defined as the charge divided by the capacitance...