What Happens to Voltage When a Dielectric is Inserted into a Charged Capacitor?

[whoknows12345]

A very large parallel plate capacitor has two plates, each with an area of 9m^2, and a separation of 3nm between them.

If the capacitor was then disconnected from the battery (12V), and then a dielectric of k=4 inserted between the plates, fnd the new values for the following (after system has reached equilibrium)

C= 0.1062 F
Q= .3186 C
V= ____ V

I did V=Q/C = 3V, but I got it wrong. I don't see what I did wrong really.

 

[Cyrus]

Take the dielectric constant of k=4 into consideration.

 

[whoknows12345]

so it would be 3v/4? and that would give me the answer?

 

[berkeman]

It would help if you could post the exact text of the question. You've left some parts out, and that is probably where the error happens. Can you please post the whole question verbatim?