Question about charged capacitors and inserting a dielectric into one

[Kaushik]

Homework Statement

The parallel combination of two air filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage, $V$. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is?

Relevant Equations

$Q = CV$
$C_{eq}= C_1 + C_2$ for parallel
$\frac{1}{C_{eq}}= \frac{1}{C_1} + \frac{1}{C_2}$ for series

First when it is connected to the battery, the capacitors start accumulating charges such that the potential difference equals that of the battery. Then the current stops flowing.
$Q_1 = CV$
$Q_2 = nCV$
Where 1 and 2 represent the capacitor with capacitance C and nC respectively

Then, when we remove the battery and add a dielectric between the capacitor of capacitance $C$ ($C_{new} = KC$), the potential difference across it reduces. Now, the capacitor 2 starts behaving like a battery, providing charge to the capacitor 1.
After this, how can I proceed?

 

[BvU]

providing charge to the capacitor 1

And when does this 'provision' come to an end ?

 

[Kaushik]

And when does this 'provision' come to an end ?

When the potential difference between capacitor 1 and capacitor 2 become equal?

 

[BvU]

Correct.
And what is conserved during this provision process ?
Set up a set of equations before and after to find the new $V$.