Answer

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**Hint:**In order to solve this question, we are first going to take the charges on the two capacitors before and after putting the dielectric medium in between the capacitor plates. Then , using the law of conservation of charge for the charges initially and finally, the potential difference is calculated.

**Formula used:**

The formula for the least count of any instrument is the ratio of the range of the instrument to the total number of the divisions. It is given by the equation:

\[L.C. = \dfrac{{Range}}{{total\,number\,of\,divisions}}\]

**Complete step-by-step solution:**

Initially, the capacitance of the first capacitor is given as \[C\]

The charge on the capacitor is taken as: \[{q_1} = CV\]

The capacitance of the second capacitor is given as \[2C\]

Hence the charge on this capacitor is given as \[{q_2} = 2CV\]

Now, when this capacitor is completely filled with a material

The dielectric constant of the material is\[K\]and let the voltage be \[V'\]

Then, finally the charge on the capacitor is given as \[{q_1} = KCV'\]

And that on the second capacitor is given as \[{q_2} = 2CV'\]

Now as according to the law of conservation of charge, the total charge initially and finally remains the same, so,

\[CV + 2CV = KCV' + 2CV'\]

Finding the value of the final voltage\[V'\], we get

\[V' = \dfrac{{3V}}{{K + 2}}\]

Hence, the potential differences across the capacitors will become \[V' = \dfrac{{3V}}{{K + 2}}\]

**Note:**The charge conservation is the principle that the total electric charge in an isolated system never changes. The net quantity of electric charge, the amount of positive charge minus the amount of negative charge in the universe, is always conserved. The charge conservation has been used in this question.

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