Dielectric Constant Calculation: 110 & 240 µC

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The discussion revolves around calculating the dielectric constant of a slab inserted into an air-filled parallel plate capacitor. Initially, the capacitor acquires a charge of 110 micro Coulombs, and after inserting the dielectric, the charge increases to 240 micro Coulombs. The relevant equations involve the relationships between capacitance (C), voltage (V), and charge (Q), with the dielectric constant (K) being a key factor. The problem requires establishing two equations based on the two scenarios and solving for K, while noting that the voltage remains constant during the process. Understanding these relationships is crucial for determining the dielectric constant accurately.
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When a certain air filled parallel plate capacitor is connected cross a battery, it acquires a charge (on each plate) of 110 micro Columbs. While the battery connection is maintained, a dielectric slab is inserted into and fills the region between the plates. This results in the accumulation of an additional charge of 240 micro Columbs on each plate.

What is the dielectric constant of the dielectric slab?

so I have Q1=110 micro Columbs and Q2= 240 micro Columbs;E= 8.85419e-12 C^2/N*m^2; and Air= 1.00054 which is believe is K1

I just don't know what equation to use or how to go about this problem in general.
 
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What is the relation between C, V and Q.
In the problem which quantity remains constant in both the cases?
Write down two equations for two cases and then solve for K.
 

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