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Question: Two parallel plate capacitors of capacitances \[C\] and \[2C\] are connected in parallel and charged...

Two parallel plate capacitors of capacitances CC and 2C2C are connected in parallel and charged to a potential difference VV. The battery is then disconnected and the region between the plates of the capacitor CCis completely filled with a material of dielectric constant KK. The potential differences across the capacitors now becomes...........

Explanation

Solution

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.=Rangetotal number of divisionsL.C. = \dfrac{{Range}}{{total\,number\,of\,divisions}}

Complete step-by-step solution:
Initially, the capacitance of the first capacitor is given as CC
The charge on the capacitor is taken as: q1=CV{q_1} = CV
The capacitance of the second capacitor is given as 2C2C
Hence the charge on this capacitor is given as q2=2CV{q_2} = 2CV
Now, when this capacitor is completely filled with a material
The dielectric constant of the material isKKand let the voltage be V′V'
Then, finally the charge on the capacitor is given as q1=KCV′{q_1} = KCV'
And that on the second capacitor is given as q2=2CV′{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′CV + 2CV = KCV' + 2CV'
Finding the value of the final voltageV′V', we get
V′=3VK+2V' = \dfrac{{3V}}{{K + 2}}
Hence, the potential differences across the capacitors will become V′=3VK+2V' = \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.