Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the region between the plates of capacitor C is completely filled with a material of dielectric constant K. The potential difference across the capacitors now becomes…. $\frac{xV}{K + 2}$ , the value of x is

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answer is 3.

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Detailed Solution

Total capacitance = 2C + C = 3C
Total change a = 3CV
When the capacitors is filled with a dielectric, it’s capacitance become KC
Therefore after the capacitor is filled with dielectric the capacitance of the combination
C¹ = KC + 2C = C(K+2)
Charged remains same that is Q = 3CV
$\frac{Q}{C^{1}} = \frac{3 CV}{(K + 2) C} = \frac{3 V}{(K + 2)}$