The plates of a parallel plate capacitor are charged up to 100 V. Now, after removing the battery, a 2 mm thick plate is inserted between the plates. Then, to maintain the same potential difference, the distance between the capacitor plates is increased by 1.6 mm. The dielectric constant of the plate is
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a
5
b
1.25
c
4
d
2.5
answer is A.
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Detailed Solution
As battery is disconnected, So charge will remain the same. It is given that final potential is the same. So final capacitance should be equal to initial capacitance, i.e., ε0Ad=ε0A(1.6+d)−t(1−1/K)or K=5