The plates of a parallel plate capacitor are charged upto $100 V$ . now, after removing the battery, a $2 m m$ thick plate is inserted between the plates. Then, to maintain the same potential difference, the distance between the capacitor plates is increased by $1.6 m m$ . The dielectric constant of the plate is :

5

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1.25

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4

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2.5

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Solution

The correct option is A $5$
As battery is disconnected, so charge will remain the same.
It is given that final potential is the same.
So final capacitance should be $C_{1} = C_{2}$
$\frac{ε_{0} A}{d} = \frac{ε_{0} A}{(1.6 + d) - t (1 - 1 / K)} \Rightarrow K = 5$

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The correct option is A 5As battery is disconnected, so charge will remain the same. It is given that final potential is the same. So final capacitance should be C1=C2ε0Ad=ε0A(1.6+d)−t(1−1/K)⇒K=5

The correct option is A $5$
As battery is disconnected, so charge will remain the same.
It is given that final potential is the same.
So final capacitance should be $C_{1} = C_{2}$
$\frac{ε_{0} A}{d} = \frac{ε_{0} A}{(1.6 + d) - t (1 - 1 / K)} \Rightarrow K = 5$