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  4. The plates of a parallel plate capacitor are charged to a potential of 200V . Now, a dielectric slab of thickness 4mm is inserted between its plates and to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by 3.2mm . The dielectric constant of the slab is

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Q. The plates of a parallel plate capacitor are charged to a potential of $200V$ . Now, a dielectric slab of thickness $4mm$ is inserted between its plates and to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2mm$ . The dielectric constant of the slab is

NTA AbhyasNTA Abhyas 2022

Solution:

$\frac{\epsilon _{0} A}{d} \, = \, \frac{\epsilon _{0} A}{d^{′} - t + \frac{t}{K}}$
$\Rightarrow d=d^{′}-t+\frac{t}{K}$
$\Rightarrow d^{′}-d=t\left(1 - \frac{1}{K}\right)$
Here, $d^{′}-d=3.2 \, mm, \, t=4 \, mm$
$\therefore 3.2=4\left(1 - \frac{1}{K}\right)$
$\Rightarrow \frac{3 .2}{4}=1-\frac{1}{K}$
$\Rightarrow 1-\frac{1}{K}=\frac{4}{5}$
$\therefore K=5$