The plates of a parallel plate capacitor are | vedclass

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(A) Let the initial distance between the plates be $d$. The potential difference in air is $V = E_0 d = \frac{\sigma}{\varepsilon_0} d$. When a dielec

Vedclass · Accepted Answer

$5$

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(A) Let the initial distance between the plates be $d$. The potential difference in air is $V = E_0 d = \frac{\sigma}{\varepsilon_0} d$. When a dielectric slab of thickness $t = 2\;mm$ is inserted,the new potential difference $V'$ with the increased distance $d' = d + 1.6\;mm$ is given by $V' = \frac{\sigma}{\varepsilon_0} (d' - t + \frac{t}{K})$. Since the potential difference remains the same,$V = V'$,therefore: $d = (d + 1.6) - 2 + \frac{2}{K}$. $d = d + 1.6 - 2 + \frac{2}{K}$. $0 = -0.4 + \frac{2}{K}$. $0.4 = \frac{2}{K}$. $K = \frac{2}{0.4} = 5$. The dielectric constant of the plate is $5$.

The plates of a parallel plate capacitor are charged up to $100\;V$. $A$ $2\;mm$ thick plate is inserted between the plates. Then,to maintain the same potential difference,the distance between the plates is increased by $1.6\;mm$. The dielectric constant of the plate is:

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