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(A) Let the initial distance between the plates be $d$ and the electric field be $E$. The potential difference is $V = Ed$.When a dielectric slab of t

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$5$

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(A) Let the initial distance between the plates be $d$ and the electric field be $E$. The potential difference is $V = Ed$.When a dielectric slab of thickness $t = 2 \, mm$ and dielectric constant $K$ is inserted,and the plate separation is increased by $\Delta d = 1.6 \, mm$,the new distance becomes $d' = d + 1.6 \, mm$.The potential difference remains the same,so $V = E(d' - t) + \frac{E}{K}t$.Substituting the values: $Ed = E(d + 1.6 - 2) + \frac{E}{K}(2)$.$Ed = E(d - 0.4) + \frac{2E}{K}$.$Ed = Ed - 0.4E + \frac{2E}{K}$.$0.4E = \frac{2E}{K}$.$K = \frac{2}{0.4} = 5$.

The plates of a parallel plate capacitor are charged up to $100 \, V$. $A$ $2 \, mm$ thick plate is inserted between the plates. 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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