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(A) Let the initial distance between the plates be $d$. The initial potential difference is $V = E \cdot d$,where $E$ is the electric field.When a die

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$5$

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(A) Let the initial distance between the plates be $d$. The initial potential difference is $V = E \cdot d$,where $E$ is the electric field.When a dielectric slab of thickness $t = 2 \, mm$ is inserted,the new distance becomes $d' = d + 1.6 \, mm$.The potential difference across the capacitor with the dielectric is given by $V = E(d' - t) + \frac{E}{K} \cdot t$,where $K$ is the dielectric constant.Since the potential difference $V$ remains the same,we equate the two expressions:$E \cdot d = E(d + 1.6 - 2) + \frac{E}{K} \cdot 2$Dividing by $E$:$d = d - 0.4 + \frac{2}{K}$$0.4 = \frac{2}{K}$$K = \frac{2}{0.4} = 5$Thus,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. 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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