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(A) Let the initial distance between the plates be $d$. The initial potential difference is $V = E_0 d$,where $E_0$ is the electric field between the

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$5$

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(A) Let the initial distance between the plates be $d$. The initial potential difference is $V = E_0 d$,where $E_0$ is the electric field between the plates.When a dielectric slab of thickness $t = 2 \, mm$ and dielectric constant $K$ is inserted,the new distance between the plates becomes $d' = d + 1.6 \, mm$.The potential difference $V$ remains constant. The new potential difference is given by $V = E_0(d' - t) + E_0 \frac{t}{K}$.Since $V$ is the same,we have $E_0 d = E_0(d + 1.6 - 2) + E_0 \frac{2}{K}$.Dividing by $E_0$,we get $d = d - 0.4 + \frac{2}{K}$.This simplifies to $0.4 = \frac{2}{K}$.Therefore,$K = \frac{2}{0.4} = 5$.

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