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

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$5$

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(A) Let the initial distance between the plates be $d$. The 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,and the plate separation is increased by $\Delta d = 1.6\,mm$,the new separation is $d' = d + 1.6\,mm$.The potential difference $V$ remains the same. The electric field in the air gap is $E_0$ and in the dielectric is $E_0/K$.The new potential difference is $V = E_0(d' - t) + \frac{E_0}{K}t$.Since $V = E_0 d$,we have $E_0 d = E_0(d + 1.6 - 2) + \frac{E_0}{K}(2)$.Dividing by $E_0$,we get $d = d - 0.4 + \frac{2}{K}$.$0.4 = \frac{2}{K}$.$K = \frac{2}{0.4} = 5$.

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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