A parallel plate capacitor is maintained at a certain potential difference. When a dielectric slab of thickness 3 mm is introduced between the plates, the plate separation had to be increased by 2 mm in order to maintain the same potential difference between the plates. The dielectric constant of the slab is

The capacitance before the introduction of the slab is $\mathrm{C}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}$

If Q is the charge on the plates, the potential difference is $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}=\frac{\mathrm{Qd}}{{\mathrm{\epsilon }}_0\mathrm{A}}$  ……(1)

Let d' be the new separation between the plates. When a slab of thickness t and dielectric constant K is introduced, the new capacitance is 

${\mathrm{C}}^{\mathrm{\prime }}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{{\mathrm{d}}^{\mathrm{\prime }}-\mathrm{t}\left(1-\frac{1}{\mathrm{K}}\right)}$ 

Since charge Q remains the same, the new potential difference is ${\mathrm{V}}^{\mathrm{\prime }}=\frac{\mathrm{Q}}{{\mathrm{C}}^{\mathrm{\prime }}}=\frac{\mathrm{Q}\left[{\mathrm{d}}^{\mathrm{\prime }}-\mathrm{t}\left(1-\frac{1}{\mathrm{K}}\right)\right]}{{\mathrm{\epsilon }}_0\mathrm{A}}$   ……..(2)

Given V' = V. Equating Eqs. (1) and (2), we get

$\mathrm{d}={\mathrm{d}}^{\mathrm{\prime }}-\mathrm{t}\left(1-\frac{1}{\mathrm{K}}\right)\text{ or }{\mathrm{d}}^{\mathrm{\prime }}-\mathrm{d}=\mathrm{t}\left(1-\frac{1}{\mathrm{K}}\right)$

Given d' = d = 2 mm and t = 3 mm. Thus

$2=3\left(1-\frac{1}{\mathrm{K}}\right)$

which gives K = 3.