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 $$\begin{gathered} C=\frac{ {\epsilon }_{0}A}{d}. If Q is the charge on the plates, the potential difference is \\ V=\frac{Q}{C}=\frac{Qd}{{\epsilon }_{0}A} \left(1\right) \\ Let d\text{'} be the new separation between the plates. When a slab of thickness t and dielectric cons\tan t K is introduced, the new capaci\tan ce is \\ C\text{'}=\frac{{\epsilon }_{0}A}{d\text{'}-t\left(1-{\displaystyle \frac{1}{K}}\right)} \\ Since charge Q remains the same, the new potential difference is \\ V\text{'}=\frac{Q}{C\text{'}}=\frac{Q\left[d\text{'}-t\left(1-\frac{1}{K}\right)\right]}{{\epsilon }_{0}A} \left(2\right) \\ Given V\text{'} = V. Equating Eqs. \left(1\right) and \left(2\right), we get \\ d = d\text{'} - t \left(1-\frac{1}{K}\right) or d\text{'} - d = t\left(1-\frac{1}{K}\right) \\ Given d\text{'}-d=2 mm and t = 3 mm. Thus \\ 2=3\left(1-\frac{1}{K}\right) \\ which gives K = 3. \end{gathered}$$