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 C=ε0AdIf Q is the charge on the plates, the potential difference is V=QC=Qdε0A  ……(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 C′=ε0Ad′−t1−1K Since charge Q remains the same, the new potential difference is V′=QC′=Qd′−t1−1Kε0A   ……..(2)Given V' = V. Equating Eqs. (1) and (2), we getd=d′−t1−1K or d′−d=t1−1KGiven d' = d = 2 mm and t = 3 mm. Thus2=31−1Kwhich gives K = 3.