A parallel plate air capacitor is charged up to 100 ~V . A plate 2 ~mm thick is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by 1.6 ~mm . The dielectric constant of the thick plate is

Step-by-step solution

If d is the initial distance between the plates then capacitance is given by C = kA ₀ ~d When a plate of thickness t =2 ~mm is inserted between the plates the capacitance, C₁= k A ₀ (d-t+ t k ) If distance between the plates is increased by x =1.6 ~mm Capacitance become, C ₂= kA ₀ ( ~d + x - t + t k ) But C ₂= C aligned & d + x - t + t k = d & t - x = t k & 2-1.6= 2 k & 0.4= 2 k k =5 aligned