The two plates of a parallel plate capacitor are 4 mm apart. A slab of dielectric constant 3 and thickness 3 mm is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the capacitance of the capacitor becomes 2/3rd of its original value. What is the new distance between the plates?

Step-by-step solution

Here, distance between parallel plates d = 4 mm = 0.004 m, K = 3, thickness t = 3 mm = 0.003 m and d 1 =? ∴ C = ε 0 A d and C 1 = ε 0 A d 1 − t ( 1 − 1 K ) since C 1 = 2 3 C (given) ∴ ε 0 A d 1 − t ( 1 − 1 K ) = 2 3 ε 0 A d 1 d 1 − t ( 1 − 1 K ) = 2 3 d 1 d 1 − 0.003 ( 1 − 1 3 ) = 2 3 × 0.004 1 d 1 − 0.003 × 2 3 = 1 0.006 1 d 1 − 0.002 = 1 0.006 d 1 − 0.002 = 0.006 d 1 = 0.006 + 0.002 = 0.008 m = 8 mm .