Q107 / 180 · NEET 2020 (Phase 2) · Physics

A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of the same area but the thickness, d2, is inserted between the plates as shown in figure having dielectric constant K(=4). The ratio of new capacitance to its original capacitance will be,

Figure

  • A
    2:1
  • Correct answer
    B
    8:5
  • C
    6:5
  • D
    4:1

Solution

First, recall the formula of capacity of a parallel plate capacitor at free space in terms of cross-section area and separation between plates,C0=ϵ0Ad, when  a dielectric slabs of thickness, t and dielectric constant k, inserted between plates then-new capacitance becomes, 

Ck=ϵ0Ad-t+tk⇒Ck=ϵ0Ad-d2+d8, here t=d2.

⇒Ck=85ϵ0Ad⇒Ck=85C0⇒CkC0=85

Hence, the correct answer is option: 'B'.

More Capacitance Questions