Question

The plates of a parallel capacitor are charged up to 100 V. If 2 mm thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by 1.6mm the dielectric constant of the plate is 

(A) 5 

(B) 4 

(C) 1.25 

(D) 2.5

Answer 1

The correct option (A) 5

Explanation:

C = (A∈₀/d) initial

C' = [(A∈₀)/{d' – t + (t/k)}]  after putting slab of thickness t

here d' = d + 1.6

as Q = CV

hence if potential difference remains same then capacitance must remain same

∴ C = C'

∴ (A∈₀/d) = [(A∈₀)/{d' – t + (t/k)}]

 i.e. d = d' – t + (t/k)

 as d' = d + 1.6

∴ d = d + 1.6 – 2 + (2/k)

∴ (2/k) = 0.4

 ∴ k = (2/0.4) = 5