Question

A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be 

(A) 200 % 

(B) 33.3 % 

(C) 400 % 

(D) 66.6 %

Answer 1

The correct option (D) 66.6 %

Explanation:

initially  C = (A∈₀/d)    (1)

when it is filled with dielectric of thickness t then

 C' = [(A∈₀)/{d – t + (t / k)}]

given : k = 5 & it is half filled, means t = (d/2)

∴ C' = [(A∈₀)/{d – (d/2) + {d/(2 × 5)}] = [(A∈₀)/{(d/2) + (d/10)}]

 = [(A∈₀)/(6d/10)] × (10/6)

∴ C' = (5/3) (A∈₀/d) = (5/3)C   from (1)

∴ % increase in capacitance = [(C' – C)/C] × 100

= [{(5/3)C – C}/C] × 100

= (2/3) × 100 = 66.66%