An air-filled parallel plate capacitor of plate area 2 cm² and plate separation 4 mm has capacitance C₀. It is filled with two slabs of same area but thickness 1 mm and 3 mm of dielectric constants K and 3K, respectively. Now its capacitance changes to C. The ratio C’/C is

Correct option is A

Correct answer is A
Explanation:
The original capacitance of the air-filled capacitor is:
C = (ε0 × A) / d
Given: A = 2 cm2 = 2 × 10-4 m2, d = 4 mm = 4 × 10-3 m.
So, C = (ε0 × 2 × 10-4) / (4 × 10-3) = (ε0 × 5 × 10-2).
When the capacitor is filled with two slabs, the effective capacitance is calculated as:
1/C' = (d1 / K1 ε0 A) + (d2 / K2 ε0 A)
Where:
d1 = 1 mm = 1 × 10-3 m, K1 = K
d2 = 3 mm = 3 × 10-3 m, K2 = 3K
Substitute the values:
1/C' = ((1 × 10-3) / (K × ε0 × 2 × 10-4)) + ((3 × 10-3) / (3K × ε0 × 2 × 10-4))
1/C' = (5/Kε0) + (5/Kε0) = 10 / Kε0
So, C' = (Kε0 × 10)/10 = 2K × ε0 × 5 × 10-2.
Now, the ratio C'/C is:
(C'/C) = (2K × ε0 × 5 × 10-2) / (ε0 × 5 × 10-2)
(C'/C) = 2K