Test: Effect of Dielectric on Capacitance (NCERT) - Grade 12 MCQ

The material suitable for using as a dielectric must have high dielectric strength X and large dielectric constant K.

Here C₁ = ε₀A/d = 10pF …….(i)
(∵ K = 4)
= 8 × 10 (using (i)) 
= 80 pF

Capacitance of a parallel capacitor with air is C = ε₀A/d​
Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium is C ′ = Kε₀A/d​
where K is the dielectric constant of a medium
or C′/C ​= K or K = 15/3​ = 5 or K = ε₀/ε ​
or ε = Kε₀ ​= 5 × 8.854 × 10⁻¹² = 0.44 × 10⁻¹⁰C²N ⁻¹ m⁻²

As capacitance C₀ = ε₀A/d
∴ after inserting copper plate C = ε₀A/d−b

as Q = cv
initially c = 5μF, v = 1v
∴ Q = 5 × 10^–6 c = 5μc
when dielectric is introduced, let capacitance be c'
∴ c' = [(A∈₀)/{d – t + (t/k)}]
here k = 4, t = 4 cm, d = 6 cm
∴ c' = [{(A∈₀)/d}/{{d – t + (t/k)}/d}] ------ dividing by d
As [(A∈₀)/d] = initially capacitance = c = 5μF
∴ c' = [c/{{6 – 4 + (4/4)}/6}] = [c/(3/6)] = 2c = 10μF
∴ Q' = c'v = 10 × 10^–6 × 1 = 10μc
∴ addition charge flowing = Q' – Q = 10 – 5 = 5μc