Question

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10⁻³ m² and the separation between the plates is 2 mm.

1. Calculate the capacitance of the capacitor. 
2. If this capacitor is connected to 100 V supply, what would be the charge on each plate? 
3. How would charge on the plates be affected if a 2 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected?

Answer 1

Data: k = 1 (air), A = 6 × 10⁻³ m²,

d = 2 mm = 2 × 10⁻³ m²,

V = 100 V,

t = 2 mm = d,

k₁ = 6,

ε₀ = 8.85 × 10⁻¹² F/m

a) The capacitance of the air capacitor,

C₀ = `(epsilon_0 A)/d`

= `((8.85 xx 10^-12)(6 xx 10^-3))/((2 xx 10^-3))`

= 26.55 × 10⁻¹² F

= 26.55 pF

b) Q₀ = C₀V

= (26.55 × 10⁻¹²)(100)

= 26.55 × 10⁻¹⁰ C

= 2.655 nC

c) The relative permittivity dielectric k₁ entirely fills the space between the plates (∵ t = d), resulting in C = k₁C₀ as the new capacitance.

V remains the same while the supply is connected.

∴ Q = CV = kC₀V = kQ₀ = 6(2.655 nF) = 15.93 nC

Therefore, the charge on the plates increases.