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.

i) Calculate the capacitance of the capacitor.

ii) If this capacitor is connected to 100 V supply, what would be the charge on each plate?

iii) 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 m,V = 100V, t = 2 mm = d, k₁ = 6,
ε₀ = 8.85 × 10^-12 F/m
(i) The capacitance of the air capacitor, C₀ = \(\frac{\varepsilon_{0} A}{d}\)
= \(\frac{\left(8.85 \times 10^{-12}\right)\left(6 \times 10^{-3}\right)}{2 \times 10^{-3}}\)
= 26.55 × 10^-12 F = 26.55 pF

(ii) Q₀ = C₀V = (26.55 × 10^-12)(100)
= 26.55 × 10^-10 C = 2.655 nC

(iii) The dielectric of relative permittivity k₁ completely fills the space between the plates (∵t = d), so that the new capacitance is C = k₁C₀.
With the supply still connected, V remains the same.
∴ Q = CV = kC₀V = kQ₀ =6(2.655 nF) = 15.93 nC
Therefore, the charge on the plates increases.