Question

In a parallel plate capacitor with air between the plates, each plate has an area of 5 × 10^–3 m² and the separation between the plates is 2.5 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 2.5 mm thick mica sheet of K = 8 is inserted between the plates while the voltage supply remains connected?

Answer 1

(i) Capacitance, C = \(\cfrac{\varepsilon_0A}d\)

(ii) Charge Q = CV

= 17.7 × 10^–12 × 100

= 17.7 × 10^–10 C

(iii) New charge, Q = KQ

= 8 × 17.7 × 10^–10

= 1.416 × 10^–8 C