An air-filled capacitor consists of two parallel plates, each with an area of 8 cm^2 , separated by a distance 2.9 mm. A 22 V potential difference is applied to these plates. What is the magnitude of the surface charge density on each plate?

The magnitude of the surface charge density on each plate is 6.714 x 10⁻⁸ C/m²

Explanation:

Given;

area of the parallel plates, A = 8 cm² = 8 x 10⁻⁴ m²

distance between the plates, d = 2.9 mm = 0.0029 m

potential difference applied to the plates, V = 22 V

The electric field between the plates is given by;

The surface charge density is given by;

σ = ε₀E

σ = (8.85 x 10⁻¹²)(7586.207)

σ = 6.714 x 10⁻⁸ C/m²

Therefore, the magnitude of the surface charge density on each plate is 6.714 x 10⁻⁸ C/m²

when it gains or loses electrons

the part of the wave he is travelling on is the crest