Answer:
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;
[tex]E = \frac{V}{d} \\\\E = \frac{22}{2.9*10^{-3}}\\\\E = 7586.207 \ V/m[/tex]
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
