Question

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?

Answer 1

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;

$E = \frac{V}{d} \\\\E = \frac{22}{2.9*10^{-3}}\\\\E = 7586.207 \ V/m$

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