A charged particle with a charge of -2.0 10⁻⁶ C is positioned close to a nonconducting plate that has a surface charge density of 4.0 10⁻⁶ C m ⁻² . Determine the force of attraction between the particle and the plate.

Step-by-step solution

The electric field produced by a large nonconducting plate with surface charge density is given by: E = 2 ₀ The force experienced by a charged particle of charge q in this electric field is: F = |q|E = |q| 2 ₀ Substituting the given values |q| = 2.0 10⁻⁶ C , = 4.0 10⁻⁶ C m ⁻² , and ₀ = 8.85 10⁻¹² C ^2 N ⁻¹ m ⁻² : F = 2.0 10⁻⁶ 4.0 10⁻⁶ 2 8.85 10⁻¹² F = 8.0 10⁻¹² 17.7 10⁻¹² F 0.45 N