Three identical charges, each of 2.0 10⁻⁶ C , are held fixed at the three corners of an equilateral triangle of side 5 cm . Determine the magnitude of the Coulomb force experienced by one of the charges due to the other two.

Step-by-step solution

Let the charges be placed at the vertices of the equilateral triangle. The force exerted by one charge on another at a distance a is given by Coulomb's law: F = 1 4 ₀ q^2 a^2 Substituting the given values q = 2.0 10⁻⁶ C and a = 5 10⁻² m : F = 9 10^9 (2.0 10⁻⁶)^2 (5 10⁻²)^2 F = 9 10^9 4.0 10⁻¹² 25 10⁻⁴ = 36 10⁻³ 25 10⁻⁴ = 14.4 N The charge at any vertex experiences two such forces from the other two charges. Since the triangle is equilateral, the angle between these two forces is 60^ . The magnitude of the resultant