Three identical 22.0*10^-6 C charges are on the corners of an equilateral triangle of side 0.111 m. What is the magnitude of the net force on q1 (60 degree interior angles)?

Answer 1

Force due to q2 = K q1 q2 /d² = Vector F1

Force due to q3 = K q1 q3 /d² = Vector F2

q1 = q2 = q3 = 22*10 ^-6 C

d = 0.111 m

Total force on q1 = Vector F1 + Vector F2

Angle between F1 and F2 = 60 degree

Resultant Force

F² = F1 ² + F2 ² + 2F1 * F2 * Cos 60 ........ Eq 1,

Plug in F1 = F2 = 9*10⁹ *22*10^-6 * 22*10^-6/ ) / (0.11)² and Cos 60 = 0.5

Get F as answer