Question

Charge 1 with a mass of 4 grams and a charge of +8 micro-coulombs is initially resting on a horizontal surface with friction. Charge 2 of +18 micro-coulombs is held in place 23 cm directly above charge 1. A third charge of -6 micro-coulombs is initially placed far from charge 1 directly to the left of charge 1. It is slowly brought closer to charge 1, at what distance will it begin to move charge 1 and overcome friction if the coefficient of static friction is 1.31? Answer in cm.

Answer 1

r₁₂ = distance between charge 1 and charge 2 = 23 cm = 0.23 m

F₂ = force by charge 2 on charge 1 = k q₁ q₂/r²₁₂ = (9 x 10⁹) (8 x 10^-6) (18 x 10^-6)/(0.23)² = 24.5 N

m = mass of charge 1 = 4 g = 4 x 10^-3 kg

F_n = normal force on charge 1

Along the vertical direction , force equation is given as

F_n = F₂ + mg

F_n = 24.5 + (4 x 10^-3) (9.8) = 24.54 N

_s = coefficient of static friction = 1.31

static frictional force acting on the charge q1 is given as

F_f = _s   F_n = (1.31) (24.54) = 32.15 N

r = distance between charge q₁ and charge q₃

F₃ = force by charge 3 on charge 1 = k q₁ q₃/r²

To overcome the frictional force ,

F₃ = F_f

k q₁ q₃/r² = 32.15

(9 x 10⁹) (8 x 10^-6) (6 x 10^-6)/r² = 32.15

r = 0.116 m = 11.6 cm