Question

D | Question 5 18 pts A charge of +1.7 micro-coulombs lies on a horizontal surface. It has a mass of 3 kg and the coefficient of static friction between it and the surface is 1.6. Another charge of -2.6 micro-coulombs lies on the horizontal surface 8 cm to the right of the charge. A third charge which is negative is located 20 cm above the positive charge on the surface. What is the smallest magnitude this charge needs to be in order for the positive charge to overcome static friction such that the negative charge on the surface begins to move it? Answer in micro-coulombs. 93 qi 9 g2

Answer 1

for the charge to move , the net force on it must be greater than or equal to the static frictional force

F_net = f_s

m = mass of charge = 3 kg

$\mu$_s = Coefficient of static friction = 1.6

static frictional force on the charge is given as

f_s = $\mu$_s mg   

so

F_net = f_s = $\mu$_s mg   

F_net = $\mu$_s mg   

F_net =(1.6) (3 x 9.8)

F_net = 47.04 N

r₂ = distance of charge q₂ from q₁ = 0.08 m

r₃ = distance of charge q₃ from q₁ = 0.20 m

F₂ = force by charge q₂ on q₁ = k q₁ q₂/r₂² = (9 x 10⁹) (1.7 x 10^-6) (2.6 x 10^-6)/(0.08)² = 6.22 N

F₃ = force by charge q₃ on q₁ = k q₁ q₃/r₃² = (9 x 10⁹) (1.7 x 10^-6) q₃/(0.20)²

net force using Pythagorean theorem is given as

F_net = sqrt(F²₂ + F²₃ )

47.04 = sqrt((6.22)²+ F²₃ )

F₃ = 46.6 N

we know that

F₃ = (9 x 10⁹) (1.7 x 10^-6) q₃/(0.20)²

46.6 = (9 x 10⁹) (1.7 x 10^-6) q₃/(0.20)²

q₃ = 121.8 x 10^-6 C

magnitude of charge = 121.8 microcoulombs