Answer:
Q_total = 2 10⁻⁶ C
Explanation:
Let's apply the conditions of static equilibrium to this case, in the adjoint we can see a diagram of the forces.
X axis
Tₓ - Fe = 0
Tₓ = Fe
Y axis
[tex]T_{y}[/tex] - W = 0
T_{y} = W
Let's use trigonometry to find the stress components, the angle is measured with respect to the vertical
sin 10 = Tₓ / T
cos 10 = T_{y} / T
Tₓ = T sin 10
T_{y} = T cos 10
we substitute
T sin 10 = Fe
T cos 10 = W
T = mg / cos10
T = 0.120 9.8 / cos 10
T = 1,194 N
Fe = 1,194 sin 10
Fe = 0.2073 N
the electric force is
F = k q₁q₂ / r²
in this case, as the cans touch, they have the same charge and the distance r is searched for by trigonometry
sin 10 = y / L
y = L sin 10
y = 0.30 sin 10
y = 0.052 m
this is the distance from the vertical to one can the distance between the two cans is
y_total = 2y
y_totlal = 2 0.052 = 0.104 m
Fe = F = k q² / r²
q = √ (F r² / k)
q = √ (0.2073 0.104²/9 10⁹) = √ (24.913 10⁻¹⁴)
q = 4.99 10⁻⁷ C
This is the charge of a can, as the transfer is carried out by contact, the flannel transfers half of its charge to the cans and these when separating face one keeps half of the transferred charge, therefore the total charge of the flannel is
Q_total = 4 q
Q_total = 19.97 10⁻⁷ C
Q_total = 2 10⁻⁶ C
