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Sagot :
Answer:
a) F _total = 15.08 10⁻³ N , b) m = 1,539 10⁻³ kg
Explanation:
a) To solve this problem we can use Coulomb's law to find the force created by each charge
F =[tex]k \frac{\ q_{1} q_{2} }{r_{12}^{2} }[/tex]
and the total force is
F = F₁₅ + F₂₅ + F₃₅ + F₄₅
The bold are vectros, In the exercise they indicate the charges
q₁ = q₂= q₃ = q₄ = 1 10⁻⁶ C
q₅ = 2 10⁻⁶ C
Let's set a reference system where the sphere is on the z axis and the other charges on the xy plane, let's write the coordinates of card charge
sphere (subscript 5)
x₅ = 0
y₅ = 0
z₅ = 2 m
charge 1
x = 0.5 m
y = 0.5 m
z = 0
charge 2
x = -0.5 m
y = 0.5 m
z = 0
charge 3
x = -0.5 m
y = -0.5 m
z = 0
charge 4
x = 0.5 m
y = -0.5 m
z = 0
With these values we can calculate the distance between each charge and the sphere
charge 1 and sphere
r₁₅² = (x₅ -x₁)² + (y₅ - y₁)² + (z₅ -z₁)²
substitute
r₁₅² = (0- 0.5)² + (0 - 0.5)² + (2 -0)²
r₁₅² = 0.5² + 0.5² + 2²
r₁₅² = 4.5
we can see that for the other charges the result is the same since being squared always gives positive
r₁₅ = r₂₅ = r₃₅ = r₄₅
the force created by the card charge on the sphere is the projection on the Z axis of the total force
Let's find the angle with respect to the Z axis
tan φ = r / z
where r is the magnitude of the vector in the xy plane
r = [tex]\sqrt{0.5^{2} + 0.5^{2} }[/tex]
r = 0.7071 m
φ = tan⁻¹ r / z
φ = tan⁻¹ (0.7071 / 2)
φ = 19.5
consequently the total force is
F_total = 4 F cos 19.5
F _total = [tex]4 \frac{k \ q_{1} q_{5} }{r_{15}^{2}}[/tex] cos 19.5
let's calculate
F_toal = 4 9 10⁹ 1 10⁻⁶ 2 10⁻⁶ /4.5 cos 19.5
F _total = 15.08 10⁻³ N
b) For this part indicate that the sphere is in equilibrium with the weight
F_total - W = 0
W = mg
F_total = mg
m = F_total / g
m = 15.08 10³ / 9.8
m = 1,539 10⁻³ kg
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