Answer:
q1 = 1.68 10⁻⁶ C and q2 = 0.56 10⁻⁶ C
Explanation:
Let's write Newton's second law for the two spheres, let's start by looking for the stress components with trigonometry
sin θ = Tx / T
Tx = T sin θ
Cos θ = Ty / T
Ty = cos θ
Case 1 q1 = q2
Fe -Tx = 0
Ty -W = 0
Case 2 the charge (q) is the same for the two spheres,
Fe-Tx = 0
Ty -W = 0
T cos θ = mg
T = mg / cos θ
k q q / r² - T sin 30 = 0
k q2 / r² - mg / cos 30 sin 30 = 0
k q2 / r² = mg tan 30
Let's look for the separation of the spheres, using trigonometry
sin 30 = a / L
a = L sin 30
a = 0.500 sin 30
a = 0.25
This is the length from a sphere to the midpoint of the triangle, whereby the distance between the two spheres is double
r = 2 a = 2 0.25
r = 0.500 m
Let's calculate the charge
k q² / r² = mg tan 30
q2 = mg r² /k tan 30
q2 = 8 10⁻³ 9.8 0.5² /8.99 10⁹ tan 30
q2 = 1.26 10⁻¹² C2
Now let's work in case 1 q1≠q2
We have the same equations, but the angle T = 20º
a = L sin 20
r = 2 a
k q1 q2 / r2 = mg tan 20
Let's calculate the charges
a = 0.500 sin 20
a = 0.171 m
r = 0.342 m
q1q2 = mg r² / k tan 20
q1q2 = 8 10⁻³ 9.8 0.342² / 8.99 10⁹ tan 20
q1q2 = 0.371 10⁻¹² C2
Let's write the system of equations
q2 = 1.26 10⁻¹² (1)
q1 q2 = 0.371 10⁻¹²
q1 + q2 = Q
When analyzing this system of equations, we see that the load (q) of the spheres should be half of the total load
q = Q / 2
With (1)
Q² / 4 = 1.26 10⁻¹²
Q = √ 4 1.26 10⁻¹²
Q = 2.24 10⁻⁶ C
We solve the other two equations
(Q -q2) q2 = 0.371 10-12
Q q2 - q22 = 0.371 10-12
q2 2 -2.24 10-6 q2 + 0.371 10-12
We solve the second degree equation
q2 = {2.24 10⁻⁶ + - √ [(2.24 10⁻⁶) 2 - 4 1 0.371 10⁻¹²]} / 2
q2 = (2.24 10⁻⁶ + - √ 3,534 10⁻¹²) / 2
q2 = 2.24 10⁻⁶ + - 1.88 10⁻⁶) 2
q2 ’= 2.16 10⁻⁶ C
q2 ’’ = 0.56 10⁻⁶ C
We calculate q1
q1 + q2 = 2.24 10⁻⁶
q1 ’= 2.24 10⁻⁶ - q2’
q1 ’’ = 2.24 10-⁻⁶ -q2 ’’
q1 ’= 0.08 10⁻⁶ C
q1 ’’ = 1.68 10⁻⁶ C
As they indicate that q2 <q1 the correct solution is
q1 = 1.68 10⁻⁶ C
q2 = 0.56 10⁻⁶ C
