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Sagot :
Answer:
the value of the charge is q₃ = [tex]- \frac{4}{9} q[/tex]- 4/9 q, in the position x= l/3
Explanation:
The forces in this system are given by Coulomb's law
F = [tex]k \frac{q_1 q_2}{ r^2}[/tex]
As the forces are vector, we must add them as a vector, also let us take that charges of the same sign repel and charges of the opposite sign attract, therefore for the system to be in equilibrium the third charge must be of the opposite sign, that is, NEGATIVE .
Let's analyze the situation for card charge, let's use the indexes 1 for the charge q₁ = + q located at the origin (x₁ = 0), index 2 for the face q₂ = + 4q located at x₂ = l and the index 3 for the third charge
Let's find the location of charge 3 so that it is in balance
∑ F = F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
we seek every force
F₁₃ = k [tex]\frac{q_1q_3 }{x_{13}^2}[/tex]
F₂₃ = k [tex]\frac{q_2q_3}{ x_{23}^2}[/tex]
the distances are
x₁₃ = x-0 = x
x₂₃ = l -x
we substitute
k \frac{q_1q_3 }{x^2} = k \frac{q_2q_3 }{(l-x)^2}
we solve
(l-x)² = [tex]\frac{q_2}{q_1}[/tex] x²
l² - 2lx + x² = \frac{4q}{q} x²
3x² + 2l x - l² = 0
we solve the quadratic equation
x = [-2l ± [tex]\sqrt{4l^2 + 4\ 3\ l^2}[/tex] / (2 3)
x = [-2l + 2l [tex]\sqrt{1+3}[/tex]] / 6 = 2l [-1 ± 2] / 6
x₁ = -l
x₂ = l / 3
as charge 3 must be between the two charges the correct answer is
x = l / 3
with charge 3 in this location it remains in equilibrium regardless of the value of its NEGATIVE charge.
now let's analyze the force on the other charges
charge 1 so that it is in balance
∑F = F₁₃ - F₁₂ = 0
F₁₃ = F₁₂
[tex]k \frac{q_1q_3}{(l/3)^2} = k \frac{q_1 q_2}{ l^2}[/tex]
l² q₃ = q₂ (l/3)²
q₃ = q₂ / 9
q₃ = 4q / 9
q₃ = [tex]\frac{4}{9} q[/tex]
in summary the value of the charge is q₃ = [tex]- \frac{4}{9} q[/tex]- 4/9 q
in the position x= l/3
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