Respuesta :
Answer:
q₁ = +12.19 μC
q₂ = +55.81 μC
Explanation:
From Coulomb's law,
[tex]F = k\frac{q_{1}q_{2}}{r^{2}}[/tex]
Where,
F is the electrostatic force
k is the constant of proportionality 9 × 10⁹ Nm²/C²
q₁ and q₂ are the point charges
r is the distance between the charges
[tex]84 = 9 * 10^{9}\frac{q_{1}q_{2}}{0.270^{2}}[/tex]
[tex]q_{1}q_{2} = \frac{84 * 0.270^{2}}{9 * 10^{9}}[/tex]
q₁q₂ = 6.804 × 10 ⁻¹⁰ C
q₁q₂ = 0.6804 × 10 ⁻⁹ C -------(1)
Also, the total charge of the system is 68.0 μC
⇒ q₁ + q₂ = 68.0 μC
For simplicity, q₁ + q₂ = 68 × 10⁻⁶ C --------(2)
Solving equation 1 and 2 simultaneously using substitution method. From equation 2,
q₂ = 68 × 10⁻⁶ - q₁ ---------(3)
Substituting equation (3) into (1)
q₁(68 × 10⁻⁶ - q₁) = 0.6804 × 10 ⁻⁹
68 × 10⁻⁶q₁ - q₁² = 0.6804 × 10 ⁻⁹
q₁² - 68 × 10⁻⁶q₁ + 0.6804 × 10 ⁻⁹ = 0
solving the quadratic equation using quadratic formula,
a = 1, b = - 68 × 10⁻⁶, c = 0.6804 × 10 ⁻⁹
[tex]q_{1} = \frac{-b \± \sqrt{b^{2} - 4ac} }{2a}[/tex]
[tex]q_{1} = \frac{-(- 68 * 10^{-6}) \± \sqrt{(- 68 * 10^{-6})^{2} - (4*1*0.6804*10^{-9}})}{2*1}[/tex]
q₁ = +12.19 μC or +55.81 μC
Testing for the two q₁ to get q₂
When q₁ = +12.19 μC
12.19 μC + q₂ = 68.0 μC
q₂ = 68.0 μC - 12.19 μC
q₂ = +55.81 μC
When q₁ = +55.81 μC
55.81 μC + q₂ = 68.0 μC
q₂ = 68.0 μC - 55.81 μC
q₂ = +12.19 μC
We have two possible solutions for the question.
When q₁ = +12.19 μC, q₂ = +55.81 μC and
When q₁ = +55.81 μC, q₂ = +12.19 μC
Considering the conditions given in the system,
(1) Two positive point charges q₁ and q₂
(2) Charge q₂ is greater than charge q₁, q₂>q₁
The solution that satisfies the condition is q₁ = +12.19 μC, q₂ = +55.81 μC
Value of two positive point charges of system consists which experiences electrostatic force as q1 is 12.91 μC and q2 is 55.81 μC.
What is electrostatics force?
Electrostatics force is the force of attraction of repulsion between two bodies.
According to the Coulombs law, the force of attraction of repulsion between charged two bodies is directly proportional to the product of charges of them and inversely proportional to the square of distance between them. It can be given as,
[tex]F=k\dfrac{q_1\times q_2}{r^2}[/tex]
Here, (k) is the coulombs constant, (q1 and q2) is the charges of two bodies and (r) is the distance between the two charges.
Given information-
The total charge of the system is 68.0 μC.
The electrostatic force, each charge experiences, is of magnitude 84.0 N.
The separation between the two charges is 0.270 m.
As the total charge of the system is 68.0 μC. thus,
[tex]q_1+q_2=68\\q_1=68-q_2[/tex]
Let the above equation is equation 1.
As, the electrostatic force, each charge experiences, is of magnitude 84.0 and the separation between the two charges is 0.270 m. thus by the Coulombs law,
[tex]84=9\times10^{9}\times\dfrac{q_1\times q_2}{(0.27)^2}[/tex]
Put the value of [tex]q_1[/tex] from equation 1 into the above equation as,
[tex]84=9\times10^{9}\times\dfrac{(68-q_2)\times q_2}{(0.27)^2}\\q_2^2-68q_2-6.804\times10^{-10}=0[/tex]
Solving the above quadratic equation, we get.
[tex]q_2=12.19,\;\;\; 55.81\rm \mu C[/tex]
Put this value of [tex]q_2[/tex] in equation 1, we get,
[tex]q_1=55.81, \;\;\;12.19\rm \mu C[/tex]
Now the given have that [tex]q2>q1[/tex]. Thus, the value of q1 is 12.91 μC and q2 is 55.81 μC.
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