PLEASE HELP Due today!
1 answer:
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To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.
The intensity depending on the power is defined as
Where
P = Power
r = Radius
Replacing the values that we have,
The definition of intensity tells us that,
Where,
Magnetic field
Permeability constant
c = Speed velocity
Then replacing with our values we have,
Re-arrange to find the magnetic Field B_0
Therefore the amplitude of the magnetic field of this light is
Answer:
q₁ = +12.19 μC
q₂ = +55.81 μC
Explanation:
From Coulomb's law,
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
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
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