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Answer:
The force magnitude is 1.75 μN acting outward from the origin towards x₂
Explanation:
The given parameters, are;
The location of q₁ = The origin
The location of q₂ = 14.7 mm from q₁
The repulsive force exerted on q₂ by q₁ = 2.62 μN
The location the particle q₂ is located to = 18.0 mm from q₁
By Coulomb's law, we have;
Where;
k = Coulomb constant ≈ 8.99 × 10⁹ kg·m³/(s²·C²)
r = The distance between the particles
F = The force acting between the particles
When r = 14.7 mm F = 2.62 μN
∴ q₂ × q₁ = r² × F/k = (14.7×10^(-3))²×2.62×10^(-6)/(8.9875517923^9) ≈ 1.48 × 10⁻¹⁸
q₂ × q₁ = 1.48 × 10⁻¹⁸ C²
When the distance is increased to 18.0 mm, we have;
F = (8.9875517923^9) × 1.4796647 × 10^(-18)/((18×10^(-3))²) = 1.75 × 10⁻⁶ N
∴ F = 1.75 μN.
Therefore;
The force magnitude is 1.75 μN outward from the origin towards x₂.
Answer:
Explanation:
The difference between parallel and anti parallel alignment of the z component of an electron's spin magnetic dipole moment is given by
where μ= dipole moment B= strength of magnetic field and θ= angle between direction of magnetic field and dipole
here θ_2 and θ_1 are 180 and 0° respectively.
here μ is bohr magnetron or magnetic moment of an electron = 9.27×10^{-24}
put this value we get