Three equal point charges of 2 µC are in free space at (0,0,0) ,(2,0,0)
And (0,2,0) ,respectively find the force on Q4 = 5µCat (2,2,0)
Answers
Explanation:
hope I'm right. and hope it will help
Concept:
The force between the two charges can be calculated as,
F₁₂ = (q₁ q₂)(R₂-R₁)/ 4π∈₀ |R₂-R₁|³
Where q₁, and q₂ are the charges and R₂ - R₁ is the distance between the charges in the vector form.
Given:
Three equal point charges of 2 µC are at (0,0,0) ,(2,0,0) and (0,2,0).
Find:
The force applied on Q₄= 5µC at (2,2,0).
Solution:
The force between the charge at (0,0,0) and Q₄= 5µC at (2,2,0),
F₁₄ = (q₁ q₄)(R₄-R₁)/ 4π∈₀ |R₄-R₁|³
F₁₄ = (2×10⁻¹²×5×10⁻¹²)(2i+2j+0k)/ 4π∈₀ (2√2)³
F₁₄ = 7.95×10⁻³ (i+j) J
Similarly, the force between the charge at (2,0,0)and Q₄= 5µC at (2,2,0),
F₂₄ = (q₂ q₄)(R₄-R₂)/ 4π∈₀ |R₄-R₂|³
F₂₄ = (2×10⁻¹²×5×10⁻¹²)(0i-2j+0k)/ 4π∈₀ (2)³
F₂₄ = 22.5× 10⁻³ j N
Similarly, the force between the charge at (0,2,0) and Q₄= 5µC at (2,2,0),
F₃₄ = (q₃ q₄)(R₄-R₃)/ 4π∈₀ |R₄-R₃|³
F₃₄ = (2×10⁻¹²×5×10⁻¹²)(2i+0j+0k)/ 4π∈₀ (2)³
F₃₄ = 22.5× 10⁻³ i N
So, the total force is
F = F₁₄ + F₂₄ + F₃₄ = 7.95×10⁻³ (i+j) J + 22.5× 10⁻³ j J + 22.5× 10⁻³ i J
F = 30.45 × 10⁻³ (i+j) J = 3.045 × 10⁻² (i+j) N
Hence, the total force on the charge Q₄= 5µC at (2,2,0) is 3.045 × 10⁻² (i+j) N.
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