Physics, asked by divyakamala, 4 months ago

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

Answered by akshaykumarvp2002
5

Explanation:

hope I'm right. and hope it will help

Attachments:
Answered by soniatiwari214
0

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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