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The position vector of two 1nC charges are (1,1,-1)m and (2,3,1)m respectively.Then magnitude of coulombian force between two charges is.?
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The position vector of two 1nC charges are (1,1,-1)m and (2,3,1)m resp...
Introduction
To find the magnitude of the Coulombian force between two point charges, we can use Coulomb's law, which states that the force between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
Given Data
- Charge 1 (q1) = 1 nC = 1 × 10^-9 C
- Charge 2 (q2) = 1 nC = 1 × 10^-9 C
- Position of Charge 1 (r1) = (1, 1, -1) m
- Position of Charge 2 (r2) = (2, 3, 1) m
Distance Calculation
The distance (r) between the two charges can be calculated using the distance formula:
- r = √[(x2 - x1)² + (y2 - y1)² + (z2 - z1)²]
- Here, (x1, y1, z1) = (1, 1, -1) and (x2, y2, z2) = (2, 3, 1)
Calculating:
- r = √[(2 - 1)² + (3 - 1)² + (1 + 1)²]
- r = √[1 + 4 + 4] = √9 = 3 m
Coulomb's Law
The formula for the magnitude of the force (F) is:
- F = k * |q1 * q2| / r²
Where:
- k = Coulomb's constant ≈ 8.99 × 10^9 N m²/C²
Substituting the values:
- F = (8.99 × 10^9) * |(1 × 10^-9) * (1 × 10^-9)| / (3)²
- F = (8.99 × 10^9) * (1 × 10^-18) / 9
- F = (8.99 / 9) × 10^-9 N
- F ≈ 1.0 × 10^-9 N
Conclusion
The magnitude of the Coulombian force between the two 1 nC charges is approximately 1.0 × 10^-9 N.
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The position vector of two 1nC charges are (1,1,-1)m and (2,3,1)m respectively.Then magnitude of coulombian force between two charges is.?
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