Since the second charge is 2 m away and is labeled q₂ = - 4 Coulombs. The third charge is 1 m away and is labeled q₃ = + 3 Coulombs, the electrical force between q₂ and q₃ is -1.1 × 10¹¹ N
To answer the question, we need to find the electrical force of attraction between two charges and this is given by Coulomb's law
Coulomb's law
This states that the electrical force of attraction between two charges, F is directly proportional to the product of the charges q and Q and inversely proportional to the square of their distance apart, d.
So, mathematically F = kqQ/d²
Now, given that there are three charges are on a line. A positive charge is on the far left labeled q₁ = + 6 Coulombs. The second charge is 2 m away and is labeled q₂ = - 4 Coulombs. The third charge is 1 m away and is labeled q₃ = + 3 Coulombs.
The electrical force of attraction between q₂ and q₃
So, the electrical force of attraction between q₂ and q₃ is F = kq₂q₃/d² where
- k = 8.99 × 10⁹ Nm²/C²,
- q₂ = - 4 C,
- q₃ = + 3 C and
- d = 1 m (since q₂ and q₃ are 1 m apart)
Substituting the values of the variables into the equation, we have
F = kq₂q₃/d²
F = 8.99 × 10⁹ Nm²/C² × (-4 C) × (+ 3 C)/(1 m)²
F = 8.99 × 10⁹ Nm²/C² × (-12 C²)/(1 m)²
F = -107.88 × 10⁹ Nm²/1 m²
F = -1.0788 × 10¹¹ N
F ≅ -1.1 × 10¹¹ N
Since the second charge is 2 m away and is labeled q₂ = - 4 Coulombs. The third charge is 1 m away and is labeled q₃ = + 3 Coulombs, the electrical force between q₂ and q₃ is -1.1 × 10¹¹ N
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