Answer:
The magnitude of the force between the charges is approximately -2,858.77 N
Explanation:
The given details are;
The magnitude of the given charges;
q₁ = -0.00125 C, q₂ = +0.00333 C
The distance between the charges, r = 3.62 m
The magnitude of the electric force between the charges, 'F', between two charged particles or objects is given as follows;
[tex]F = \dfrac{k \cdot q_1 \cdot q_2}{r^2}[/tex]
Where;
k = Constant = 9 × 10⁹ N·m²/C²
q₁ = The magnitude of charge on the first charged object = -0.00125 C
q₂ = The magnitude of charge on the second charged object = +0.00333 C
r = The distance between the charges = 3.62 m
Plugging in the values, gives;
[tex]F = \dfrac{9 \times 10^9 \ \dfrac{N \cdot m^2}{C^2} \times (-0.00125 \, C) \times (+0.00333 \, C)}{\left( 3.62 \, m \right)^2} \approx -2,858.77 \, N[/tex]
The magnitude of the force between the charges, F ≈ -2,858.77 N
