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Two balloons are charged with an identical quantity and type of charge: -0.0025 C. They are held apart at a separation distance of 8 m. Determine the magnitude of the electrical force of repulsion between them.

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hamzaahmeds

Answer:

F = 878.9 N

Explanation:

The electrostatic force of attraction or repulsion is given by Coulomb's Law as follows:

F = kq₁q₂/r²

where,

F = Force pf repulsion between balloons = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = q₂ = magnitudes of 1st and 2nd charge = 0.0025 C

r = distance between balloons = 8 m

Therefore,

F = (9 x 10⁹ N.m²/C²)(0.0025 C)(0.0025 C)/(8 m)²

F = 878.9 N

The magnitude of the electrical force of repulsion between them is F = 878.9 N

Calculation of the magnitude of the electrical force:

Here we use Coulomb's Law

F = kq₁q₂/r²

Here,

F = Force of repulsion

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = q₂ = magnitudes of 1st and 2nd charge = 0.0025 C

r = distance between balloons = 8 m

So,

F = (9 x 10⁹ N.m²/C²)(0.0025 C)(0.0025 C)/(8 m)²

F = 878.9 N

Hence, The magnitude of the electrical force of repulsion between them is F = 878.9 N

Learn more about force here: https://brainly.com/question/24758639

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