Answer:
1438.4 N
Explanation:
Applying,
F = kqq'/r²................... Equation 1
Where F = Electrostatic force on the balloons, k = coulomb's constant, q = charge on the first balloon, q' = charge on the second balloon, r = distance between the two ballons.
From the question,
Given: q = q' = 4.0×10⁻⁵ C, r = 0.1 m
Constant: k = 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = ( 4.0×10⁻⁵×4.0×10⁻⁵×8.99×10⁹)/0.1²
F = 14.384/0.01
F = 1438.4 N
The magnitude of the repulsive force will be 1438.4 N.Opposite charges repel each other by a force known as the repulsive force.
The forces repel each other is known as the repulsive force.According to Coulomb's law, like charges repel each other.In Magnetism repulsive force between magnets of same orientation occurs.
The given data in the problem is;
q₁,q₂ is the charges =4.0 x 10 -5C 78
d is the distance = 0.10 m
F is the magnitude of the repulsive force=?
The magnitude of the repulsive force will be;
[tex]\rm F =\frac{ Kq_1q_2}{r^2} \\\\ \rm F =\frac{ 8.99 \times 1^9 \times 4.0 \times 10^-5 \times 4 \times 10^-5}{(0.10)^2} \\\\ \rm F =1438.4 N.[/tex]
Hence the magnitude of the repulsive force will be 1438.4 N.
To learn more about the repulsive force refer to the link;
https://brainly.com/question/807785
