Two small nonconducting spheres have a total charge of (a) When placed 28.0 cm apart, the force each exerts on the other is 12.0 N and is repulsive. What is the charge on each

q = 1 x 10⁻⁵ C = 10 μC

Explanation:

The repulsive  force between the charges is given by Coulumb's Law:

where,

F = Electrostatic Force = 12 N

k = Coulomb's Constat = 9 x 10⁹ Nm²/C²

r = distance between charges = 28 cm = 0.28 m

Since the values or charges are not given. We assume that both charges have same mahnitude. Therefore,

q₁ = q₂ = q = charge on each sphere = ?

Therefore,

q = 1 x 10⁻⁵ C = 10 μC

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft