Objects A and B are both positively charged. Both have a mass of 118 g, but A has twice the charge of B. When A and B are placed 10 cm apart, B experiences an electric force of 0.48 N.

(a) How large is the force on A?

(b) What are the charges qA and qB?

(c) If the objects are released, what is the initial acceleration of A?

a) Fa = 1.26 N b) 0.52 μC c) 10.68 m/s²

Explanation:

a) According to Coulomb's law, the electric force on A due to B, is of equal magnitude than the one on B due to A, so the electric force on A is also 0.48 N in magnitude.

Assuming that the charges are at the same level, the electric force on A is completely horizontal, so it has no component in the vertical direction.

In the vertical direction, the only force acting on the mass A is gravity, which is always downward, and it has the following value:

Fg = m*g = mA*g = 0.118 kg*9.8m/s² = 1.16 N

The total force on A, will be just the vector sum of these both forces, so we get:

F = √ (0.48) ² + (1.16) ² = 1.26 N

b) According to Coulomb's Law, the electric force on any of the charges can be expressed as follows:

F = k*qa*qb / (rab) ² (1)

Replacing by the information provided, we have:

F = 9*10⁹N*m²/C²*2*q² / (0.1) ²m² = 0.48 N

Solving for q:

q = 0.52 μC

c) According 2nd Newton's Law, once released the charge will undergo an acceleration due to the total force present, as follows:

Fa = m*a

⇒ a = F/m = 1.26 N / 0.118 kg = 10.68 m/s²