Answer:
a) Fa = 1.26 N b) 0.52 μC c) 10.68 m/s²
Step-by-step 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