Answer:
a) 1.6 mN b) -1.6 mN c) -1.6 mN d) 1.6 mN
Explanation:
The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:
F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)
The direction of the force, is along the line that joins the charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.
a) So, for the force on q₂, we have:
F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)
b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:
F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)
c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:
F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)
F₂₁ = 1.6 mN (going in the positive x direction, towards q₂)
is the angle between the direction of the force and the horizontal (x-axis)
, therefore the y-component of the force is