This question is incomplete, the complete question is;
Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is
|FI = |QQ'I / d²
where K = 1/4π∈0, and
∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.
Consider two point charges located on the x-axis:
one charge, q₁ = -18.5 nC, is located at
x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?
Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N
Explanation:
Given that;
Q₁ = -18.5 nC Q₃ = 51 nC Q₂ = 30.5 nC
x₁ = - 1.715m x₃ = - 1.085m x₂ = 0
Now
x - component of Net force on charge Q₃ is
(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²
(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹
(Fnet3)x = -3.3287 × 10⁻⁵ N
