Question

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separated by a distance d is

|F|=K|QQ′|d2,

where K=1/4πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q1 = -10.0 nC , is located at x1 = -1.695 m ; the second charge, q2 = 30.5 nC , is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.245 m ?

Your answer may be positive or negative, depending on the direction of the force.

Answer 1

The net force exerted by the two charges is 10.97 x 10⁻⁵ N along negative x-direction.

Step-by-step explanation:

K=1/4πϵ0, where ϵ0=8.854×10−12C2

K = 9x10⁹

The electric force on point charge q₃ due to charge q₂ is

F₃₂ = kq₃q₂ / (1.245)²

= (9x10⁹ * 49.5x10⁻⁹ * 30.5x10⁻⁹) / (1.245)²

= 13,587.75 x 10⁻⁹ / 1.55

= 8.76629 x 10⁻⁵ N

The electric force on point charge q₃ due to charge q₁ is

F₃₁ = kq₃q₁ / (1.695 - 1.245)²

= (9x10⁹ * 49.5x10⁻⁹ * 10.0x10⁻⁹) / 0.2025

= 2.2000 x 10⁻⁵ N

The net electric force on point charge q₃ is

F₃ = -F₃₁ - F₃₂

= - 8.76629 x 10⁻⁵ N - 2.2000 x 10⁻⁵ N

= 10.97 x 10⁻⁵ N along negative x-direction