Answer:
F = - 1.68 10⁻⁴ N
, it is directed to the left of the x-axis
Explanation:
Coulomb's law is
F = k q₁ q₂ / r²
Where K is the Coulomb constant that value 8.99 10⁹ N m²/ C², q are the electric charges and r is the distance between them. Let's apply to our problem for each pair of charges
Let's reduce the magnitudes to the SI system
q₁ = 3.0 nc (1C / 10 9 nC) = 3.0 10⁻⁹ C
x₁ = 2.0 cm (1m / 100cm) = 2.0 10⁻² m
q₂ = -6.0 nC = -6.0 10⁻⁹ C
x₂ = 4.0 cm = 4.0 10⁻² m
q₃ = 5.0 nC = 5.0 10⁻⁹ C
x3 = 0 m
Charges q1 and q3
r = x₁ -x₃
r = 2.0 10⁻² -0
r = 2.0 10⁻² m
F₁₃ = 8.99 10⁹ 3.0 10⁻⁹ 5.0 10⁻⁹ / (2.0 10⁻²)²
F₁₃ = 33.7 10⁻⁵ N
As the charges are of the same sign, the force is repulsive, therefore it is directed to the left of the x-axis
Charges q2 and q3
r = r₂ –r₃
r = 4.0 10⁻² - 0 = 4.0 10⁻² m
F₂₃ = 8.99 10⁹ 6.0 10⁻⁹ 5.0 10⁻⁹ / (4.0 10⁻²)²
F₂₃ = 16.86 10⁻⁵ N
As the charges are of different sign, the force is attractive, therefore it is directed to the right of the x-axis
The force is a vector magnitude, so each component must be added independently, in this case all the forces are on the x-axis, let's take the right direction as positive
F = F₂₃ - F₁₃
F = 16.86 10⁻⁵ - 33.7 10⁻⁵
F = - 16.84 10⁻⁵ N
F = - 1.68 10⁻⁴ N
The negative sign means that it is directed to the left of the x-axis
