Q = magnitude of charge at each of the three locations A, B and C = 2 x 10⁻⁶ C
r₁ = distance of charge at origin from charge at B = 50 - 0 = 50 cm = 0.50 m
r₂ = distance of charge at origin from charge at C = 100 - 0 = 100 cm = 1 m
F₁ = magnitude of force by charge at B on charge at origin
F₂ = magnitude of force by charge at C on charge at origin
Magnitude of force by charge at B on charge at origin
[tex]F_{1}=\frac{kQ^{2}}{r_{1}^{2}}[/tex]
inserting the values
[tex]F_{1}=\frac{(9\times 10^{9})(2 \times 10^{-6})^{2}}{0.5^{2}}[/tex]
F₁ = 0.144 N
Magnitude of force by charge at C on charge at origin
[tex]F_{2}=\frac{kQ^{2}}{r_{2}^{2}}[/tex]
inserting the values
[tex]F_{2}=\frac{(9\times 10^{9})(2 \times 10^{-6})^{2}}{1^{2}}[/tex]
F₂ = 0.036 N
Net force on the charge at the origin is given as
F = F₁ + F₂
F = 0.144 + 0.036
F = 0.18 N
from the diagram , direction of net force is towards left or negative x-direction.
The magnitude of the electrostatic force which acts on the charge at the origin is 0.18 N.
The magnitude of the electrostatic force between the charges is determined from Coulomb's law as shown below;
[tex]F = \frac{kq^2}{r^2}[/tex]
[tex]F_{12} = \frac{(9\times 10^9) \times (2 \times 10^{-6})^2}{0.5^2 } \\\\F_{12} = 0.144 \ N[/tex]
[tex]F_{13} = \frac{(9\times 10^9) \times (2 \times 10^{-6})^2}{1^2 } \\\\F_{13} = 0.036 \ N[/tex]
F(net) = 0.144 + 0.036
F(net) = 0.18 N
Learn more about electrostatic force here:https://brainly.com/question/17692887
