Fnet = 1.125x10⁻²²N.
To solve this problem we are going to use Coulomb's Law equation as follow:
[tex]F = K\frac{ |q_{1}q_{2} |}{r^{2}}[/tex] Where:
K is a constant of proportionality.
q₁ and q₂ are charges that could be positive or negative.
r is the distance between charges.
To calculate the net electrostatic force on q₂ due the other two charges, we have to calculate the force exerted by q₁ on q₂, and the force exerted by q₃ on q₂. Then, the net electrostatic force on q₂ due the other charges is Fnet = F₁₂ + F₃₂.
First, the charges are located on the + X axis of a coordinate system.
The force exerted by q₁ on q₂
[tex]F_{12}=9.0x10^{-9}\frac{N.m^{2}}{C^{2}}\frac{(2x10^{-9}C)(1x10^{-9c})}{(2x10^{-2})^{2}} =4.50x10^{-23}N[/tex]
Both charges are positive and exist repulsion between them. Then, q₂ experiences a repulsion force due q₁ towards the side + X which means F₁₂ is positive
The force exerted by q₃ on q₂.
[tex]F_{32}=9.0x10^{-9}\frac{N.m^{2}}{C^{2}}\frac{(3x10^{-9}C)(1x10^{-9c})}{(4x10^{-2})^{2}} =6.75x10^{-23}N[/tex]
The charges has different signs and exist atraction between them. Then, q₂ experiences an atraction force due q₃ towards the side + X which means F₃₂ is positive.
The net electrostatic force on q₂ due the other charges is:
Fnet = 4.50x10⁻²³N + 6.75x10⁻²³N = 1.125x10⁻²²N
