Answer:
f(t) = 28,7 [N]
Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.
The net force on +q₂ is the sum of the force of +q₁ on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)
The two forces have the same direction to the right of charge q₂, we have to add them
Then
f(t) = f₁₂ + f₃₂
f₁₂ = K * ( q₁*q₂ ) / (0,1)²
q₁ = + 8 μC then q₁ = 8*10⁻⁶ C
q₂ = + 3,5 μC then q₂ = 3,5 *10⁻⁶ C
K = 9*10⁹ [ N*m² /C²]
f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻² [ N*m² /C²]* C*C/m²
f₁₂ = 252*10⁻¹ [N]
f₁₂ = 25,2 [N]
f₃₂ = 9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²
f₃₂ = 78,75*10⁻³/ 2,25*10⁻²
f₃₂ = 35 *10⁻¹
f₃₂ = 3,5 [N]
f(t) = 28,7 [N]