Christopher B. answered 02/23/22
Experienced Physics Teacher/Tutor with Engineering Background
Hey Gabriel,
What you want to do here is say that x = distance from the 9μC charge, which makes 4-x = distance from the 4μC charge. Now you have two equations for the force on this third charge (one caused by the 9μC charge, the other by the 4μC charge), which you can set equal to each other.
Be careful when you are squaring the distances in the denominator, (4-x)2 should yield a pretty ugly denominator on that side of the equation, and then you should be looking at a quadratic equation -- when you get 2 solutions, only one should be between the 2 charges (ie: x < 4m).
Good luck.
Gabriel Z.
Also How would I go about doing this problem, Calculate the magnitude of the force, in Newtons, on a positive 5μC charge at the midpoint between the two charges.02/23/22

Christopher B.
02/23/22
Gabriel Z.
Yes, they want to know the magnitude of the force on a 5muC charge at 2m02/23/22
Gabriel Z.
So for that would I do, (8.99*10^-6)* (9*10^-6)*(5*10^-6)/2^2+(8.99*10^-6)*(4*10^-6)(5*10^-6)/2^2 ?02/23/22
Gabriel Z.
Do I want to multiply each charge by the middle charge (5muC)? (8.99*10^9 * (9-10^-6)(5*10^-6)/(4-x)^2 = (8.99*10^9)* (4*10^-6)(5*10^-6)/(4-x)^202/23/22