Net electric force on neg charge next to a pos charge, inbetween a elec field

[hopeless123]

Homework Statement

Two point charges, Q1 = -6.9 µC and Q2 = 1.1 µC are located between two oppositely charged parallel plates, as shown in Fig. 16-65. The two charges are separated by a distance of x = 0.38 m. Assume that the electric field produced by the charged plates is uniform and equal to E = 72000 N/C. Calculate the net electrostatic force on Q1 and give its direction.

Homework Equations

Felectric = K (Q1 * Q2) / (r^2)

Felectric = qE

The Attempt at a Solution

I'm unsure how to approach from this point. I found the force between the two charges.

F = (9x10^9) [(6.9 x 10^-6)(1.1 x 10^-6) / (.38 ^2)] = .4730609418 N

Then i figured that because it is a constant electric field I could use this equation.
Since the problem states it wants the net charge on Q1, i used that charge for this equation.
F = qE = (6.9 x 10^-6)(72000) = .4968 N

Now I'm trying to find the netforce and whatever I put in seems to be wrong. Electric fields flow from positive to negative so I figured adding these two values would be the correct answer. The force of attraction to Q2 to the right and the force of the electric field to the right. What am I doing wrong?

 

[Alewhey]

Homework Statement

Two point charges, Q1 = -6.9 µC and Q2 = 1.1 µC are located between two oppositely charged parallel plates, as shown in Fig. 16-65. The two charges are separated by a distance of x = 0.38 m. Assume that the electric field produced by the charged plates is uniform and equal to E = 72000 N/C. Calculate the net electrostatic force on Q1 and give its direction.

Homework Equations

Felectric = K (Q1 * Q2) / (r^2)

Felectric = qE

The Attempt at a Solution

I'm unsure how to approach from this point. I found the force between the two charges.

F = (9x10^9) [(6.9 x 10^-6)(1.1 x 10^-6) / (.38 ^2)] = .4730609418 N

Then i figured that because it is a constant electric field I could use this equation.
Since the problem states it wants the net charge on Q1, i used that charge for this equation.
F = qE = (6.9 x 10^-6)(72000) = .4968 N

Now I'm trying to find the netforce and whatever I put in seems to be wrong. Electric fields flow from positive to negative so I figured adding these two values would be the correct answer. The force of attraction to Q2 to the right and the force of the electric field to the right. What am I doing wrong?

You have the right equations and the right idea - E fields can be summed linearly - however, Q1 has -ve charge and so will be attracted to the +ve plate, and also to (+vely charged) Q2. So the forces are in opposite directions and one should subtract them to find the net force.