Two point charges, Q1 = -6.7 µC and Q2 = 1.2 µC are located between two oppositely charged parallel plates. The two charges are separated by a distance of x = 0.32 m. Assume that the electric field produced by the charged plates is uniform and equal to E = 75000 N/C. Calculate the net electrostatic force on Q1 and give its direction.

I do not see the geometry diagram. Are you sure you have the signs correct? Try subtracting instead of adding. That seems the most likely error. Your general idea is fine.

(75000*-6.7*10^-6)-(9*10^9*-6.7*10^-6*1.2*10^-6)/0.32^2

Your calculation for the net electrostatic force on Q1 seems to be correct. However, the value you provided for the electric field, E = 75000 N/C, does not match the given information of the problem.

According to the problem, the electric field produced by the charged plates is given as E = 75000 N/C. Based on this assumption, let's recalculate the net electrostatic force on Q1.

ForceQ1 = Force P + Force Q2

Let's calculate each term separately:

Force P = EQ1
= 75000 N/C × (-6.7 µC)
= -0.5025 N

Force Q2 = kQ1Q2 / x^2
= (9 × 10^9 N·m^2/C^2) × (-6.7 µC) × (1.2 µC) / (0.32 m)^2
= -0.020205 N

Now, calculate the net electrostatic force on Q1 by adding both forces:

Net Force = Force P + Force Q2
= -0.5025 N + (-0.020205 N)
= -0.522705 N

So, the net electrostatic force on Q1 is approximately -0.52 N. The negative sign indicates that the direction of the force is opposite to the direction of the electric field.

Your calculations seem mostly correct, but there is a small mistake in your equation for the electrostatic force between Q1 and Q2. The correct equation should be:

FQ2 = kQ1Q2 / x^2

Plugging in the given values, we have:

FQ2 = (9 x 10^9 Nm^2/C^2) * (-6.7 x 10^-6 C) * (1.2 x 10^-6 C) / (0.32 m)^2

Calculating this expression gives:

FQ2 = -0.315 N

Now let's calculate the net electrostatic force on Q1 by summing up the forces from the electric field (ForceP) and the force from Q2 (FQ2):

ForceQ1 = ForceP + FQ2

ForceP = EQ1 = (75000 N/C) * (-6.7 x 10^-6 C) = -0.5025 N

Summing up the forces:

ForceQ1 = -0.5025 N + (-0.315 N)
ForceQ1 = -0.8175 N

So the net electrostatic force on Q1 is approximately -0.8175 N. The negative sign indicates that the force is attractive, meaning it pulls Q1 towards Q2.