Two point charges, Q1 = -7.0 and Q2 = 2.0 are located between two oppositely charged parallel plates, as shown in the figure . The two charges are separated by a distance of x = 0.33m . Assume that the electric field produced by the charged plates is uniform and equal to E = 6.9×104 .

Calculate the magnitude of the net electrostatic force on Q1 .

-3.81*10EXPO4

To calculate the magnitude of the net electrostatic force on Q1, we can use the formula for the electric force between two point charges. The formula is:

F = k * |Q1 * Q2| / r^2

Where:
- F is the force between the charges.
- k is the electrostatic constant, approximately equal to 9 × 10^9 Nm^2/C^2.
- Q1 and Q2 are the magnitudes of the charges.
- r is the distance between the charges.

In this case, we have Q1 = -7.0 (in Coulombs) and Q2 = 2.0 (in Coulombs). The distance between the charges is x = 0.33m.

First, we calculate the force between Q1 and Q2:

F1 = k * |Q1 * Q2| / r^2

F1 = (9 × 10^9 Nm^2/C^2) * |(-7.0 C) * (2.0 C)| / (0.33m)^2

Next, we need to consider the force due to the electric field produced by the charged plates (F2). The electric field strength (E) is given as E = 6.9 × 10^4 N/C. The force is given by:

F2 = Q1 * E

F2 = (-7.0 C) * (6.9 × 10^4 N/C)

The net electrostatic force on Q1 is given by the sum of F1 and F2:

Net Force = F1 + F2

Finally, we substitute the calculated values into the formula to find the magnitude of the net electrostatic force on Q1.