Calculate the x- and y-components of the net electric field

[masamune]

Four charges q1 = q3 = -q and q2 = q4 = +q, where q = 9 µC, are fixed at the corners of a square with sides a = 1.3 m.

(a) Calculate the x- and y-components of the net electric field at the midpoint M of the bottom side of the square.
(b) Find the total force exerted on q4 by the charges q1, q2, and q3:
(c) Find the force on a test charge Q = -0.4 µC placed at the midpoint M' of the top side of the square:

I figured out that the Y-component of the net electric field at the midpoint M of the bottom side of the square is zero because the fields cancel out leaving only a field with an X component.
To figure out the X component, I tried to use the law of superposition and add the electric fields like vectors. I used E=F/q and then E=(kQ)/r^2 on each of the charges to the left and right of the midpoint. Then, since the electric fields are moving in opposite directions, I subtracted the two fields to get zero, but apparently the X component of the field isn't zero.

For part b, I drew the force vector diagram, so I have 3 vectors with one going left, the other going down and another going to the upper right. I don't know what to do next so any help would be appreciated here.

For part c, it's a similar situation with part A and so the Y component is zero again. This time however, I used coulomb's law and the equation kQq/r^2. I plugged in numbers and got 0.0767. However I believed that the forces on the test charge were in the same direction and so I multiplied by 2 to get 0.1534 . This wasn't right either. Sorry for the long post and any help you can give me would be very awesome.

 

[genxhis]

For part A, the horizontal component is nonzero because one of the charges attracts while the other repels. If each charge is on opposite sides of the point under consideration, then they effectively act in the same direction. For part B, consider breaking each of these vectors down into component form and then summing them component-wise. For part C, consider first that the electric field found in part A is similar to the one acting at the upper midpoint. Now take into account the magnitude of test charge Q to find the force acting on it.

 

[M98Ranger]

(a) To calculate the x- and y-components of the net electric field at the midpoint M of the bottom side of the square, we can use the law of superposition and add the electric fields from each charge at that point. Since the electric fields are vectors, we need to add them like vectors, taking into account their direction and magnitude. The x-component of the net electric field will be the sum of the x-components of the individual electric fields, and the y-component will be the sum of the y-components.

Let's first calculate the electric field at M due to q1 and q2. The distance from q1 to M is a/2 = 0.65 m, and the distance from q2 to M is √(a^2 + a^2)/2 = 0.92 m. Using the equation E = kq/r^2, we can calculate the magnitude of the electric field at M due to each charge:

E1 = (9x10^9 Nm^2/C^2)(9x10^-6 C)/(0.65 m)^2 = 199.53 N/C
E2 = (9x10^9 Nm^2/C^2)(9x10^-6 C)/(0.92 m)^2 = 109.99 N/C

The x-component of the electric field at M due to q1 is Ex1 = E1cos(45°) = 199.53 N/C x cos(45°) = 140.96 N/C. Similarly, the x-component of the electric field at M due to q2 is Ex2 = E2cos(45°) = 109.99 N/C x cos(45°) = 77.78 N/C.

Since the electric fields from q1 and q2 are in opposite directions, we need to subtract their x-components to find the net electric field at M in the x-direction:

Ex = Ex1 - Ex2 = 140.96 N/C - 77.78 N/C = 63.18 N/C

Similarly, the y-component of the net electric field at M can be calculated by subtracting the y-components of the individual electric fields:

Ey = Ey1 - Ey2 = 140.96 N/C - 77.78 N/C = 63.18 N/C

Therefore, the net electric field at M is Ex = Ey =