Complete Question:
A point charge is placed at each corner of a square with side length a. The charges all have the same magnitude q.Two of the charges are positive and two are negative, as shown in the attached figure.(a) What is the direction of the net electric field at the center of the square ?(i) leftward direction(ii) rightward direction(iii)upward direction(iv) downward direction(b) What is the magnitude of the net electric field at the center of the square due to the four charges in terms of q and a.
Answer:
a) iv) downward direction.
b) 4*[tex]\sqrt{2}[/tex]*k*q / a²
Explanation:
a) In order to find the net electric field due to the 4 charges in the center of the square, we need first to find the distance r from any of the charges (due to symmetry) to the center of the square.
Applying simple trigonometry, we have:
r² = (a/2)² + (a/2)² = 2*a²/4 = a²/2
So, due to symmetry, the magnitude of the field due to any of the charges in the center of the square, applying Coulomb´s Law and the definition of E, is as follows:
E = k*q/r² = k*q/(a²/2)
Now, which is the direction of the E?
By definition, it´s the direction that would follow a positive test charge located in the center of the square, due to the force exerted by the 4 charges on it.
As the electrostatic force is a linear one, we can apply superposition, adding the fields due to the different charges as they were the only present, as follows:
1) Field due to positive charge in the top left corner: It follows the straight line joining the center with the bottom right charge, aiming to it.
2)Field due to positive charge in the top right corner: It follows the straight line joining the center with the bottom left charge, aiming to it.
3) Field due to the negative charge in the bottom right corner: It follows the straight line joining the center with the bottom right charge, aiming to it.
4) Field due to the negative charge in the bottom left corner:It follows the straight line joining the center with the bottom left charge, aiming to it.
So, due to the symmetry we have actually 2 fields adding themselves in each direction, (1) and (3) towards the bottom right, and (2) and (4) towards the bottom left.
If we decompose these fields in their horizontal and vertical components, we can see that the horizontal components compensate each other, so the only component remaining is the vertical one, that adds the 4 vertical components.
Each one of this field component is just the projection, on a vertical axis going through the center, of any of the fields that we have just found.
Due to the figure is a square, the angle between the line joining the center with any of the corners, and the vertical, is 45º.
So, any of the vertical components of the field can be calculated as follows:
Ey = k*q/(a²/2)* cos 45º = (k*q/(a²/2))*(√2/2) = k*q/a²
By symmetry, the total field is the sum of the 4 identical components, all aiming downward, as follows:
Et = 4* Ey = (4*√2*k*q)/a²
