Four point charges are placed at the corners of a square as shown in the figure. (The top left corner has a +4 mu-C charge, the top right corner has a +1 mu-C charge, the bottom left corner has a -3 mu-C charge, and the right bottom corner has a +4 mu-C charge.) Each side of the square has length 2.0 m. Determine the magnitude of the electric field at the point P, the center of the square. Please explain the process to get the right answer. The mu-C that I was referring to in the problem is my shorthand of
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Expert Chegg Question Answer:
The top left and bottom right corners will cancel each other out at the center, so we only need to worry about the 1 μC and -3 μC charges.
SInce the square is 2 meters in length for each side, the distance to the center of the square from each corner is found by the pythagorean theorem
d = √[(1)2 + (1)2]
d = 1.414 m
Now, the formula for the electric field is E = kq/r2
So E for the 1 μC charge is
E = (9 X 109)(1 X 10-6)/(1.414)2
E = 4500 N/C
This E field will have equal x and y components found by since the field is at a 45o angle to the center
Fx and Fy = 4500(cos 45) = 3182 N/C
Since the charge is positive, and the 1 μC charge is at the top right, the x component pushes downward and the y component pushes left.
For the -3 μC charge, we get
E = (9 X 109)(3 X 10-6)/(1.414)2
E = 13500 N/C
This E field will also have equal x and y components found by since the field is at a 45o angle to the center
Fx and Fy = 13500(cos 45) = 9546 N/C
Since the charge is negative, and the -3 μC charge is at the bottom left, the x component pulls downward and the y component pulls left.
The sum of the x and y components will be equal and found by 3182 + 9546 = 12728 N/C
The final step is to add the total x and y components using the pythagorean theorem
Total E = √[(12728)2 + (12728)2 ] = 18000 N/C (The units can also be listed as V/m, its the same as N/C)