Physics Help?

Question

Calculate the magnitude and direction of the electric field at the center of a square 55.0 cm on a side if one corner is occupied by a +31.5 µC charge and the other three are occupied by -23.1 µC

Answer 1

Well, I have another way to do it, draw the square with the charges, the electric field is a vector, so for the positive charge, the electric field will be repulsive, for the other charges, the negatives, the electric field will be atractive.

Note : Let's supposed the positive charge is on the left upper corner

So, first of all, let's calculate the magnitude of the electric field for every charge :

For +31.5 µC :

the distance from the corner of the square to the center :

D = 55 / 100*sqrt(2) / 2

9*10^9*31.5*10^-6 / D^2 = 4.7*10^5 N/C

For the negative charges : -23.1 µC

The electric field will be the same, so let's calculate for one of them :

9*10^9*23.1*10^-6 / D^2 = 3.5*10^5 N/C

for each negative charge the electric field : 3.5*10^-5.

But if you have drown the electric field vectors, you will realize, that the electric field by two negative charges, cancel each other.

I supposed that the posititve charge was on the upper left corner so then, you will have :

3.5*10^5 N/C + 4.7*10^5 N/C

why plus ? because one is repulsive, the other one is atractive, so they are in the same direction making 45º with horizontal.

If we supposed the center is the origin, then the magnitude and direction of the electric field will be :

magnitude : 8.2*10^5 (N/C)

direction : 8.2*cos(45)*10^5 i ( + x)
and : 8.2*sen(45)*10^5 -j ( -y)

direction = 5.8*10^5 i - 5.8*10^5 j

Hope that helped

Answer 2

two charge on opposite corner is neutral.(my english isn't good I wish you'v undrestand)
the distance of each cornet to center of square is= (squart2)/2*0.55~1.42*0.275=0.3905m
E=kq/d
Ea=(9*10^9*31.5*10^-6)/ 0.3905~ 725992.13
Eb=(9*10^9*23.1*10^-6)/0.3905~532394.37
and both are in a same direction:
725992.13+532394.37=1258386.5