answer) we know the formula
E=kq/r2
first we have to find the electric field due to all the charges in x direction and then also in y direction
let a small positive charge in the middle as shown in the figure,
so first in x direction
E1x=kq1/r2
here r=a2/ 2 ( from the figure)
so E1x=kq1/(a2/2)2cos45o=(2kq1/a2)*cos45o
E2x=2q2/a2)*cos45o
E3x=(-2kq3/a2)cos45o
E4x=(-2kq4/a2)cos450
then in y direction
E1y=(-2kq1/a2)sin45o
E2y=(2q2/a2)sin45
E3y=2kq3/a2)sin45
E4y=(-2kq4/a2)sin45
so adding all the electric field in x direction
Ex=2kcos45/a2( q1+q2-q3-q4)=2kcos45/a2( 7.74nC-18.9nC+18.9nC-7.74nC)=0
so Ex=0
now adding all electric field in y direction
Ey=2ksin45/a2( -q1+q2+q3-q4)=2*8.99*109sin45/0.0382(-7.74*10-9-18.9*10-9+18.9*10-9-7.74*10-9)=1.36*105C
so the answer is 1.36*105 N/Cor 1.4*105 N/C
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