This can be solved easily if we focus on the physics first. SO what the physics here?
Well let's see at the top left corner and bottom right corner of the square (#q_2 and q_4#). Both charges are at equal distance from the center, thus the net field at center is equivalent to a single charge q of #-10^8 C# at the bottom right corner. Similar arguments for #q_1 and q_3 # lead to the conclusion that #q_1 and q_3 # can replaced by a single charge of #10^-8 C# at the top right corner.
Now let's finf the distance of separation #r#.
#r = a/2 sqrt(2); r^2 = a^2/2#
The field magnitude is given by:
#|E_q| = [kq/r^2]_(r^2=a^2/2) = 2 (kq)/a^2#
and for the #q=2q; |E_(2q)| = 2|E_q| = 4 (kq)/a^2#
#vec(E_("tot")) = E_(q)[(color(blue)(cos(-45)i + sin(-45)j)) +2(color(red)(cos(45) i + sin(45)j)) + (color(green)(cos(225) i + sin(225)j)) + 2(color(purple)(cos(135) i + sin(135)j))] = #
#vec(E_("Net")) = E_(q)[(color(blue)(sqrt(2)/2i - sqrt(2)/2j)) +2(color(red)(sqrt(2)/2 i + sqrt(2)/2)j)) + (color(green)(-sqrt(2)/2 i - sqrt(2)/2j)) + 2(color(purple)(-sqrt(2)/2 i + sqrt(2)/2j))]# the i component cancel out and we are left with: #vec(E_("Net")) =E_(q) * sqrt(2)j #
Compute #E_(q)=2(kq)/a^2; k = 8.99xx10^9; q=10^-8; a^2=(5/100)^2#
#E_(q)=2*(8.99xx10^9 * 10^-8)/(5/100)^2 = 7.19xx10^4 N/C#
#vec(E_("Net")) =7.19xx10^4 * sqrt(2)j = 1.02xx10^5j #
