Problems with calculating a net field of two charges, help?

Question

Positive charges with magnitudes 3.50 mC and 6.75 mC are located at the origin and at the point (10.00, 0) meters on the x axis, respectively. What are the strength (N/C) and direction (degrees) of the field they generate at the point (0, 10.00) meters on the y axis?

Answer 1

♣ Coulomb says:

vector E=q*r/(4π*ε0*|r|^3), where |r| is distance from a charge q to test point, and r is this vector, ε0=8.854e-12;  

♦ thus our E=E1+E2, where

E1=3.50e-3*(0, 10)/ (4π*ε0*10^3),

E2=6.75e-3*(-10, 10)/ (4π*ε0*(√2*10)^3),

Draw a pic!

♠ thus E= (6.75e-3*(-10)/(√2*10)^3, 3.50e-3*10/10^3 +6.75e-3*10/(√2*10)^3)/ (4π*8.854e-12) =(-214491, 529062) V/m;  

at last |E|=√(214491^2 + 529062^2) = 570888 N/C;

angle = atan(529062/ -214491) = 112°;