Answer
$|E|=6.59\times10^7\frac{N}{C}$
$\theta=56.3^o$
Work Step by Step
$E_{Ay}=k\frac{q}{r^2}=(9.0\times10^9\frac{Nm^2}{C^2})\frac{26\mu C}{(0.08m)^2}=3.66\times10^7\frac{N}{C}$
$E_{Ax}=0\frac{N}{C}$
$E_B=k\frac{q}{r^2}=(9.0\times10^9\frac{Nm^2}{C^2})\frac{26\mu C}{(0.08m)^2}=3.66\times10^7\frac{N}{C}$
$E_{By}=1.83\times10^7\frac{N}{C}$
$E_{Bx}=\sqrt{(3.66\times10^7\frac{N}{C})^2-(1.83\times10^7\frac{N}{C})^2}$
$=3.17\times10^7\frac{N}{C}$
$E_y=5.48\times10^7\frac{N}{C}$
$E_x=3.66\times10^7\frac{N}{C}$
$|E|=\sqrt{(5.48\times10^7\frac{N}{C})^2+(3.66\times10^7\frac{N}{C})^2}$
$=6.59\times10^7\frac{N}{C}$
$\theta=\arctan\Bigg({\frac{5.48\times10^7\frac{N}{C}}{3.66\times10^7\frac{N}{C}}}\Bigg)=56.3^o$
