Answer
$\theta=-45^o$
$|E|=3.76\times10^4\frac{N}{C}$
Work Step by Step
$Q_1$__$_r$___$Q_2$
$\hspace{1mm}|$ $\hspace{1.2cm}|$
$\hspace{1mm}|$ $\hspace{1.2cm}|$
$Q_3$______$Q_4$
$E=\frac{F}{q}=k\frac{Q}{r^2}$
$|E_{2,3}|=\frac{Q}{r^2}=(9.0\times10^9\frac{Nm^2}{C^2})\frac{3.25\mu C}{(1.22m)^2}=1.97\times10^4\frac{N}{C}$
$|E_1|=\frac{Q}{r^2}=(9.0\times10^9\frac{Nm^2}{C^2})\frac{3.25\mu C}{(1.73m)^2}=9.77\times10^3\frac{N}{C}$
$\sqrt{2E_{1x}^2}=E_1$
$E_{1x}=E_{1y}$
$E_{1x}=\sqrt{\frac{E_1^2}{2}}=\sqrt{\frac{0.5(9.77\times10^3\frac{N}{C})^2}{2}}=6.91\times10^3\frac{N}{C}$
$E_x=E_y$
$\theta=-45^o$
$|E|=\sqrt{2E_x^2}=\sqrt{2(6.91\times10^3\frac{N}{C}+1.97\times10^4\frac{N}{C})^2}$
$=3.76\times10^4\frac{N}{C}$
