Answer
$(17.1~\mu V/m)~sin~[(2.0\times 10^{14}~rad/s)~t~]$
Work Step by Step
$\sum~E_h = (10.0~\mu V/m)+(5.00~\mu V/m)~cos~45^{\circ}+(5.00~\mu V/m)~cos~(-45^{\circ}) = 17.1~\mu V/m$
$\sum~E_v = (0)+(5.00~\mu V/m)~sin~45^{\circ}+(5.00~\mu V/m)~sin~(-45^{\circ}) = 17.1~\mu V/m = 0$
Note that in the resultant, $\phi = 0$, since $\sum ~E_h \gt 0$ and $\sum~E_v = 0$
We can write the resultant at $P$:
$(17.1~\mu V/m)~sin~[(2.0\times 10^{14}~rad/s)~t~]$
