Answer
The magnitude of the electric field due to the two point charges is $~402~N/C$
Work Step by Step
We can find the distance $r$ from each point charge to the point $(0.50~m,0.50~m)$:
$r = \sqrt{(0.50~m)^2+(0.50~m)^2}$
$r = 0.707~m$
We can find the x-component of the net electric field:
$E_x = \frac{(9.0\times 10^9~N~m^2/C^2)(20.0\times 10^{-9}~C)}{(0.707~m)^2}~(cos~45^{\circ}) - \frac{(9.0\times 10^9~N~m^2/C^2)(10.0\times 10^{-9}~C)}{(0.707~m)^2}~(cos~45^{\circ})$
$E_x = 127.3~N/C$
The vertical component of the electric field at the point $(0.50~m,0.50~m)$ is the sum of the vertical components of the electric field due to each point charge:
$E_y = \frac{(9.0\times 10^9~N~m^2/C^2)(20.0\times 10^{-9}~C)}{(0.707~m)^2}~(sin~45^{\circ}) + \frac{(9.0\times 10^9~N~m^2/C^2)(10.0\times 10^{-9}~C)}{(0.707~m)^2}~(sin~45^{\circ})$
$E_y = 381.8~N/C$
We can find the magnitude of the electric field due to the two point charges:
$E = \sqrt{E_x^2+E_y^2}$
$E = \sqrt{(127.3~N/C)^2+(381.8~N/C)^2}$
$E = 402~N/C$
The magnitude of the electric field due to the two point charges is $~402~N/C$.
