College Physics (4th Edition)

$E = 12.7~N/C$
We can find the magnitude of the horizontal component of the electric field due to the charge at the bottom corner: $E_x = \frac{kq}{r^2}$ $E_x = \frac{(9.0\times 10^9~N~m^2/C^2)(1.00\times 10^{-9}~C)}{(1.0~m)^2}$ $E_x = 9.0~N/C$ We can find the magnitude of the vertical component of the electric field due to the charge at the top corner: $E_y = \frac{kq}{r^2}$ $E_y = \frac{(9.0\times 10^9~N~m^2/C^2)(1.00\times 10^{-9}~C)}{(1.0~m)^2}$ $E_y = 9.0~N/C$ We can find the magnitude of the electric field at point D due to the two point charges: $E = \sqrt{E_x^2+E_y^2}$ $E = \sqrt{(9.0~N/C)^2+(9.0~N/C)^2}$ $E = 12.7~N/C$