## College Physics (4th Edition)

$E = 9.79\times 10^5~N/C$
We can find the distance $r$ from the charge at the opposite corner to point A: $r = \sqrt{(0.300~m)^2+(0.300~m)^2}$ $r = 0.4243~m$ There is a non-zero horizontal component of the electric field due to the charge at the opposite corner from point A. We can find the x-component of the electric field: $E_x = \frac{kq}{r^2}~cos~45^{\circ}$ $E_x = \frac{(9.0\times 10^9~N~m^2/C^2)(7.00\times 10^{-6}~C)}{(0.4243~m)^2}~(cos~45^{\circ})$ $E_x = 2.474\times 10^5~N/C$ The vertical component of the electric field at point A is the sum of the vertical components of the electric field due to each point charge: $E_y = \frac{(9.0\times 10^9~N~m^2/C^2)(7.00\times 10^{-6}~C)}{(0.300~m)^2} + \frac{(9.0\times 10^9~N~m^2/C^2)(7.00\times 10^{-6}~C)}{(0.4243~m)^2}~(sin~45^{\circ})$ $E_y = 9.474\times 10^5~N/C$ We can find the magnitude of the electric field due to the two point charges: $E = \sqrt{E_x^2+E_y^2}$ $E = \sqrt{(2.474\times 10^5~N/C)^2+(9.474\times 10^5~N/C)^2}$ $E = 9.79\times 10^5~N/C$