## College Physics (4th Edition)

The magnitude of the electric field due to the two point charges is $~402~N/C$
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 $~402~N/C$.