Answer:
Er = 231.76 V/m, 27.23° to the left of E1
Explanation:
To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:
E1:
[tex]E_1_x = 0V/m\\E_1_y=100V/m[/tex]
E2:
Keep in mind that the x component of electric field E2 is directed to the left.
[tex]E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m[/tex]
∑x: [tex]E_1_x+E_2_x = 0V/m - 106.07V/m = -106.07V/m[/tex]
∑y: [tex]E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m[/tex]
The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.
[tex]||E_r||= \sqrt{(-106.07V/m)^2+(206.07V/m)^2} = 231.76 V/m\\\\\alpha = arctan(\frac{206.7 V/m}{-106.07 V/m}) = 117.24degrees[/tex]
or 27.23° to the left of E1.
