Answer:
The resultant field will have a magnitude of 241.71 V/m, 30.28° to the left of E1.
Explanation:
To find the resultant electric fields, you simply need to add the vectors representing both electric field E1 and electric field E2. You can do this by using the component method, where you add the x-component and y-component of each vector:
E1 = 99 V/m, 0° from the y-axis
E1x = 0 V/m
E1y = 99 V/m, up
E2 = 164 V/m, 48° from y-axis
E2x = 164*sin(48°) V/m, to the left
E2y = 164*cos(48°) V/m, up
[tex]Ex: E_{1_{x}} + E_{2_{x}} = 0 V/m - 164 *sin(48) V/m= -121.875 V/m\\Ey: E_{1_{y}} + E_{2_{y}} = 99 V/m + 164 *cos(48) V/m = 208.74 V/m\\[/tex]
To find the magnitude of the resultant vector, we use the pythagorean theorem. To find the direction, we use trigonometry.
[tex]E_r = \sqrt{E_x^2 + E_y^2}= \sqrt{(-121.875V/m)^2 + (208.74V/m)^2} = 241.71 V/m[/tex]
The direction from the y-axis will be:
[tex]\beta = arctan(\frac{-121.875 V/m}{208.74 V/m}) = 30.28[/tex]° to the left of E1.
