Question

The total electric field E consists of the vector sum of two parts. One part has a magnitude of E1 = 1012 N/C and points at an angle θ1 = 32° above the +x axis. The other part has a magnitude of E2 = 1667 N/C and points at an angle θ2 = 58° above the +x axis. Find the magnitude and direction of the total field. Specify the directional angle relative to the +x axis. magnitude N/C direction ° counterclockwise from the +x-axis

Answer 1

here,

the first electric field , E1 = 1012 N/C * ( cos(32) i + sin(32) j)

E1 = (858 i + 536.3 j) N/C

the second electric field , E2 = 1667 N/C * ( cos(58) i + sin(58) j)

E2 = (883.4 i + 1414 j) N/C

the total electric field , E = E1 + E2

E = (858 i + 536.3 j) N/C + (883.4 i + 1414 j) N/C

E = (1741.4 i + 1950.3 j) N/C

the magnitude of total electric field , |E| = sqrt(1741.4^2 + 1950.3^2) N/C

|E| = 2614.6 N/C

the direction of total electric field , theta = arctan(1950.3 /1741.4)

theta = 48.2 degree counterclockwise from +x axis