The total electric field consists of the vector sum of two parts. one part has a magnitude of e1 = 1300 n/c and points at an angle θ1 = 35o above the +x axis. the other part has a magnitude of e2 = 1700 n/c and points at an angle θ2 = 55o above the +x axis. find the magnitude and direction of the total field. specify the directional angle relative to the x axis.

2954.6 N/C, 46.36 degree from positive  axis

Explanation:

E1 = 1300 N/C, θ1 = 35 degree

E2 = 1700 N/C, θ2 = 55 degree

Now write the electric fields in vector form

E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j

E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j

Resultant electric field

E = E1 + E2

E =  1064.9 i + 745.6 j + 975.08 i + 1392.6 j

E = 2039.08 i + 2138.2 j

Magnitude of E

E = sqrt (2039.08^2 + 2138.2^2)

E = 2954.6 N/C

Let it makes an angle Φ from X axis

tan Φ = 2138.2 / 2039.08 = 1.049

Φ = 46.36 degree from positive X axis.

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: