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3 years ago

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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 ang

le 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.

1 answer:

V125BC [204]3 years ago

5 0

Answer:

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.

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Explanation:

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In this first part of the problem, we have to resolve the force into its two components, along the x and the y direction.

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First of all, we have to find the horizontal component of the pulling force, which is given by

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