The total electric field consists of the vector sum of two parts. One part has a magnitude of E 1= 1309.0 N/C and points at an angle = 15.2 o above the x axis. The other part has a magnitude of E 2 = 1924.5 N/C and points at an angle = 40.8 o above the x axis. Find the direction angle of the total field. (Specify the directional angle in degree and relative to the x axis.)
The total electric field consists of the vector sum of two parts. One part has a magnitude of E 1= 1309.0 N/C and points at an angle = 15.2 o above the x axis. The other part has a magnitude of E 2 = 1924.5 N/C and points at an angle = 40.8 o above the x axis. Find the direction angle of the total field. (Specify the directional angle in degree and relative to the x axis.)
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter25: Gauss’s Law
Section: Chapter Questions
Problem 31PQ
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The total electric field consists of the vector sum of two parts. One part has a magnitude of E 1= 1309.0 N/C and points at an angle = 15.2 o above the x axis. The other part has a magnitude of E 2 = 1924.5 N/C and points at an angle = 40.8 o above the x axis. Find the direction angle of the total field.
(Specify the directional angle in degree and relative to the x axis.)
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