Magnitude and direction of Electric field

AI Thread Summary
The magnitude of the electric field E = (-11i + 14j) N/C is calculated as approximately 17.8 N/C using the formula SQRT((-11)^2 + (14)^2). For the direction, the angle is initially found using the inverse tangent function, resulting in -51.8 degrees, indicating the angle is below the x-axis. To express the angle counterclockwise from the positive x-axis, it is necessary to adjust this value, leading to a correct angle of 308.2 degrees. The discussion emphasizes the importance of recognizing the quadrant of the vector when determining the angle. Understanding how to apply trigonometric functions and quadrant considerations is crucial for accurate calculations.
vanitymdl
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Homework Statement



Calculate the magnitude of the electric field E = (-11i^+14j^)N/C


∣∣E⃗ ∣∣ = _______ N/C


Calculate the direction (relative to the +x-axis ) of the electric field E = (-11i^+14j^)N/C.

θ =_________∘ counterclockwise from the +x-axis

Homework Equations





The Attempt at a Solution



So magnitude would be SQRT(-11^2 + 14^2) = SQRT(317) = 17.8 N/C

For direction you'd take the inverse tan so:

theta = ATAN(14/-11) = -51.8 degrees

But that means the angle is 51.8 degrees BELOW the x-axis, so counter-clockwise it'd be

360-51.8 = 308.2 degrees
BUT I'm getting this part wrong and I don't know why? what am I doing wrong
 
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From the expression for the field vector in component form, it looks like the angle lies in the second quadrant.
 
So should i just subtract 51.8 from 180 degrees?
 
vanitymdl said:
So should i just subtract 51.8 from 180 degrees?
Yes.
 
Thank you!
 
How do you know to find theta using arctan(14/11)? and then to subtract that from 180?
 
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