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Magnitude and direction of Electric field

  • Thread starter vanitymdl
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  • #1
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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
 

Answers and Replies

  • #2
19,799
4,048
From the expression for the field vector in component form, it looks like the angle lies in the second quadrant.
 
  • #3
64
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So should i just subtract 51.8 from 180 degrees?
 
  • #4
19,799
4,048
  • #5
64
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Thank you!
 
  • #6
1
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How do you know to find theta using arctan(14/11)? and then to subtract that from 180?
 

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