Finding magnitude and direction of electric field vector (-11i+14j) N/C

[vanitymdl]

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

 

[Chestermiller]

From the expression for the field vector in component form, it looks like the angle lies in the second quadrant.

 

[vanitymdl]

So should i just subtract 51.8 from 180 degrees?

 

[Chestermiller]

So should i just subtract 51.8 from 180 degrees?

Yes.

 

[vanitymdl]

Thank you!

 

[casisas]

How do you know to find theta using arctan(14/11)? and then to subtract that from 180?