How to find angle Electric field makes to x-axis?

[Dana Fishel]

Homework Statement

Find the angle that

makes with the positive x axis:
(Consider vectors above the x-axis to make positive angles and vectors below the x-axis to make negative angles.)

Homework Equations

pheta = arctan (Ex/Ey)
tan(pheta) = Ex/Ey

The Attempt at a Solution

I already found that Ex= -2 N/C and Ey= -2 and IEI = 2.83 N/C
So I tried pheta = arctan (-2/-2) = 45 degrees but it's still marked incorrect.

 

[TSny]

Hello and welcome to PF!

Try to decide which quadrant of the x-y plane the E vector points in. Note the signs of the x and y components.

The arctan function of many calculators will only return angles between -90 and + 90 degrees (i.e., 1st and 4th quadrants).

 

[Dana Fishel]

I wasn't sure about this so I tried +45 degrees and -45 degrees, but those are wrong. Would subtracting from 180 or 360 help?

 

[TSny]

Did you make a sketch showing the E vector in the x-y plane? Can you see from the sketch the answer for the angle? To get the answer, you shouldn't have to use your calculator.

 

[Dana Fishel]

I drew one, but can't get it to upload. So since both values are negative, it would be in quadrant 3, but I'm still confused about finding pheta.
Since Ex and Ey make a triangle with the y and x-axis and we want to use tan, is pheta 90 degrees?

 

[Dana Fishel]

Never mind. I tried doing pheta = arcos of (-2/2.83) = -135degrees (since quadrant 3)
and that's correct, but I don't really get why it works...

Is it because the angle you get is the outside angle to the triangle? (Sorry this sounds really weird without the picture.)

Thank you for your help though!

 

[gneill]

The arctan function cannot tell how the individual signs of any vector components are distributed between the numerator or denominator of the ratio that forms its argument. It only sees the result of any such division.

A positive valued argument may indicate a vector in the first quadrant (two positive values for the ratio) or in the third quadrant (two negatives yielding a positive valued argument). A negative argument could mean a vector in either the 2nd or 4th quadrant.

Because of this ambiguity which cannot be avoided with a single function argument, the arctan function is constrained to only return angles in the 1st and 4th quadrants. It's up to you to sort out the actual quadrant and translate the result accordingly.

Many calculators and programming languages also have the atan2() function which takes two arguments, one for the numerator and one for the denominator of the ratio. It then has access to the individual signs of the vector components and will return a result assigned to the correct quadrant.

 

[TSny]

Suppose you had a vector with x component +2 and y component +2. If you use the arctan function on your calculator to find the angle, you get $arctan(\frac{+2}{+2}) = arctan(1)= 45^o$. This is correct for the angle as measured counterclockwise from the positive x axis.

In your case, you have a vector with x component -2 and y component -2. Your calculator gives $arctan(\frac{-2}{-2}) = arctan(1)= 45^o$. This is not the correct angle as measured from the positive x axis. You can see why your calculator can't handle this correctly since it naturally treats $\frac{-2}{-2}$ as identical to $\frac{+2}{+2}$. In other words, your calculator's arctan function can't tell the difference between a vector in the first quadrant and a vector in the third quadrant. Likewise, it can't distinguish between a vector in the 4th quadrant and vector in the 2nd quadrant.

Generally, your calculator will give the correct answer for vectors in the 1st and 4th quadrants. For the 2nd and 3rd quadrants, adding 180 degrees to what your calculator gives you will give the correct answer for the angle as measured counterclockwise from the positive x-axis. For example, in your case, your calculator gave an answer of 45 degrees. But a rough sketch of the vector shows that it is in the third quadrant. If you add 180 to 45 you get 225 degrees as measured counterclockwise from the positive x axis; or, as they say, 225 degrees "above the x axis". You can see that this is the same as 135 degrees "below" the x-axis (i.e., 135 degrees clockwise from the positive x axis.)

To be safe, it is always a good idea to draw a sketch of the vector. Then you can construct your own right triangle on the sketch and use simple trig on the triangle to find the direction of the vector.