# Homework Help: Magnitude and angle of an electric field vector

1. Feb 14, 2017

### Jrlinton

1. The problem statement, all variables and given/known data
The electric potential at points in an xy plane is given by V = (1.5 V/m2)x2 -(2.9 V/m2)y2. What are (a) the magnitude of the electric field at the point (3.9 m, 2.9 m) and (b) the angle that the field there makes with the positive x direction.

2. Relevant equations

3. The attempt at a solution
Okay, Part a I have completed correctly by multiplying the derivative of the difference potential and the negative of direction component for both x and y.
So for x we get
1.5*-2=-3(x)
And y
2.9*-2=-5.8(y)
Inserting the given values for x and y I get that
x=-11.7
y=-16.82
Using pythagorean theorem I get that
r=20.4891 V/m

The problem I have (and feel silly about) is coming up with the angle in part b
I think it should simply be:
arctan(-16.82/-11.7)+180°
=235.178° or -124.823° ⇐This was incorrect

2. Feb 14, 2017

### haruspex

It is -(2.9 V/m2)y2, not +(2.9 V/m2)y2

3. Feb 14, 2017

### Staff: Mentor

The arctan function can't tell what signs are associated with the numerator and denominator of the argument that you pass to it, so it can't sort out the right quadrant for the angle by itself. That's up to you. Sketch the vector using the x and y components to find the right quadrant and then adjust the arctan result accordingly.

Alternatively, some calculators provide an atan2(y,x) function that deals with the signs of the arguments for you and correctly places the angle. Others provide rectangular to polar conversion that essentially do the same thing for the angle part of the conversion.

4. Feb 14, 2017

### haruspex

That is all true, but the ambiguity is a rotation through 180 degrees. In the present problem, jr has a result in the third quadrant when it should be in the second.

5. Feb 14, 2017

### Jrlinton

I understand that the function cannot determine the quadrant of the angle and just assumes that it is in the first two quadrants. I also understand that this particular would be in the third quadrant.
I suppose that what I am having trouble with is if it is asking for θ being the blue, yellow or orange angle. I understand that using the arctan function will only give me the absolute value of the purple angle of 55.2° to the x axis. It would actually be -55.2° to the axis. So correct me if I am wrong but each angle above would be:
θorange=180+55.2=235.2°
θyellow=-124.2°
θblue=55.2°

6. Feb 14, 2017

### Jrlinton

Thanks haruspex. I missed that negative sign.

7. Feb 14, 2017

### Staff: Mentor

Thanks. I jumped to a conclusion before confirming the reason for jumping!

I see the underlying issue now.