Finding the angle of an electric field with respect to an axis

[CentreShifter]

Homework Statement

The electric potential at points in an xy plane is given by V = (2.7 V/m²)x² -(4.4 V/m²)y². What are (a) the magnitude of the electric field at the point (3.4 m, 1.6 m) and (b) the angle that the field there makes with the positive x direction.

Homework Equations

I have already taken the partial derivatives of V with respect to x and y to get the components to the field and then the magnitude. E_x=18.36 V/m and E_y=-14.08 V/m

The Attempt at a Solution

The magnitude of the field at the given point is 23.137 V/m. I absolutely cannot for the life of me figure out how to calculate that angle, and I know it's in front of my face. I've tried arctan, but it just doesn't work.

 

[Delphi51]

Surely it has to be arctan(Ex/Ey) = -37.5 degrees.

 

[CentreShifter]

That's what I thought! Apparently it's not, hence my frustration.

 

[Nabeshin]

E_x=18.36 V/m and E_y=-14.08 V/m

Erm, are you sure about Ey?

 

[CentreShifter]

Erm, are you sure about Ey?

If I'm wrong it's the sign. I know this because I have verified the magnitude is correct.

 

[Nabeshin]

If I'm wrong it's the sign. I know this because I have verified the magnitude is correct.

Where did the 14.08 come from?

 

[CentreShifter]

Where did the 14.08 come from?

I used the definition of the derivative: $$\frac{f(a,b+h)-f(a,b)}{h}$$. h was .001.

 

[Nabeshin]

Sorry my mistake. I agree with Delphi51 ^^