# Homework Help: Finding the angle of an electric field with respect to an axis

1. Feb 20, 2009

### CentreShifter

1. The problem statement, all variables and given/known data

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.

2. Relevant 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. Ex=18.36 V/m and Ey=-14.08 V/m

3. 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.

2. Feb 20, 2009

### Delphi51

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

3. Feb 20, 2009

### CentreShifter

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

4. Feb 20, 2009

### Nabeshin

Erm, are you sure about Ey?

5. Feb 20, 2009

### CentreShifter

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

6. Feb 20, 2009

### Nabeshin

Where did the 14.08 come from?

7. Feb 20, 2009

### CentreShifter

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

8. Feb 20, 2009

### Nabeshin

Sorry my mistake. I agree with Delphi51 ^^

Last edited: Feb 20, 2009