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

• CentreShifter
In summary, the electric potential at points in an xy plane is given by V = (2.7 V/m2)x2 -(4.4 V/m2)y2. The magnitude of the electric field at the point (3.4 m, 1.6 m) is 23.137 V/m. The angle that the field there makes with the positive x direction is -37.5 degrees. The components of the field are Ex=18.36 V/m and Ey=-14.08 V/m. The values were obtained by taking the partial derivatives of V with respect to x and y and then taking the magnitude.
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. Ex=18.36 V/m and Ey=-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.

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

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

CentreShifter said:
Ex=18.36 V/m and Ey=-14.08 V/m

Erm, are you sure about Ey?

Nabeshin said:
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.

CentreShifter said:
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?

Nabeshin said:
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.

Sorry my mistake. I agree with Delphi51 ^^

Last edited:

## What is an electric field?

An electric field is a physical field that surrounds electrically charged particles and exerts a force on other electrically charged particles within the field. It is represented by a vector that points in the direction of the force that a positive test charge would experience if placed in the field.

## How do you find the angle of an electric field?

The angle of an electric field can be found by using the inverse tangent function (arctan) of the ratio of the vertical and horizontal components of the electric field vector. This angle represents the direction of the field with respect to a chosen axis, usually the x-axis.

## Why is finding the angle of an electric field important?

Knowing the angle of an electric field is important because it allows us to understand the direction in which a charged particle will move when placed in the field. This information is crucial in many scientific and technological applications, such as designing electronic circuits and predicting the behavior of particles in particle accelerators.

## What units are used to measure the angle of an electric field?

The angle of an electric field is typically measured in degrees (°) or radians (rad). Degrees are a more common unit for everyday use, while radians are often used in mathematical calculations and equations.

## Can the angle of an electric field be negative?

Yes, the angle of an electric field can be negative. This indicates that the field is pointing in the opposite direction of the chosen axis. However, it is common to use the absolute value of the angle when calculating the magnitude of the electric field.

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