# How do we know if an electric field is negative or positive?

1. Apr 19, 2013

### x86

1. The problem statement, all variables and given/known data
Note: this is actually my own question, not something from a book. So if I am wrong about some terminology please let me know

In physics questions, they will tell you an electric field is |Some Value|. Since |Some Value| (lets call it X) is always positive, how do we know if the field is negative or positive?

2. Relevant equations
F = q * E

3. The attempt at a solution
For example
An electric field has a force of x N/C and a charge of Y Coulombs is placed in the field.

If Y > 0, we know it is a positively charged particle
If Y < 0, we know it is a negatively charged particle

Now, if the question asks how to find the direction the particle moves in; how is this possible to determine?

Let's assume Y > 0 and it is a cation

If the electric field is positively charged, the direction of the cation will be AWAY from the electric field.

If the electric field is negatively charged, the direction of the cation will be TOWARDS the electric field.

Therefore, it is impossible to determine the direction (as far as I understand).

So, how can I determine the magnitude of the particle, if I don't know the charge of the field?

i.e. say E = 3 * 10^9 and q = 3 * 10^-9

Then F = 3 * 3 = 9 N

But in what direction will it be?

We also don't know where the field is positioned, so how is it possible to supply a vector?

EDIT:
Thank you Doc Al. That clears up the confusion for me

Last edited: Apr 19, 2013
2. Apr 19, 2013

### Staff: Mentor

The electric field is a vector. It has a direction, not a "charge". If the field points to the right, the force it exerts on a positive charge will be to the right.

3. Apr 19, 2013

### vela

Staff Emeritus
The electric field doesn't have a force of $x$ N/C. Its magnitude is $x$ N/C. (I'm assuming here that $x>0$.)

The phrase "magnitude of the particle" makes no sense.

You seem to be confusing the electric charge with the electric field it produces.