# Ideas of field

1. Sep 14, 2010

### jwxie

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

Here is an interesting reading to my question.
http://coraifeartaigh.wordpress.com/2010/02/27/what-is-a-field/

The author wrote:
The reason he said each charge will experience a different force is assuming different r (and not a constant r)?

Another question is to understand the purpose of a field (in mathematics and physics).

So a field is use to describe the effect of a test particle that places in space?

Any help is appreciated! Thank you.

2. Sep 14, 2010

### Uku

The test particle will experience a force which depends on the distance from the point-charge and on it's own charge magnitude.

"Each charge you bring up to A will experience a different force" is valid if the charges are different in magnitude.

There is no "purpose" for a field. It just exists. Rather, a test particle is used to describe the field.

3. Sep 14, 2010

### collinsmark

Hello jwxie,
I admire your desired to learn the meaning behind the equations! But I don't think the author was talking about different 'r' values. Instead, I think the author was talking about different q values, where q is the electric charge of B. In other words, if we only talk about forces and forces alone, the force on (B) caused by charge (A) is

$$F = k \frac{q_A q_B}{r^2}$$

So when the author says, "force experienced by any charge B due to A will also depend on the magnitude of B" I think he's talking about different values of qB.
Well, I think the author is trying to say that the purpose of defining a field in the first place is to determine the effects that charge (A) has on the space around it, while removing the dependency of qB.

While it's an otherwise well written article, I think it might be a little unclear on this point. Let's call charge (B) the test charge. Then the magnitude of the electric field caused by charge (A) is

$$E = F/q_B = \frac{k\frac{q_A q_B}{r^2}}{q_B} = k\frac{q_A}{r^2}.$$

Notice the magnitude of the electric field caused by (A) is not a function of the charge of test charge qB, which can be anything. I think that is what the author is trying to say.

[Edit: 'Looks like Uku beat me to the point.]

Last edited: Sep 14, 2010
4. Sep 14, 2010

### jwxie

Ha, Great helps from both of you. Appreciated.

I am not so bright when it comes to understanding the math and physics concepts. It takes me quite a while to understand every thing. Well that's how an ordinary physics major is. I wished I was a genius in math and physics LOL.

Thank you!