# Sign of differential length in work

1. Feb 24, 2017

### jonjacson

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

Calculate the work done by a force, against an electric field, to bring a charged particle (2 coulomb) from the point (2,0,0) to (0,0,0). Also calculate the work from (0,0,0) to (0,2,0).

Finally calculate the work done going directly from (2,0,0) to (0,2,0) and compare both results.

The field is E= (2x, -4y)

2. Relevant equations

dW= - q F dr

3. The attempt at a solution

The only trouble I have is with the signs.

I got 24 Joule in the last part, and -8 and 16 in the first two. I guess the -8 is wrong and it should be 8, so the sum is 24, as it should be in a conservative field.

My attempt was:

a) from (2,0,0) to (0,0,0)

Here I think the differential of lenght is -dx, not dx, that is causing all the trouble.

W= - (+2) * ∫ 2x * (-dx)= 4 * ∫x dx = 4 (x2/2) and this must be evaluated from 2, lower limit, to 0 upper limit. So I get:

4 ( 0 - 4/2) = -8

According to the book, the Work is:

W=-2 ∫2x * dx ,

So he gets:

+ 8 Joules.

CLearly he is using a differential of lenght positive, but x goes from 2 to 0, clearly from right to left, and clearly is just the opposite of the field that is 2x, which means is positive.

So what I am doing wrong?

In the work formula there is always a minus to show the applied force is opposite to the field, so once you have introduced that minus sign at the start of the formula, you don't need to specify in which direction the differential of lenght goes?

2. Feb 24, 2017

### haruspex

You need to be consistent with assignments of positive directions. When you write that the work is ∫F.dx you are assuming the F and dx are being taken as positive in the same direction. The force is F, not -F, and the infinitesimal displacement is dx, not -dx. Each of those might take a negative value, but that's another matter. E.g. If the integration range is 1 to 0 then the dx's will be negative.

3. Feb 24, 2017

### jonjacson

So it does not matter the "absolute" value, it is the relative value right?

For example
Ups, I made a mistake, what is given is the field not the force. So we have:

dW= Fdr =-qEdr , where the minus sign means the work is done by a force applied against the field, assuming a possitive charge. After integration this work is positive.
If the force is on the same direction as the field E, we simply have to change the sign to get:

dW= q E dr, and the work will be positive.After integration this work is negative, meaning the work has been done by the original system of charges.

Is this correct?

4. Feb 24, 2017

### rude man

Differential work dW done by the (externally applied) force F is F⋅ds = -qE⋅ds = -2(2x)(-dx) = 4x dx.
dx is by definition positive. It's a magnitude. If you want to go negative it's -dx as I've done above. If I write "-2" the "2" is still positive.
So it's W = Σ(4x dx) = ∫4x dx from 0 to 2, not 2 to 0, = +8.
My best answer is the above: -qE⋅ds = -2(2x)(-dx) = 4x dx, dx > 0 always.

Nice thing about physics is that usually common sense tells you the sign! Keep that in mind and be careful with the grammar. For example, distinguish between work done by an E field vs. work done by an external force against an E field, etc.

5. Feb 24, 2017

### jonjacson

Maybe an image will help.

Why do you say we have to evaluate the integral from 0 to 2 if the exercise is asking from 2 to 0?

6. Feb 24, 2017

### haruspex

No.
You don't have to.

7. Feb 24, 2017

### rude man

Why do you say we have to evaluate the integral from 0 to 2 if the exercise is asking from 2 to 0?
You can integrate from 2 to 0 but then you have to integrate -2(2x)(dx) = -4∫xdx from 2 to 0 = +8 again.
dx is always +; the necessary sign is provided by the limits of integration going from 2 to 0 instead of the other way around.

8. Feb 24, 2017

### haruspex

No, you don't.
No, it isn't.

If I integrate ∫x=01x.dx I get 1/2. If I integrate ∫x=10x.dx I get -1/2. It is unnecessary to turn the second into - ∫x=01x.dx or Into -∫x=10x.|dx|.

Last edited: Feb 24, 2017
9. Feb 24, 2017

### jonjacson

I don't know if I get it. It looks like in this case signs do not represent absolute spatial orientation, they are an artificial criterion to show if two quantities are parallel or antiparallel right?

10. Feb 24, 2017

### haruspex

Which post are you replying to?

11. Feb 24, 2017

### jonjacson

It was a generic sentence.

To make it clear, there are four signs in this equation:

-the first sign, that is there because of a definition
-the sign of the charge
-the sign of the force and of the differential of path

The first, and the last two are related right?

If E is positive on the x axis, and an applied force against E moves the particle on the negative axis direction the work will be positive, supplied to the system.

If E is positive, and there is work done on the particle that moves to the right, that energy is provided by the field E and the work is negative.

But, what if it is not clear from the equations if the net effect is positive or negative?

12. Feb 24, 2017

### haruspex

It is unnecessary to insert a sign relating to the path. It should arise naturally from the integration bounds.
Can you give an example?

13. Feb 24, 2017

### jonjacson

THanks for that. So inside the integral there is no need to add any artificial sign.

What about the first sign? If the force is not applied against the field, but instead we are talking about the field itself there is no minus sign at the start right?

Well I imagine a complex tridimensional field, and a complicated path where I can't see a priori if the work is going to be negative or positive.

14. Feb 24, 2017

### haruspex

You don't need to do anything special according to the direction of the force or route of the path.
$W = \int_S\vec F.\vec {ds}$
If we ignore mass and/or acceleration:
$\vec F = -q\vec E$
$W = -q\int_S\vec E.\vec {ds}$
In the present problem, the first leg is from (2,0,0) to (0,0,0):
$W_1 = -q\int_{x=2}^0 E_x.dx = q\int_{x=0}^2 E_x.dx$

15. Feb 24, 2017

### jonjacson

I find really weird seeing that E is pointing to the right in the x axis, the path is pointing to the left and you don't get an extra minus sign when you multiply them.

Since those two quantities are vectors they have a magnitude, but they also have a direction. ds for me is equal to -dx i.
Being i the unit vector in the positive x axis.

If you write E and ds explicitly you have:

E= (2x,...)
ds= (-dx,...)

That is basic geometry, if you multiply them you get an extra minus sign.

In other words, I would understand what you just wrote if the x axis were positive to the left, and negative to the right of the point (2,0,0), but that is not the case. Maybe I should go to bed and tomorrow look for it.

16. Feb 24, 2017

### haruspex

I assume you mean $\vec{ds}=-\vec i dx$. (ds typically stands for $|\vec {ds}|$.)
If so, that is wrong. By definition, $\vec {ds} = (dx, dy, ....) = \vec i dx+\vec j dy+...$.
dx is an increase in x. If the path is going right to left then that increase is negative, i.e. dx takes a negative value.

17. Feb 24, 2017

### jonjacson

Yes, I meant that.

Now it is clear. Thanks!

18. Feb 24, 2017

Excellent.