Calculating Work Done by Electrical Force on a Moving Charge

[assaftolko]

Ok i always thought that when you reffer to Fdr then dr is a small displacement vector that always points in the direction of movement. Of course i know that the general position vector r is taken from some origin to where the moved object is at, i just thought that for this line integral dr has somewhat different role.. Thanks a lot to both of you!

 

[vela]

When you wrote Fdr, what did you mean? Did you mean $F\,dr$ or $\vec{F}\cdot d\vec{r}$? If it's the former, dr isn't a vector. If it's the latter, it is the small displacement vector that points in the direction of movement.

 

[assaftolko]

I meant both of them in some way.. I don't quite get the transition dr (vector) does from being an entity that points in the direction of motion to: -dr (scalar)... Its not so clear to me what is the role of this minus, which is present in addition to the minus sign generated from these 2 vectors (force and dr) being 180 deg oriented

 

[vela]

I explained why the minus sign appears back in post 19.

 

[assaftolko]

I explained why the minus sign appears back in post 19.

From what I know: dr, when not written as a vector, represents the size of the vector dr, and size is a positive quantity. When you write -dr, as I see it, you are saying that the vector dr is pointing in the negative direction of the r^ axis, my analogue is that of forces in simple Newtonian problems: mg is a vector quantity and if I write it as: -mg in my y-axis force equations then that's because I decided that the positive y-axis is going up.

That's the kind of analogy I'm trying to do here, because I basiclly don't understand something you said earlier, that: |dr|= -dr... this equation is hard for me to grasp... and by the way - why did you write ||\vec{dr}||? What does this 2 bars symbol represent?

 

[vela]

In the integral
$$\int_1^0 dx,$$ what's the sign of dx?

 

[assaftolko]

In the integral
$$\int_1^0 dx,$$ what's the sign of dx?

I don't quite get how to answer this question even... dx is a very small displacement in the x^ axis direction... so my instinct tells me it's positive, otherwise it would have been written as -dx

I'm sorry I'm being such a hard-*** but it's not because I don't want to understand...

 

[vela]

The integral is essentially a sum. By claiming dx>0 in that integral, you're saying by adding up a bunch of positive quantities, you can get a negative answer. You can't simply ignore the limits and assume dx is always positive.

 

[assaftolko]

The integral is essentially a sum. By claiming dx>0 in that integral, you're saying by adding up a bunch of positive quantities, you can get a negative answer. You can't simply ignore the limits and assume dx is always positive.

Ok so what you're saying is that in this integral you wrote here: it's an "inhereted property" embedded inside dx that it's negative, no different than for example the equation x-6=-7

 

[assaftolko]

A problem that emphasizes my difficulty:

A ball with a mass of 0.04kg is moving on a frictionless rail that is given by y=3x^2. The forces involved are the normal force, gravity and an external force F(\vec{r}) = (5x^2y+e^-0.2x , x-0.3y^2). I need to calculate this force's work from yA=12m to yB=0

Here, we also have a case where the bottom integration limit will be bigger than the upper one, but in the official solution they wrote with respect to integration for the y compoment:

\int^{0}_{12}(x-0.3y^2)\bulletdy, and not -dy... So why this problem is different?

 

[vela]

It's because in those solutions, the dot product is found using $\vec{F}\cdot d\vec{r} = F_x\,dx +F_y\,dy$. You don't discard the signs of the individual components when you use this formula. In your approach to the electric potential problem, however, you're using $\vec{F}\cdot d\vec{r} = \|\vec{F}\| \|d\vec{r}\|\cos \theta$. The negative sign arises because of the absolute value. The norm of the vector $d\vec{r}$ is given by $\|d\vec{r}\| = \lvert dr\rvert = -dr$ because $dr<0$.

Say we want to calculate
$$\int_A^B \vec{F}\cdot d\vec{r}$$ where $\vec{F} = \hat{i}$ and points A and B lie, respectively, at (1,0) and (0,0).

We can parameterize the route $\vec{r}(\lambda) = x(\lambda)\,\hat{i}+0\,\hat{j}$ using $x(\lambda) = 1-\lambda$, where the parameter $\lambda$ runs from 0 to 1. Then we have $d\vec{r} = -d\lambda\,\hat{i}$. Here $d\lambda>0$ because the integral goes from $\lambda=0$ to $\lambda=1$. This makes sense because $d\vec{r}$ should point in the -x direction when we move from x=1 to x=0.

If we calculate the dot product in terms of components, we get $\vec{F}\cdot d\vec{r} = (-d\lambda)\,(\hat{i}\cdot \hat{i}) = -d\lambda$. The integral then becomes
$$\int_{\lambda=0}^{\lambda=1} -d\lambda = -1.$$ So far so good. If we use the other method to calculate the dot product, we have
$$\vec{F}\cdot d\vec{r} = \|\vec{F}\| \|d\vec{r}\| \cos \theta = 1\times\lvert d\lambda\rvert\times\cos 180^\circ = -\lvert d\lambda\rvert = -d\lambda,$$ so we'll end up with the same answer, as we should.

Now suppose instead we use the parameterization $x(\lambda) = \lambda$ where $\lambda$ goes from 1 to 0. This time, we'll get $d\vec{r} = d\lambda\,\hat{i}$. If we calculate the dot product in terms of components, we get $\vec{F}\cdot d\vec{r} = d\lambda\,(\hat{i}\cdot \hat{i}) = d\lambda$. The integral then becomes
$$\int_{\lambda=1}^{\lambda=0} d\lambda = -1.$$ There's no need to take into account the signs of the components. It all works out automatically.

If we use the other method to calculate the dot product, we have
$$\int_{\lambda=1}^{\lambda=0} \vec{F}\cdot d\vec{r} =
\int_{\lambda=1}^{\lambda=0} \|\vec{F}\| \|d\vec{r}\| \cos 180^\circ =
\int_{\lambda=1}^{\lambda=0} -\lvert d\lambda\rvert.$$ From this, you can clearly see we must have $\lvert d\lambda\rvert = -d\lambda$ otherwise we get the wrong sign on the final result. But this exactly what we should expect because $d\lambda<0$ with this parameterization of the path.

 

[rude man]

OR:

d s = dx i + dy j but y = 3x² so
d s = (dy/6x) i + dy j
F = F_x i + F_y j so
F * d s = F_xdy/6x + F_ydy

the y component of this dot-product being F_ydy
and the integral of which is W_y = ∫F_ydy with lower limit 12 and upper limit 0. (Of course, total work W = W_x + W_y).

This is based strictly and entirely on vector algebra and the definition of work as the integrated dot-product of force and displacement. There is just no room for further interpretations.

 

[assaftolko]

It's because in those solutions, the dot product is found using $\vec{F}\cdot d\vec{r} = F_x\,dx +F_y\,dy$. You don't discard the signs of the individual components when you use this formula. In your approach to the electric potential problem, however, you're using $\vec{F}\cdot d\vec{r} = \|\vec{F}\| \|d\vec{r}\|\cos \theta$. The negative sign arises because of the absolute value. The norm of the vector $d\vec{r}$ is given by $\|d\vec{r}\| = \lvert dr\rvert = -dr$ because $dr<0$.

Say we want to calculate
$$\int_A^B \vec{F}\cdot d\vec{r}$$ where $\vec{F} = \hat{i}$ and points A and B lie, respectively, at (1,0) and (0,0).

We can parameterize the route $\vec{r}(\lambda) = x(\lambda)\,\hat{i}+0\,\hat{j}$ using $x(\lambda) = 1-\lambda$, where the parameter $\lambda$ runs from 0 to 1. Then we have $d\vec{r} = -d\lambda\,\hat{i}$. Here $d\lambda>0$ because the integral goes from $\lambda=0$ to $\lambda=1$. This makes sense because $d\vec{r}$ should point in the -x direction when we move from x=1 to x=0.

If we calculate the dot product in terms of components, we get $\vec{F}\cdot d\vec{r} = (-d\lambda)\,(\hat{i}\cdot \hat{i}) = -d\lambda$. The integral then becomes
$$\int_{\lambda=0}^{\lambda=1} -d\lambda = -1.$$ So far so good. If we use the other method to calculate the dot product, we have
$$\vec{F}\cdot d\vec{r} = \|\vec{F}\| \|d\vec{r}\| \cos \theta = 1\times\lvert d\lambda\rvert\times\cos 180^\circ = -\lvert d\lambda\rvert = -d\lambda,$$ so we'll end up with the same answer, as we should.

Now suppose instead we use the parameterization $x(\lambda) = \lambda$ where $\lambda$ goes from 1 to 0. This time, we'll get $d\vec{r} = d\lambda\,\hat{i}$. If we calculate the dot product in terms of components, we get $\vec{F}\cdot d\vec{r} = d\lambda\,(\hat{i}\cdot \hat{i}) = d\lambda$. The integral then becomes
$$\int_{\lambda=1}^{\lambda=0} d\lambda = -1.$$ There's no need to take into account the signs of the components. It all works out automatically.

If we use the other method to calculate the dot product, we have
$$\int_{\lambda=1}^{\lambda=0} \vec{F}\cdot d\vec{r} =
\int_{\lambda=1}^{\lambda=0} \|\vec{F}\| \|d\vec{r}\| \cos 180^\circ =
\int_{\lambda=1}^{\lambda=0} -\lvert d\lambda\rvert.$$ From this, you can clearly see we must have $\lvert d\lambda\rvert = -d\lambda$ otherwise we get the wrong sign on the final result. But this exactly what we should expect because $d\lambda<0$ with this parameterization of the path.

First of all let me just say that I'm very grateful you're taking the time and effort with your explanations!

In your simple example of ∫dx from 1 to 0: Of course we do know that in calculus when I'm facing this integral I won't change the sign of dx to be -dx right? This integration should come out negative: -1. So my fundamental problem (I'm not a physics or engineering major so I have no idea what do you mean by parameterization with this lambda: I'm a med student) is to understand why do we know it's essential to put here -dr and not dr. Also: I still would like to know what does dr stand for? I always thought that dr or |\vec{dr}| are 2 ways of writing the same thing: the size of the vector \vec{dr}. It's now clear to me from your posts that they are not the same, so It's crucial for me to understand what does dr (as it's written here) stand for.

 

[vela]

Reread post 25. dr is the infinitesimal version of $\Delta r$.

 

[assaftolko]

Ok I think I got it, thanks a lot!