Line Integral, Dot Product Confusion

[Lost1ne]

From my interpretation of this problem (image attached), the force applied to the point charge is equal and opposite to the repulsive Coulomb force that that point charge is experiencing due to the presence of the other point charge so that the point charge may be moved at a constant velocity. I agree that the equation should yield a positive value, and I agree that the equation is valid, but I'm still a bit confused.

I don't see why this equation would hold if I used this evaluation of the dot product: \vec F \cdot d \vec r = | \vec F | * | d \vec r | * cos(Θ). (My LaTeX failed, but I hope you can see what I mean.) Our applied force vector and charge displacement vector are in the same direction. With that being said, as cos(0) = 0, I wouldn't see why our integral would have the negative sign on the left of the equality. However, removing this negative sign would of course change our answer, resulting in a negative value which would be an incorrect answer. Why does this approach with this interpretation of the dot product not seem to work (or, more likely, where does my thinking go wrong)?

 

[Delta2]

From my interpretation of this problem (image attached), the force applied to the point charge is equal and opposite to the repulsive Coulomb force that that point charge is experiencing due to the presence of the other point charge so that the point charge may be moved at a constant velocity. I agree that the equation should yield a positive value, and I agree that the equation is valid, but I'm still a bit confused.

I don't see why this equation would hold if I used this evaluation of the dot product: \vec F \cdot d \vec r = | \vec F | * | d \vec r | * cos(Θ). (My LaTeX failed, but I hope you can see what I mean.) Our applied force vector and charge displacement vector are in the same direction. With that being said, as cos(0) = 0, I wouldn't see why our integral would have the negative sign on the left of the equality. However, removing this negative sign would of course change our answer, resulting in a negative value which would be an incorrect answer. Why does this approach with this interpretation of the dot product not seem to work (or, more likely, where does my thinking go wrong)?

The $\theta$ angle between $\vec{F}$ and $d\vec{r}$ is actually $\pi$ not 0 and $\cos\pi=-1$. $\vec{F}$ points inwards and $d\vec{r}$ points outwards.

 

[Lost1ne]

The $\theta$ angle between $\vec{F}$ and $d\vec{r}$ is actually $\pi$ not 0 and $\cos\pi=-1$. $\vec{F}$ points inwards and $d\vec{r}$ points outwards.

Okay. In that case, here's the other part of my post that I deleted:

If our infinitesimal vector is defined to point in the positive direction of our coordinate system, then the negative sign would appear as the cos(pi radians) = -1. (That is, if a "dr" was defined to be outwards, away from our "destination charge" and anti-parallel to our "actual displacement. This would be similar to claiming that a vector (dx, dy) always points towards the first quadrant in the XY plane.) This seems a bit weird and introduces a notion of "actual displacement" (of our object of interest) versus the vector (dx, dy, dz) (or whatever it's equivalent is in other coordinate systems) that is seen in our work integral, but the math, at least so far, would work.

I guess my question is more of the interpretation of the differential vector that appears in the work integral. Is this vector truly our infinitesimal displacement vector that points in the direction of the displacement of the object, or is it a vector that is constrained to point in the "positive" direction of our chosen coordinate system? If it's the first, which is what I thought initially, then why aren't things working out?

 

[Delta2]

Okay. In that case, here's the other part of my post that I deleted:

If our infinitesimal vector is defined to point in the positive direction of our coordinate system, then the negative sign would appear as the cos(pi radians) = -1. (That is, if a "dr" was defined to be outwards, away from our "destination charge" and anti-parallel to our "actual displacement. This would be similar to claiming that a vector (dx, dy) always points towards the first quadrant in the XY plane.) This seems a bit weird and introduces a notion of "actual displacement" (of our object of interest) versus the vector (dx, dy, dz) (or whatever it's equivalent is in other coordinate systems) that is seen in our work integral, but the math, at least so far, would work.

I guess my question is more of the interpretation of the differential vector that appears in the work integral. Is this vector truly our infinitesimal displacement vector that points in the direction of the displacement of the object, or is it a vector that is constrained to point in the "positive" direction of our chosen coordinate system? If it's the first, which is what I thought initially, then why aren't things working out?

No $d\vec{r}$ is the vector of displacement, it doesn't always point outwards. From where did you get the impression that it always points outwards or towards the positive direction of the coordinate system?

 

[Lost1ne]

No $d\vec{r}$ is the vector of displacement, it doesn't always point outwards. From where did you get the impression that it always points outwards or towards the positive direction of the coordinate system?

Yeah, it seemed like a very weird interpretation and felt like me forcing myself to change my understanding of things just for the math to make sense. That's what caused me to delete it.

So if my initial understanding about that differential displacement vector is correct, then perhaps I'm just interpreting the problem wrong? It seems to me like the displacement of our charge that we are applying the force to and that applied force vector are in the same direction. If that's the case, as cos(0) = +1, I'm having trouble seeing where the negative sign would appear in the left part of our equation, following the dot product definition of | \vec F | * | d \vec r | * cos(Θ).

I think I'm misinterpreting something about the line integral in general (not just as it pertains to this specific problem) as even in the case of a horizontal spring mass system, extending the mass rightwards with the "zero" x-position defined at zero stretch (and the positive direction pointing rightwards), the work done by the spring force on the block as the block moves from x = +1 meter to x = 0 meters would follow \int_1^0 -kx \, dx , providing the correct answer but still having a negative sign that shouldn't appear if the "magnitude definition" of the dot product is followed, as the spring force and mass displacement are in the same direction.

 

[robphy]

Sorry I'm too lazy to TeXify the discussion here...
but I sense that this old post might be helpful:
https://www.physicsforums.com/threa...potential-energy-formula.815230/#post-5118595

 

[vela]

The $\theta$ angle between $\vec{F}$ and $d\vec{r}$ is actually $\pi$ not 0 and $\cos\pi=-1$. $\vec{F}$ points inwards and $d\vec{r}$ points outwards.

I disagree. $d\vec r$ points in the same direction as $\vec F$ because the charge is being pushed toward the other charge. The minus sign appears because $|d\vec r| = -dr$.

 

[Delta2]

I disagree. $d\vec r$ points in the same direction as $\vec F$ because the charge is being pushed toward the other charge. The minus sign appears because $|d\vec r| = -dr$.

Well probably you are right, now I noticed that the charge is being pushed from the infinite (from $r=\infty$ to $r=r_{12}$) , so $d\vec{r}$ points inwards and $dr$ can be taken as negative so $|d\vec{r}|=-dr$

 

[Lost1ne]

Well probably you are right, now I noticed that the charge is being pushed from the infinite (from $r=\infty$ to $r=r_{12}$) , so $d\vec{r}$ points inwards and $dr$ can be taken as negative so $|d\vec{r}|=-dr$

Yep. I’ve come to this conclusion as well, and things are making sense. Thank you.