My vague physics course

[Function]

Hello everyone

 

Another question, based on an equality from my biomedical physics course:

 

[math]\int_{A}^{B}{\vec{E}\mathop{}\!\mathrm{d}\vec{l}}=-\int_{A}^{B}{k\cdot\frac{q}{\vec{r}^2}\mathop{}\!\mathrm{d}\vec{r}}[/math]

 

Question is simple: why?

(Note that I may have written some mistakes with vectorial notation in the integrals)

 

Thanks!

 

F.

Edited October 11, 2014 by Function

[elfmotat]

Well that equation isn't true in general, so you must have some additional information about the form of the electric field - namely that it's static - to justify setting the electric field to kq/r². I can't really tell you why they're doing this without some context.

[studiot]

I am guessing that this is the movement of a charge q along a line L from A to B in an electricfield E.

 

In that case this would be an energy balance saying that the work done intergral (E.dl) equals the potential energy gained.

 

Did you miss out a dot in the first integral?

[Function]

Hmm okay... May have written a mistake (the professor had written that on the blackboard and might've written a mistake while copying...)

The context:

 

[math]V_B-V_A=-\int_{A}^{B}{\vec{E}\cdot\vec{dl}}[/math]

 

That's what's explicitely written in the 'book' itself and I accept this - it's negative because of the opposite directions.

Now what I've written on a blank page next to it:

 

[math]V_B-V_A=-\int_{A}^{B}{E\cdot dl}\cdot\cos{180^{\circ}}[/math]

[math]=\int_{A}^{B}{E\cdot dl}=-\int_{A}^{B}{\frac{q}{4r^2\pi\varepsilon_0}\cdot dr}[/math]

 

I am guessing that this is the movement of a charge q along a line L from A to B in an electricfield E.

 

In that case this would be an energy balance saying that the work done intergral (E.dl) equals the potential energy gained.

 

Did you miss out a dot in the first integral?

 

 

But "dl" implies that "l" is variable, is "l" the distance from the charge to the positive point A?

But then I wonder what r is, and why my professor switched to r, turning the integral negative... I'm pretty sure he did write both "r" and "l", but don't get the difference...

 

What do you mean with "miss out a dot"?

 

Let me quote the text that is written about the electric potential in my book (my apologies for any mistakes) (most important stuff are bold):

 

 

 

The electric potential difference between 2 points B and A, Vb-Va, is the difference in electric potential energy from the positive unit charge between the points B and A; in other words the electric potential difference between two points B and A is the work that an external force must do to bring the positive unit charge from A to B. The unit of potential difference is the Volt. The electric potential in a random point, V, is the electric potential energy, that the positive unit charge has in that specific point. For this definition, a potential on an infinite distance is being considered zero. The electric potential in a random point is thus the work done by an external force to bring a positive unit charge from infinite, to the specific point. All points with the same potential are forming an equipotential surface. Movement from charges between 2 points on an equipotential surface doesn't need any work. Field forces are perpendicular to these surfaces.

 

Edited October 11, 2014 by Function

[elfmotat]

Ah, okay.

 

So what they're doing is looking at the potential difference between two points in space for a static point charge at the origin. They define the potential difference between two points as:

 

[math]V_B-V_A=-\int_{A}^{B}{\vec{E}\cdot d \vec{l}}[/math]

 

For a point charge at the origin we have:

 

[math]\vec{E} = \frac{kq}{r^2} \hat{r}[/math]

 

[math]V_B-V_A=-kq \int_{A}^{B}{\frac{1}{r^2} \hat{r} \cdot \vec{dl}}[/math]

 

Since [math]d \vec{l}[/math] is a small change in the path from A to B, we have: [math]\hat{r} \cdot d\vec{l} = dr[/math]. Therefore:

 

[math]V(r_B)-V(r_A)=-kq \int_A^B \frac{dr}{r^2}[/math]

Edited October 11, 2014 by elfmotat

[studiot]

I am trying not to cross post with elfmotat, who I think is doing a grand job, but I would just like to reinforce one comment.

 

You need to distinguish between potential and potential difference.

This is a common confusion.

[Function]

[math]\hat{r} \cdot d\vec{l} = dr[/math]. Therefore:

 

[math]V(r_B)-V(r_A)=-kq \int_A^B \frac{dr}{r^2}[/math]

 

I understand everything until here:

why is that (the r unit vector times dl equals dr)? I've never really understood the differential concepts too good

And what's [math]V(r_B)[/math]? Is it different from [math]V_B[/math]?

I am trying not to cross post with elfmotat, who I think is doing a grand job, but I would just like to reinforce one comment.

 

You need to distinguish between potential and potential difference.

This is a common confusion.

 

Elfmotat is indeed doing a great job.

 

On the difference between the two: I am convinced of it that I know the difference thanks

Edited October 11, 2014 by Function

[elfmotat]

 

I understand everything until here:

why is that (the r unit vector times dl equals dr)? I've never really understood the differential concepts too good

 

Do you have any experience with vector calculus? I suggest looking into "line integrals." I think I remember Khan Academy having some decent videos on the subject.

 

So in general you can examine the difference between two points by taking some general path from A to B. The path doesn't have to be a straight line, it could be some curvy path. [math]\hat{r}[/math] is the unit vector in the r-direction. What [math]\hat{r} \cdot d \vec{l}[/math] is telling you is that integral only depends on differences in r. In other words the integral only depends on integrating the r-component of any path from A to B.

 

If you're familiar with spherical coordinates, For some general curve in spherical coordinate, [math]d \vec{l}[/math] will contain r, theta, and phi components:

 

[math]d \vec{l} = dr ~ \hat{r} + rd \theta ~ \hat{\theta} + rsin \theta d \phi ~ \hat{\phi}[/math]

 

from here it's easy to see that [math]\hat{r} \cdot d \vec{l}=dr[/math].

 

 

 

 

And what's [math]V(r_B)[/math]? Is it different from [math]V_B[/math]?

 

I put it in this notation to make it explicit that V only depends on r. They mean the same thing.

Edited October 11, 2014 by elfmotat

[Function]

 

Do you have any experience with vector calculus? I suggest looking into "line integrals." I think I remember Khan Academy having some decent videos on the subject.

 

So in general you can examine the difference between two points by taking some general path from A to B. The path doesn't have to be a straight line, it could be some curvy path. [math]\hat{r}[/math] is the unit vector in the r-direction. What [math]\hat{r} \cdot d \vec{l}[/math] is telling you is that integral only depends on differences in r. In other words the integral only depends on integrating the r-component of any path from A to B.

 

If you're familiar with spherical coordinates, For some general curve in spherical coordinate, [math]d \vec{l}[/math] will contain r, theta, and phi components:

 

[math]d \vec{l} = dr ~ \hat{r} + rd \theta ~ \hat{\theta} + rsin \theta d \phi ~ \hat{\phi}[/math]

 

from here it's easy to see that [math]\hat{r} \cdot d \vec{l}=dr[/math].

 

 

 

I put it in this notation to make it explicit that V only depends on r. They mean the same thing.

 

(I do understand some vector calculus (the algebraic operations))

 

I kind of understand it now. Thank you!

 

So what about the V(rb), same as Vb?

[elfmotat]

 

(I do understand some vector calculus (the algebraic operations))

 

I kind of understand it now. Thank you!

 

So what about the V(rb), same as Vb?

 

No problem. Yes, I just put V in function notation to make it clear that it's a function of r. [math]V_B - V_A = V(r_B) - V(r_A)[/math].