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Determing the direction of dl when calculating V

  • Thread starter kc8ukw
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Homework Statement



Hi. I'm wondering how one determines the direction of dl (vector) when calculating V, the electric potential. So, suppose we want to calculate the electric potential caused by a point charge q at a distance a from the point charge, if we know the electric field of a point charge.

Homework Equations



The textbook I have says to calculate the potential by Va - Vb = integral from a to b of E*dl.

The Attempt at a Solution



So, I can let b = infinity, to get Va - Vinfinity = integral from a to infinity of k*q/r2 dr, where I have said that dl = dr*rhat, since I am integrating from r = a to r = infinity. Doing the integral, and assuming Vinfinity = 0, I get the correct answer,

V = kq/a.

But suppose instead that I had integrated from infinity to a instead of the other way. So,

Vinfinity - Va = integral from infinity to a of E*dl.

Now, since I am coming from infinity toward a, I make dl = -dr*rhat. That introduces one negative sign. But the order of integration is also flipped, which introduces another. BUT, since I have Vinfinity - Va, that introduces a third negative, so I end up with

V = -kq/a

the opposite of what I got before, and the wrong answer. Where did I go wrong? I'm guessing it must have been when I said that dl = -dr*rhat - but why isn't that correct? What sets the proper direction of dl?
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi kc8ukw! welcome to pf! :smile:

(have an infinity: ∞ and an integral: ∫ :wink:)

i'm not following your argument :confused:

the potential V is defined by …

V(R) - V(∞) = ∫R E(r) dr …

what exactly are you saying?
 
  • #3
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Hi Kc8ukw,
For your particular problem, (E in the [itex]\hat{r}[/itex] direction), d[itex]\vec{l}[/itex] is in the direction of [itex]\hat{r}[/itex] (you know this already). But in spherical coordinates [itex]\hat{r}[/itex] always points radially outward. But radially outward can point, for instance, along the [itex]\hat{x}[/itex] direction or along the -[itex]\hat{x}[/itex] direction. Both of these directions, [itex]\hat{x}[/itex] and -[itex]\hat{x}[/itex], are called [itex]\hat{r}[/itex] when we are in spherical coordinates. This is because in general the direction of [itex]\hat{r}[/itex] depends on the angles θ and [itex]\phi[/itex]. So your d[itex]\vec{l}[/itex] vector here is in the direction called [itex]\hat{r}[/itex] (not -[itex]\hat{r}[/itex]!) regardless of if it points in from infinity or out to infinity.

An example from Griffiths's awesome book: Introduction to Electrodynamics might help:
Imagine two vectors [itex]\vec{A}[/itex] = [itex]\hat{y}[/itex] and [itex]\vec{B}[/itex]= -[itex]\hat{y}[/itex]. Both of these vectors are called [itex]\hat{r}[/itex] is spherical coordinates and in particular [itex]\vec{A}[/itex]+ [itex]\vec{B}[/itex] is 0 NOT 2[itex]\hat{r}[/itex]!
 
  • #4
tiny-tim
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welcome to pf!

hi finnsmama! welcome to pf! :smile:

but that wasn't yesterday, it was last year …

i don't think kc8ukw is coming back! :wink:
 
  • #5
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Hey Tiny tim,
yeah, I knew that...but people search for answers to questions all the time and this one was not answered...and I had just gone over this in class! So answering seemed the right thing to do...
 

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