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Homework Help: Electric potentials

  1. Nov 9, 2013 #1
    Problem -
    Find the potential difference between 2 pts given the electric fields at all points in the surroundings.

    Let one point be A and one point be B.
    Thus, Va - Vb=- integral from b to a(E vector. dr vector)
    Also I can write the opposite equation, that is
    Vb -Va = -integral from a to b( E vector.dr vector)

    In second case angle between dr and E vector will be exactly supplementary of 1st case. Thus taking the dot product in first case to be positive, the dot product in the second case would be negative. Also the integral from a to be of anything would be exactly the negative of the integral from b to a.

    This leads to Va -Vb=Vb-Va which is surely wrong...
    I can't really understand what I have done wrong here....
  2. jcsd
  3. Nov 9, 2013 #2
    It's not "also" but "because of the above".
    In other words, you don't have two independent effects. You just explained why the integral form a to b is the negative of the integral from b to a (along the same path).
  4. Nov 10, 2013 #3
    Let there be a positive charge 'q' at O . Let there be two positions A and B ,such that O,A,B all lie in a straight line with A closer to O .Electric field is from A to B .OA = a and OB = b .

    ’dl’ is the displacement as the test charge is moved .’dr’ is the radial displacement .The relationship between dl and dr being |dl|cosθ = dr where θ is the angle between displacement 'dl' and radial displacement 'dr'.

    Case 1 : Move test charge from A to B

    VB-VA = -∫E .dl
    VB-VA = -∫|E||dl|cos0°

    Now, |dl|cos0° = dr

    VB-VA = -∫|E|dr (Limits from a→b)

    Case 2:Move test charge from B to A

    VA-VB = -∫E .dl
    VA-VB = -∫|E||dl|cos180°

    Now, |dl|cos180° = dr

    VA-VB = -∫|E|dr (Limits from b→a)

    VA-VB = ∫|E|dr (Limits from a→b)

    The two i.e Case 1 and Case 2 give you the same result .

    Hope that helps.
    Last edited: Nov 10, 2013
  5. Nov 10, 2013 #4
    Thanks a lot Tanya! You really helped me in my understanding!

    Can you just tell me one more thing? Can you just give me a hint on how to process with the integration when the point O is shifted down by some distance while keeping the position off all other points same? What would be the direction of dr vector ? Will it be the same as before or will it be along the position vector all the time? Also generally what is the direction of dr vector? Is it always along the position vector?
    Once again, thanks a lot for your help!
  6. Nov 10, 2013 #5
    The direction of dr vector is radially outward from the origin . While moving from A to B ,suppose the test charge is at P .Then the direction of dr vector is same as that from O to P.
  7. Nov 10, 2013 #6
    This is true only if the motion is radial (outward or inward).
    The direction of [tex]d\vec{r}[/tex] is not, in general, radial. It is tangent to the trajectory. Only if the trajectory is a line passing through origin will the above be true. But in general is not along the position vector.
    Think about circular motion, for example. What is the direction of dr?
  8. Nov 10, 2013 #7
    I think it should be tangent to the circle as the difference between two adjacent position vectors would be tangent to the circle.
  9. Nov 10, 2013 #8
    Yes, you've got it. dr has the same direction as the velocity vector.
  10. Nov 10, 2013 #9
    If this is the case then how to integrate to find the potential difference if I just shift the position of O in the situation described by Tanya a bit downwards from its current potential while keeping the position of all other points same.
  11. Nov 10, 2013 #10
    Sorry I meant to say position instead of potential.
  12. Nov 10, 2013 #11
    Please do not create unnecessary confusion .

    There is a clear distinction between [itex]d\vec{l}[/itex]and [itex]d\vec{r}[/itex] .The displacement of the test charge is represented by [itex]d\vec{l}[/itex]which is tangential to the path taken ,not [itex]d\vec{r}[/itex] . [itex]d\vec{r}[/itex] points radially outward from the origin.
    Last edited: Nov 10, 2013
  13. Nov 10, 2013 #12
    I am afraid you are confusing yourself. Unless by dr or by "radial" you mean something else than the usual meaning.
    I was referring to [itex]d\vec{r}[/itex] as the differential of the position vector. Like here, in this wiki article.
    http://en.wikipedia.org/wiki/Position_(vector [Broken])
    See the second figure. Is that [itex]d\vec{r}[/itex] "radial"?

    I think the confusion may be due to the use of dr to mean either [itex]d\vec{r}[/itex] or the component dr, in various posts.

    However in the first post it is mentioned "the angle between dr and E vector" so I assumed dr is just a short way of writing [itex]d\vec{r}[/itex]. As you seem to think too, in your last post.
    And this is radial only when the displacement is radial, with no angular component.
    Last edited by a moderator: May 6, 2017
  14. Nov 11, 2013 #13
    Can you please tell me what do you mean by the component dr?
    Last edited by a moderator: May 6, 2017
  15. Nov 11, 2013 #14
    Something like dr in the expresion for the [itex]d\vec{r}[/itex] in spherical coordinates.
    Like here:

    But this is not really necessary to do your integral.
    You can express [itex]d\vec{r}[/itex] in Cartesian coordinates.
    I understand that you are integrating along a straight line, so far. So you just need the equation of that line.
    Let say that is y=ax+b.
    Then you have dy=adx and
    [tex] d\vec{r}=dx \hat{x}+ dy \hat{y}= dx \hat{x}+ a dx \hat{y}=dx (\hat{x}+ a \hat{y})[/tex]
    You could express the field in Cartesian too. The field will be something like E(x,y) but you can write it just as E(x) by using the equation of the line. And then do the integral which reduces to an integral over dx. And you know how this behaves.
    It may not be the easiest way, if the problem has some spherical symmetry. But you did not mention a specific case, just the general idea.
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