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Derivation of electric potential due to point charge

  1. Sep 9, 2011 #1
    Electric potential is the work done in moving a unit charge from infinity to a point in an electric field.

    Electric potential due to point charge:
    [itex]V=-\int \vec{E}\cdot d\vec{s}[/itex]
    [itex]V=-\int E\cdot ds cos \vartheta[/itex]
    if the stationary charge is positive and
    if the test charge is is moved from infinity to point P
    [itex]V=-\int E\cdot ds cos 180[/itex]
    [itex]V=-KQ \int \frac {1}{r^2} ds cos 180[/itex]

    now how to solve further
    as stationary charge is positive the electric field is outward i.e. from p to infinity
    and movement of charge is from infinity to P
    specially the signs and the direction of the field and the direction of ds
    and definite integration from P to infinity or from infinity to P?
    please give some imaginary picture or idea of how the test charge moves
    I am confused with it please help.

    Attached Files:

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    Last edited: Sep 9, 2011
  2. jcsd
  3. Sep 9, 2011 #2
    If the angle between E and ds is 180o, then, where in what direction is ds directed?
  4. Sep 9, 2011 #3
    from infinity towards Q
    now i have attached an image in the original post
  5. Sep 9, 2011 #4
    Yes, but how do we call this direction? It has a "special relation" to one of the variables in your equations.
  6. Sep 9, 2011 #5
    it should have relation with
    [itex] \frac {1}{r^2} [/itex]
  7. Sep 9, 2011 #6
    Yes. How do we call a straight line starting from a point a going to infinity?
  8. Sep 9, 2011 #7
    i imagined that it may be a straight line i.e. the shortest distance between infinity and P
  9. Sep 9, 2011 #8
    This straight line is called a ray, and the direction is called radial direction. On it, [itex]ds = dr[/itex]. You need to use this.
  10. Sep 10, 2011 #9
    I am confused with the signs and directions of ds ,dr and test charge and also
    how work done by external element is negative of work done by electric field.
  11. Sep 10, 2011 #10


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    I never understood, why integration in tensor calculus is obscured by some awkward notation. I guess, many textbook writers think, it's more intuitive to work with angles instead of vectors, but that's not true. To calculate a line integral, it's much more convenient to use the definition of that type of integral. Let [itex]\vec{V}(\vec{x})[/itex] be a vector field, defined in some domain of [itex]\mathbb{R}^3[/itex] and [itex]C: \lambda \in \mathbb{R} \supseteq (a,b) \mapsto \vec{x}(\lambda) \in \mathbb{R}^3[/itex] with values in the definition domain of [itex]\vec{V}[/itex]. Then the line integral over the vector field along the path [itex]C[/itex] is defined as

    [tex]\int_C \mathrm{d} \vec{x} \cdot \vec{V}(\vec{x})=\int_{a}^{b} \mathrm{d} \lambda \frac{\mathrm{d} \vec{x}(\lambda)}{\mathrm{d} \lambda} \cdot \vec{V}[\vec{x}(\lambda)].[/tex]
  12. Sep 12, 2011 #11
    The lines of forces are very simple in this case. Fhsst_electrost25.png
    Now you don't have to worry about the signs and angles here. Just simply use the formula for moving a charge through an electric field and you'll get it.
    I hope this helps.
    Thank you.
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