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Electric Potential formula derivation?

  1. Dec 29, 2014 #1
    Okay so I'm using Coloumb's law and the defining equation of electric field strength to find a proof (for my own satisfaction) for the electric potential formula:
    [tex]V_A = \frac{PE_A}{q}= \frac{Q}{4πε_0r}[/tex] (where A is the position configuration of a point charge q in an electric field)
    My derivation is as follows -
    If a graph of force of electric attraction (y axis) and distance r (x axis) from the the center of a point charge Q to a test charge q is plotted, then the area between the x axis and the curve must represent the potential energy. Since the standard definition of the potential energy at a point in an electric field is the work done in bringing a unit positive charge from infinity to that point, it follows that the area between the limits rA and ∞ represents the potential energy at point A. From the definition of an asymptote, F=0 when r=∞ .
    So, [tex]\frac{PE_A}{q} = \frac{1}{q}\int_r^∞ F dr = \frac{1}{q} \int_r^∞ \frac{Qq}{4πε_0r^2} \, dr = 0-(-\frac{Q}{4πε_0r})[/tex]
    Have I made any incorrect assumptions or used faulty reasoning? Could a better approach to this derivation be adopted?
     
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  3. Dec 29, 2014 #2

    Stephen Tashi

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    There is the tecnicality of what path you follow from "infinity" to the point in question. If you've established that the route of the path doesn't matter then you can do the calculation, as you did, on the particular path along a line going through the center of the charge Q, through the point in question, and "off to infinity".
     
  4. Dec 30, 2014 #3
    So if I replace the word "distance" with " |displacement| " (absolute values solve the +/- ve problem to preserve the equation sign), then does the derivation become clearer? I might have to do a presentation on this derivation in class so I just want it to be correct.
     
  5. Dec 30, 2014 #4

    Stephen Tashi

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    Only the component of displacement in a radial direction matters. For example, if you move the test charge in a circular orbit around the charge Q, no work is done because the component of the field tangent to circle is zero. I don't know whether you are expected to present anything about conservative fields. For your class, it might be satisfactory to define the potential as the work done to move the object in a radial direction from infinity to r.

    Your use of [itex] \frac{1}{q} [/itex] outside the integral sign and a factor of [itex] q [/itex] isn't wrong, but I think it is unnecessary. A unit test charge can be represented by factor of 1 in the integrand.
     
  6. Dec 30, 2014 #5
    Wait, so a charge ##q## in centripetal motion around another charge ##Q## has no change in its electric potential throughout the orbit( I mean when the displacement of q to Q in Euclidean space stays invariant considering all 3 dimensions, so take the eccentricity of orbit to be 0)?
    I'm 15 and I'm in my senior year of high school so you have a general idea about my audience.
     
  7. Dec 30, 2014 #6

    Stephen Tashi

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    That's interesting vocabulary - but "centripetal" motion?

    If we begin to consider motion seriously we'll get into an induced magnetic field and radiated energy - which will confuse me! It's safe to say the electric potential of a field due to a "point" charge along a circular path about the charge is constant. In your calculation, we also neglect any compications from induced fields - at least we neglect to mention them. The "unit test charge" is a theoretical construct. We can say we move it "infinitely slow". (The integration is with respect to distance, not time.)


    If the audience is just beginning calculus and just being introduced to the physical definition of "work" then you'll have your hands full just explaining the integration and its interpretation. No doubt, some audience member will ask why you don't set the limits of integration as [itex] \int_{0}^r ...[/itex]. That gets into the question of whether there is a physical interpretation of a mathematical "singularity" in a function (at distance = 0). My preferred "classical" physics model is that real charges aren't points so near once distance is near 0, you are in the interior of a charged body. In the interior, the unit test charge is getting pulled by charge distributed all around it, so the calculation of force by [itex] F = \frac{(1)Q}{4\pi\epsilon_0 r^2 } [/itex] doesn't apply in the interior.

    If the audience is already experienced with calculus and the concept of work then you can mention that if your path is wiggly line from infinity to the endpoint then the work along that path is the same as the work you computed along a path that was a radial line. One can make a "hand-wavy" argument that this should be the case because only the displacements of the path in the radial direction require doing work.
     
  8. Dec 30, 2014 #7
    The audience is used to performing calculus with simple, exponential and logarithmic functions, so I doubt I'll have trouble in explaining the integrand expression and the calculations I did to arrive at the result. But I'm sure the "wavy-line" explanation ought to ease up some trouble I might have at describing the limits :)
    Btw I've dealt with a lot of conics (and some painful axis rotations with them which I didn't really get) and some introductory concepts of relativity, so that's where some of my mathematical terminology stems from. Anyway, thanks for the help!
     
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