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Homework Help: Expression for V as a function of radius?

  1. Feb 23, 2006 #1
    I have given a cylindrical capacitor with an inner plate(radius=a) and variable radius r. I want to derive an expression, using Gauss's law, for the potential difference V between the inner plate and a point between the plates(radius=r) for the cylindrical capacitor as a function of the inner radius "a" and the variable radius "r".
    I need to compare the theoretical equation with an empirical equation obtained from a plot of the radius from axis vs. the potential difference V........
    Someone please help!!!!
     
  2. jcsd
  3. Feb 24, 2006 #2

    Tide

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    Why don't you just use Gauss' Law to find the electric field as a function of r then integrate to find the potential difference?
     
  4. Feb 24, 2006 #3
    I found that change in potential is KQ/r, but I don't know if this is a theoretical equation for V(r).
     
  5. Feb 24, 2006 #4

    Tide

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    Gauss' Law should give you a 1/r variation for the electric field so the potential should vary as ln(r).
     
  6. Feb 24, 2006 #5
    Thank you..I see... so, is the equation V(r)= ln(r/a) a theoretical one!? Mmm...I am trying to construct a plot of radius vs. V using a semi-log paper, with Yaxis for the value of r (log). But my data(from actual experiment) for the V and r vary "constantly", like, when r=2cm (from the axis), V=4Volt, and at r=3cm, V=6Volt...as we had a positive side (source of current, 12V) at the outer ring. So, my graph is just a straight line without plotting on the semi-log paper. Maybe my data is wrong!?!
     
  7. Feb 24, 2006 #6

    Tide

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    It's hard to say from what you have described whether your data is "wrong." There are a couple of factors to consider. First, the logarithmic dependence is for a cylindrical when the length (height) of the cylinder is very much larger than the diameter. Also, logarithmic variation is rather slow so that plotting over a small range of values will give the appearance of a straight line.
     
  8. Feb 25, 2006 #7
    Thank you again! My graph with semi-log is a very very slow curve, but it seems to be fairly close to a straight line. I got an equation of a line for the r(m) vs. V(v); r = 0.0052V + 0.0272, so it is also V=1/0.0052r -0.0272. I am not sure if this is the "empirical" equation, and I don't know how to compare it with V(r)= ln(r/a). How can I compare the empirical slope with the theoretical slope? I also want to get the value of the inner radius "a" from the graph, BUT how? Can I just read it? (the measured value of "a" was 3cm)
     
  9. Feb 25, 2006 #8

    Tide

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    If you got the equation from measurements then it is, by definition, empirical.

    Since you're plotting r vs. V all you have to do is "reverse" the theoretical expression for V(r). So, for example, if your theoretical equation is

    [tex]V(r) = V_0 ln \frac {r}{r_0}[/tex]

    then

    [tex]r = r_0 e^{V/V_0}[/tex]
     
  10. Feb 26, 2006 #9
    Thank you for lots of help. I think I can get the empirical formula from data if I do the 'reverse' of v. Thank You!!!
     
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