Expression for V as a function of radius?

1. Feb 23, 2006

Mimi

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........

2. Feb 24, 2006

Tide

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?

3. Feb 24, 2006

Mimi

I found that change in potential is KQ/r, but I don't know if this is a theoretical equation for V(r).

4. Feb 24, 2006

Tide

Gauss' Law should give you a 1/r variation for the electric field so the potential should vary as ln(r).

5. Feb 24, 2006

Mimi

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!?!

6. Feb 24, 2006

Tide

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.

7. Feb 25, 2006

Mimi

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)

8. Feb 25, 2006

Tide

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

$$V(r) = V_0 ln \frac {r}{r_0}$$

then

$$r = r_0 e^{V/V_0}$$

9. Feb 26, 2006

Mimi

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!!!