Calculating Capacitance for a Rolled Capacitor

[Runaway]

Homework Statement

Homework Equations

C = k *$$\epsilon$$₀ * A / d

The Attempt at a Solution

I attempted setting the equation up as:
C/2 (because rolling the capacitor doubles C) = k * $$\epsilon$$₀ * L * W / d
9.49 * 10^-8 F/2 = 3.9 * 8.85 * 10^-12C² / (N * m²) * L * 7.43cm / .00282 mm
9.49 * 10^-8 F/2 * .00282 mm = 3.9 * 8.85 * 10^-12C² / (N * m²) * L * 7.43cm
(9.49 * 10^-8 F/2 * .00282 mm)/(3.9 * 8.85 * 10^-12C² / (N * m²) *7.43cm) = L
L= 0.05218 meters

It's an online assignment and it says .05218 isn't the right answer, can someone show me where I've messed up?

 

[ehild]

Read the problem again: what should be d, the distance between the metal surfaces? Is it equal to the thickness of the metal foil or that of the paper?

ehild

 

[Runaway]

Ok, I tried:
(9.49 * 10^-8 F/2 * .0246 mm)/(3.9 * 8.85 * 10^-12C^2 / (N * m^2) *7.43cm)
Instead and got 0.4552 meters, but it still says the answer is incorrect

 

[ehild]

I don't quite understand the geometry of this capacitor. The problem says two strips of aluminium. It might look as in http://www.tpub.com/neets/book2/3f.htm

ehild

 

[Runaway]

I talked with the teacher and I was misreading the problem, it wants it to have the 9.49 E-8 F capacitance before it's rolled, not after, and the part about C being doubled when it is rolled was just put there to explain why there is a difference before and after it's rolled.
Thanks for your help ;)