Do You Need the Dielectric Constant of Copper to Find Capacitance?

[Oijl]

Homework Statement

A slab of copper of thickness b = 1.85 mm is thrust into a parallel-plate capacitor of plate area A = 2.25 cm2 and plate separation d = 5.00 mm, as shown in the figure; the slab is exactly halfway between the plates.

What is the capacitance after the slab is introduced?

Homework Equations

Q = CV
C = \kappa\epsilon_{0}L
Q=
\epsilon_{0}\oint\kappa\stackrel{\rightarrow}{E}\cdot\stackrel{\rightarrow}{dA}

The Attempt at a Solution

If I know E I can know V.
If I also know Q I can know C.
Q is given.
I find E0 and E1 with the help of Gauss' Law, from which I find E = Q / (epsilon*kappa*A), where A is the area of one surface of the capacitor.
I use E = dV to find V by integrating E, from which I get V = E0(d - b) + E1(b)

So I know V and Q, but I only know V in terms of kappa, which is not given in the problem. Furthermore, it (the dielectric constant of copper) is not given in my textbook. I know I can find it on the web, but that neither the problem nor the textbook (which is written by my university's physics department for use in the introductory physics classes here) provide it makes me pause and wonder if I'm taking the long route on this problem.

So do I HAVE to know the dielectric constant of copper for this problem, or is there a much simpler way to solve this?

 

[rl.bhat]

The copper plate of thickness b is equivalent two thin plate connected by a wire, and separated by b.
Either of this does not affect the equivalent capacitance.
Now what is the distance between the two plates of each capacitor?
How the two capacitors are connected?