Capacitor Problem involving a slab of Copper between a Capacitor

[Simon777]

Homework Statement

A slab of copper of thickness b = 1.370 mm is thrust into a parallel-plate capacitor of C = 9.00×10-11 F of gap d = 10.0 mm, as shown in the figure; it is centered exactly halfway between the plates.

If a charge q = 1.00×10-6 C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted?


I already know how to solve this problem. I just have a question about one of the steps. You need to find capacitance equivalent of c1 and c2 and since area and distance are the same, c1=c2. You end up with

C equivalent= c1/2

Now c1= (epsilon nought * Area)/ distance

so you just plug this in

Why do I only get the right answer when the distance is d-b? If you are just referring to c1, shouldn't it be half of d-b? Why use the combined distance of capacitor 1 and 2?

 

[Simon777]

Nevermind, I figured it out. I had to half d-b just like I thought. For some reason I unknowingly halved d-b and got the right answer and confused myself.