I only need the answer for the work question!!!!!! A slab of copper of thickness

Question

I only need the answer for the work question!!!!!!

A slab of copper of thickness b = 1.522 mm is thrust into a parallel-plate capacitor of C = 2.00

I only need the answer for the work question!!!!!! A slab of copper of thickness b = 1.522 mm is thrust into a parallel-plate capacitor of C = 2.00times10-11 F of gap d = 7.0 mm, as shown in the figure; it is centered exactly halfway between the plates. What is the capacitance after the slab is introduced? If a charge q = 9.00times10-6 C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? How much work is done on the slab as it is inserted?

Explanation / Answer

The conductor reduces the plate separation and in effect makes it two capacitors in series

U initial = 1/2*Q^2/C = 1/2*(6.00x10^-6)^2/4.00x10^-11 = 0.450J

After the slab is inserted the distance for each capacitor is (6.00 - 0.9231)/2 mm = 2.5385 mm

So the new capacitors C1 & C2 are equal to 4.00x10^-11*6.00/2.5385) = 9.455x10^-11F

Since the charge is maintained the energy stored in each new capacitor is

1/2*Q^2/C1 = 1/2*(6.00x10^-6)^2/9.455x10^-11 = 0.190J

So the total energy after the slab = 2*0.190 = 0.380

and the ratio = 0.45/0.38 = 1.18

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