A slab of copper of thickness b = 1.951 mm is thrust into a parallel-plate capac

Question

A slab of copper of thickness b = 1.951 mm is thrust into a parallel-plate capacitor of C = 4.00 times 10^-11 F of gap d = 8.0 mm, as shown in the figure; it is centered exactly halfway between the plates. What is the capacitance after the slab is introduced? 5.29X10^-11 F If a charge q = 3.00 times10^-6C Is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? 1.32 How much work is done on the slab as it is inserted? -2.74X10^2 J Is the slab pulled in or must it be pushed in?

Explanation / Answer

b)

Ci = initial capacitance = 4 x 10-11 F

q = charge = 3 x 10-6 C

Ui = initial energy stored = (0.5) q2/Ci = (0.5) (3 x 10-6)2 /(4 x 10-11) = 0.1125 J

Cf = final capacitance = 5.29 x 10-11 F

q = charge = 3 x 10-6 C

Uf = final energy stored = (0.5) q2/Cf = (0.5) (3 x 10-6)2 /(5.29 x 10-11) = 0.0851 J

Ratio = Ui/Uf = 0.1125/0.0851 = 1.32

c)

W = work done = difference in energy stored = Ui - Uf = 0.1125 - 0.0851 = 0.0274 A

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