A slab of copper of thickness b = 2.40 mm is thrust into a parallel-plate capaci

Question

A slab of copper of thickness b = 2.40 mm is thrust into a parallel-plate capacitor of plate area A = 2.40 cm2 and plate separation d = 5.00 mm, as shown in the figure; the slab is exactly halfway between the plates. (a) What is the capacitance after the slab is introduced? pF (b) If a charge q = 3.40 C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? (c) How much work is done on the slab as it is inserted? J (d) Is the slab sucked in or must it be pushed in?

Explanation / Answer

Given that        b = 2.40 mm        A = 2.40 cm2         d = 5.00 mm a) When slab is inserted between the parallel plates, the capacitance of the system is             C = o A / (d - b)                 = 8.85 x 10-12 x 2.40 x 10-4 / (5 - 2.4) x 10-3                 = 8.169 x 10-13 F b) Energy stored in the capacitor before slab is inserted              Ui = [Q2 / 2Ci]                  = Q2 d / 2 o A     When slab is inserted    Uf = [Q2 / 2C] = Q2 (d -b) / 2 o A Therefore        Ui / Uf = d / (d - b)                   = 5.00 / (5.00 - 2.4)                   = 1.923 c) Workdone by the slab is             U = Ui - Uf                   = Q2 b / 2 o A                   = (3.40 x 10-6 )2 x 2.4 x 10-3 / 2 x 8.85 x 10-12 x 2.40 x 10-4                   = 6.531 J

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