Copper d A slab of copper of thickness b = 2.07 mm is thrust into a parallel-pla

Question

Copper d A slab of copper of thickness b = 2.07 mm is thrust into a parallel-plate capacitor of plate area A = 2.40 cm2 and plate separation d = 5.14 mm, as shown in the figure; the slab is exactly halfway between the plates. What is the capacitance after the slab is introduced? Submit Answer Tries 0/10 If a charge q = 3.35 pC is maintained on the plates, what is the ratio of the stored energy before to that after the sao is inserted? Submit Answer Tries 0/10 How much work is done on the slab as it is inserted? Submit Answer Tries 0/10 Is the slab sucked in or must it be pushed in? The slab is sucked in The slab must be pushed in Submit Answer Tries 0/1

Explanation / Answer

Given,

b = 2.07 mm ; A = 2.4 cm^2 ; d = 5.14 mm

There are two capacitors now

C1 = C2 = e0 A/[1/2(d - b)]

C = C1 C2/(C1 + C2 ) = e0 A/(d - b)

C = 8.85 x 10^-12 x 2.4 x 10^-4/(5.14 - 2.07) x 10^-3 = 6.92 x 10^-13 F

Hence, C = 6.92 x 10^-13 F

Q = 3.35 uC

Ui = 1/2 Q^2/Ci

Uf = 1/2 Q^2/Cf

Uf/Ui = Cf/Ci = d/(d-b) = 5.14/(5.14 - 2.07) = 1.67

Hence, Uf/Ui = 1.67

W = Uf - Uf

W = 1/2 Q^2 (1/Cf - 1/Ci)

W = 1/2 (3.35 x 10^-6)^2 [1/6.92 x 10^-13 - 5.14 x 10^-3/(8.85 x 10^-12 x 2.4 x 10^-4) = -5.47 J

Hence, W = -5.47 J

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