A slab of copper of thickness b = 2.500 mm is thrust into a parallel-plate capac

Question

A slab of copper of thickness b = 2.500 mm is thrust into a parallel-plate capacitor of C = 7.00 times 10^-11 F of gap d = 1.000 times 10^1 mm, as shown in the figure: it is connected exactly halfway between the plates. A charge q = 2.00 times 10^-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? Is the slab pulled in or must it be pushed in? (You have only ONE try for this answers)

Explanation / Answer

Given,

Thickness of slab, b = 2.50 mm

Capacitance, C = 7 x 10-11 F

Gap, d = 10 mm

Charge, q = 2 x 10-6 C

Capacitance of a parallel plate capacitor is given by, C = A/d

Capacitance when copper slab is inserted, C' = A/(d-b)

C' = C x [d/(d-b)] = 7 x 10-11 x [0.01/(0.01 - 0.0025)] = 9.33 x 10-11 F

If a charge, q = 2 x 10-6 C is maintained on the plates, then

U = 0.5 q2/C = 0.5 x (2 x 10-6)2/ (7 x 10-11) = 0.029 J

U' = 0.5 q2/C' = 0.5 x (2 x 10-6)2/ (9.33 x 10-11) = 0.021 J

U/U' = 0.029 / 0.021 = 1.38

Work done on the slab is given by,

U - U' = 0.029 - 0.021 = 0.008 J

(Please rate my answer if you find it helpful, good luck...)

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