Copper slab of thickness b = 1.194 mm is Inserted at constant speed into a paral

Question

Copper slab of thickness b = 1.194 mm is Inserted at constant speed into a parallel-plate capacitor with gap d = 8.0 mm, as shown In the figure. the slab Is centered exactly halfway between the plates. If the Initial capacitance was C = 4.00E-11 F. what is the capacitance after the slab is introduced? 5 28E-11F If a change q = 6.00E-6 C is maintained on the plates as the slab is inserted, what is the ratio of the initial stored energy to the final stored energy? How much work is done by the person inserting the slab?

Explanation / Answer

The initial capacitance of the capacitor is:

C = A/d

The capacitance of the capacitor after inserting dielectric of known thickness is:

C' = A/(d-b)

Thus,

C' = C * (d/(d-b)) = 4e-11 * (8e-3/(8e-3-1.194e-3)) =4.7 e-11 F

The initial energy stored on the capacitor is:

U = 0.5 q^2/C = 0.5 * 6e-6*64e-6/4e-11 = 0.45 J

U' = 0.5 q^2/C' = 0.5 * 6e-6*6e-6/4.7 e-11 = 0.38 J

U/U' = 0.45/0.38=1.18

The amount of work done isU - U' =0.07

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