(8c25p74) A slab of copper of thickness b = 1.449 mm is thrust into a parallel-p

Question

(8c25p74) A slab of copper of thickness b = 1.449 mm is thrust into a parallel-plate capacitor of C = 7.00×10-11F of gap d = 10.0 mm, as shown in the figure; it is centered exactly halfway between the plates. What is the capacitance after the slab is introduced?
8.19×10-11 F

If a charge q = 6.00×10-6C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted?
1.17

How much work is done on the slab as it is inserted?

I just need the work part!

Explanation / Answer

Initial energy, E1 = 0.5*q^2/C
= 0.5*(6*10^-6)^2 / (7*10^-11)
= 0.257 J

since you know tha ratio, you can calculate the final energy

Work done = change in energy
= final energy - initial energy

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