Asked by Mike
A slab of copper of thickness b = 1.566 mm is thrust into a parallel-plate capacitor of C = 4.00×10-11 F of gap d = 10.0 mm, as shown in the figure; it is centered exactly halfway between the plates.
If a charge q = 3.00×10-6 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?
If a charge q = 3.00×10-6 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?
Answers
Answered by
Anonymous
when you put the slab of copper in, you are making two capacitors in series, reducing capacitance.
Find the new capacitance. Then, knowing q is the same, find V across each capacitor.
Knowing Q, V you can find energy stored. Then, compare it to the original. If energy is greater, then you had to push the plate in, if energy is less, the copper was pulled in.
Find the new capacitance. Then, knowing q is the same, find V across each capacitor.
Knowing Q, V you can find energy stored. Then, compare it to the original. If energy is greater, then you had to push the plate in, if energy is less, the copper was pulled in.
Answered by
Mike
i think i know wat to do, but i'm having trouble finding the area.
Answered by
Anonymous
How do you find the area?
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