1. Parallel plates (area A, separated by distance d) are connected to a battery,
ID: 1655215 • Letter: 1
Question
1. Parallel plates (area A, separated by distance d) are connected to a battery, and they store energy Uo. Now the battery is disconnected, and the plates are separated to a distance 5.3d, and they now store energy Uf. Find R = Uf/Uo the ratio of the energy stored at the end to the energy stored when the capacitor was first charged. NOTE: Can you explain why the energy changed?
2.A 2.05 F parallel plate capacitor with square plates is attached to a 226 V battery. If the length of each side of the plates is multiplied by 8.89, the potential difference across the battery is multiplied by 3.03 and the distance between the plates is multiplied by 7.37: find the new capacitance, in F, of this capacitor.
Explanation / Answer
1. C = e0 A / d
Energy stored, Uo = Q^2 / 2 C
as battery is disconnected so Q will remain same.
C' = e0 A / d' = C / 5.3
Uf = Q^2 / 2 C' = (5.3) (Q^2 / 2 C )
R = Uf / Uo = 5.3 ........Ans
2. C = e0 A /d = 2.05 uF
A' = 8.89^2 A = 79 A
d' = 7.37 d
C' = e0 A' / d' = (79 / 7.37) e0 A /d
C' = 10.7 C
C' = 10.7 x 2.05 uF
C' = 22 uF ........Ans
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