A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separat
ID: 1289818 • Letter: A
Question
A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant? [Answers: 4.19J, 16.8J]
The answers are given. I need help with the work, so I know how to work out the problem.
Explanation / Answer
Energy stored in capacitor = U = 8.38 J
U =(1/2)CV^2
C =(eo)A/d
C*d=(eo)A=constant
C2d2=C1d1
C2=C1d1/d2
The separation between the plates is 2.30mm . The separation is decreased to 1.15 mm.
Initial separation between the plates =d1= 2.30mm .
Final separation = d2 = 1.15 mm
(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same
Energy=U =(1/2)q^2/C
U2C2 = U1C1
U2 =U1C1 /C2
U2 =U1d2/d1
Final energy = Uf = initial energy *d2/d1
Final energy = Uf =8.38 *1.15/2.3
(A) Final energy = Uf = 4.19 J
______________________________________...
B) If the capacitor remained connected to the potential source while the separation of the plates was changed, potential remains same.
Final energy = Uf =(1/2)C!V^2
Final energy = Uf =(1/2)[C1d1/d2]V^2
Final energy = Uf =(1/2)[C1V^2]d1/d2
Final energy = Uf = initial energy *d1/d2
Final energy = Uf =8.38 *2.3/1.15
Final energy = Uf =16.76 J
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