Constants PartA A parallel-plate vacuum capacitor has 7.00 J of energy stored in
ID: 1582494 • Letter: C
Question
Constants PartA A parallel-plate vacuum capacitor has 7.00 J of energy stored in it. The separation between the plates is 3.20 mm. If the separation is decreased to 1.40 mm, what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed? A2o Submit Request Answer Part B What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed? A2 Request AnswerExplanation / Answer
A)
energy stored is U = Q^2/(2*C)
if capacitor is disconnected from the source then
charge remains same
U*c = constant
U1*C1 = U2*C2
C1 = eo*A1/d1
C2 = eo*A2/d2
A1 = A2 = A
then
U2/U1 = C1/C2
U2/7 = (eo*A1/d1) /(eo*A2/d2)= d2/d1
U2 = 7*(1.4/3.2)
U2 = 3.06 J
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B)
when source is connection
V potential difference across plates is same
U = 0.5*C*V^2
U1 / U2 = C1/C2
7/U2 = d2/d1
U2 = 7*(d1/d2) = 7*(3.2/1.4) = 16 J