A parallel-plate vacuum capacitor has 6.34 J of energy stored in it. The separat
ID: 2174086 • Letter: A
Question
A parallel-plate vacuum capacitor has 6.34 J of energy stored in it. The separation between the plates is 3.10mm . If the separation is decreased to 1.80mm , ....A) what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?
B) What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?
Explanation / Answer
Energy of capacitor U = 1/2 CV^2. If voltage is constant and you bring the plates together, V stays put but C increases. So the energy increases. U = 1/2 Q^2/C. If u disconnect the power supply, Q stays put, but C increases. So U decreases. I think u can plug in and calculate the actual values. For the next problem, what voltage do u apply? I just take it as V volts. Ed = V E = Efree/k (Efree is the free charge and k is the dielectric constant) E = Efree - Einduced (Considering magnitudes only.. vectorially, they would add up). => Einduced = (1 - (1/k))Efree => sigma induced = (1-1/k) sigma free (sigma is surface charge density) Now u can calculate sigma free easily using C = Q/V and then dividing Q by surface area. Using that, calculate sigma induced.
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