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--- 184_notes:examples:week7_energy_plate_capacitor [2021/06/15 00:50] – schram45
+++ 184_notes:examples:week7_energy_plate_capacitor [2021/06/15 01:02] (current) – [Solution] schram45
@@ Line 45: / Line 45: @@
  $U_{\text{new, disconnected}}=\frac{1}{2}\frac{Q^2}{C_{\text{new}}} = \frac{1}{2}\frac{Q^2}{2C_{\text{original}}} = \frac{1}{2}U_{\text{original}} = 0.9 \text{ J}$
-This is an interesting result! In one case, the energy doubles, and in the other case, it halves. And all we changed was whether we kept the circuit connected or not.
+This is an interesting result! In one case, the energy doubles, and in the other case, it halves. And all we changed was whether we kept the circuit connected or not. If we think about each situation it can make sense intuitively. In the first case we have the same driving potential and larger plates, which would allow more charge to be put on the plates and allowing more energy to be stored in the capacitor. In the second case, we no longer have the same driving potential when we increase the area of the plates. This just means you will have the same charge distributed over a larger area. We would expect that to decrease the amount of energy in the capacitor.