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--- 184_notes:examples:week7_energy_plate_capacitor [2020/01/28 03:00] – hallstein
+++ 184_notes:examples:week7_energy_plate_capacitor [2021/06/15 01:02] (current) – [Solution] schram45
@@ Line 15: / Line 15: @@
 ===Approximations & Assumptions===
-  * Whenever we measure energy, the capacitor is charged and and there is no longer current in the circuit.
+  * Whenever we measure energy, the capacitor is charged and and there is no longer current in the circuit: As a capacitor charges energy gets stored in the capacitor and this happens over time. We are only looking at the fully charged capacitor in this case.
-  * There is negligible resistance in the circuit, so the voltage across the capacitor is the same as in the battery.
+  * There is negligible resistance in the circuit: The voltage across the capacitor is the same as in the battery with this assumption, as no energy is lost over the wires.
-  * In the case that we disconnect the capacitor, it does not discharge at all.
+  * In the case that we disconnect the capacitor, it does not discharge at all: In reality capacitors are not perfect and lose their charge in a multitude of ways, but we will assume that is not happening in this problem.
 ===Representations===
@@ 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.