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--- 184_notes:examples:week7_energy_plate_capacitor [2018/06/19 15:36] – curdemma
+++ 184_notes:examples:week7_energy_plate_capacitor [2020/01/28 03:00] – hallstein
@@ Line 37: / Line 37: @@
 $$U = \frac{1}{2} C \Delta V^2$$
-We have a capacitance and a potential difference which correspond to the charged capacitor before changing the area. The problem statement didn't ask for this quantity, but it will be useful for comparisons: $$U_{\text{original}}=1.8 \text{ mJ}$$
+We have a capacitance and a potential difference which correspond to the charged capacitor before changing the area. The problem statement didn't ask for this quantity, but it will be useful for comparisons: $$U_{\text{original}}=1.8 \text{ J}$$
 In the first case, we simply double the area of the capacitor without changing anything else or disconnecting it. Equation (1) tells us that doubling the area will in turn double the capacitance of the capacitor to $32 \text{ mF}$. The voltage across the battery does not change. When the area of the plates is suddenly doubled, charge on the plates spreads out and allows more charge to flow onto the plates (also making current pick up again in the wire). Eventually equilibrium is reached again, current stops flowing, and the voltage across the capacitor will be $15 \text{ V}$, as before. Our calculation for the energy stored will also be the same, but with the new value for capacitance now:
-$$U_{\text{new, connected}}=\frac{1}{2} C_{\text{new}} \Delta V^2 = 3.6 \text{ mJ}$$
+$$U_{\text{new, connected}}=\frac{1}{2} C_{\text{new}} \Delta V^2 = 3.6 \text{ J}$$
 In the second case, we disconnect the capacitor before doubling the area. Since the power source is no longer in play, it is a little more difficult to know the potential difference between the plates. However, if the battery is disconnected this also means that the charge on the plates will not change. It is worth expressing the energy stored in terms of $Q$ and $C_{\text{new}}$ (instead of C and V as before), for which equation (3) is set up perfectly. We refrain from subbing in numbers until the end, so until the last expression we write $C_{\text{new}}=2C_{\text{original}}$
-$$U_{\text{new, disconnected}}=\frac{1}{2}\frac{Q^2}{C_{\text{new}}} = \frac{1}{2}\frac{Q^2}{2C_{\text{original}}} = \frac{1}{2}U_{\text{original}} = 0.9 \text{ mJ}$$
+$$U_{\text{new, disconnected}}=\frac{1}{2}\frac{Q^2}{C_{\text{new}}} = \frac{1}{2}\frac{Q^2}{2C_{\text{original}}} = \frac{1}{2}U_{\text{original}} = 0.9 \text{ J}$$
 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.