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--- 184_notes:examples:week7_charging_capacitor [2017/10/06 01:12] – [Solution] tallpaul
+++ 184_notes:examples:week7_charging_capacitor [2017/10/06 18:34] – [Solution] tallpaul
@@ Line 8: / Line 8: @@
 ===Lacking===
-  * Graphs of $I$, $Q$, $\Delta V$$.
+  * Graphs of $I$, $Q$, $\Delta V$.
 ===Approximations & Assumptions===
@@ Line 25: / Line 25: @@
 {{ 184_notes:7_cap_time_1.png?300 |Capacitor Circuit at time t=0}}
-Slightly after time $t=0$, the current begins to exist and the capacitor begins to charge. Due to the high electric field at the very beginning, the charge flows very quickly at first. Since current is simply the movement of charge, our current graph will start at a very high value. However, we expect the current to drop pretty quickly, since the large flow of charge means a quickly charging capacitor, which begins to oppose the electric field and diminish it, slowing down the flow of charge. A few moments after time $t=0$, the circuit and electric field may look like this:
+Slightly after time $t=0$, current begins to exist and the capacitor begins to charge. Due to the high electric field at the very beginning, the charge flows very quickly at first. Since current is simply the movement of charge, our current graph will start at a very high value. However, we expect the current to drop pretty quickly, since the large flow of charge means a quickly charging capacitor, which begins to oppose the electric field and diminish it, slowing down the flow of charge. A few moments after time $t=0$, the circuit and electric field may look like this:
 {{ 184_notes:7_cap_time_2.png?300 |Capacitor Circuit a Few Moments after Connection}}
@@ Line 37: / Line 37: @@
 {{ 184_notes:7_graph_i.png?400 |Current Graph}}
-Notice that we also used the charge of the capacitor in our argument for why the current drops off like it does. We basically argue that the charge graph will look like a flipped current graph: rapidly accumulating charge at first, and gradually slowing down in accumulation until equilibrium ("fully charged"-ness) is reached. This yields the shape below. We are also happy with this result, because current is defined as the rate of change of the charge (amount of charge moving through a checkpoint per unit time). More formally, $I = \frac{\text{d}Q}{\text{d}t}$. If we know that current is a decaying exponential, then we can solve the rather differential equation to find that charge is the opposite of a decaying exponential. You don't need to be able to solve this to know how the charge accumulates, but it is a nice check that we have been consistent with our reasoning.
+Notice that we also used the charge of the capacitor in our argument for why the current drops off like it does. We basically argue that the charge graph will look like a flipped current graph: rapidly accumulating charge at first, and gradually slowing down in accumulation until equilibrium ("fully charged"-ness) is reached. This yields the shape below. We are also happy with this result, because current is defined as the rate of change of the charge (amount of charge moving through a checkpoint per unit time). More formally, $I = \frac{\text{d}Q}{\text{d}t}$. If we know that current is a decaying exponential, then we can solve the differential equation to find that charge is the opposite of a decaying exponential. You don't need to be able to solve this to know how the charge accumulates, but it is a nice check that we have been consistent with our reasoning.
 {{ 184_notes:7_graph_q.png?400 |Charge Graph}}
-Lastly, we wish to produce a graph of the potential difference between the plates. You may remember from a previous project that when we find the electric field from a large plate of charge, it is pretty much constant, and depends on the charge of the plate like so:
+Lastly, we wish to produce a graph of the potential difference between the plates. We can pull from a previous in-class project in which we found the electric field from a large plate of charge. It is pretty much constant, and depends on the charge of the plate like so:
 $$E_{plate} = \frac{Q_{plate}}{2\epsilon_0 A_{plate}}$$
+Notice that we do not necessarily need to remember this equation -- the electric field of everything we have looked at so far depends linearly on the charge.
+We also remember that potential difference across a constant electric field is just the electric field times the distance. Since the distance between the plates is unchanging and the electric field simply scales with $Q$, we can expect the voltage graph to have the same shape as the charge graph:
+{{ 184_notes:7_graph_v.png?400 |Voltage Graph}}
+That's it! We can even do one more check to be sure we are happy with our result. Notice the the capacitance of a capacitor can be represented as the charge on a plate (or something else, depending on the shape of the capacitor) divided by the potential difference between the plates (or whatever): $$C = \frac{Q}{\Delta V}$$
+We know that capacitance is a constant, since it is an intrinsic property of the capacitor's structure. Therefore, $Q/\Delta V$ must be constant, so their graphs must have the same shape. This is consistent with how we reasoned above, so this check makes us happy.