Differences

This shows you the differences between two versions of the page.

--- 184_notes:examples:week7_charging_capacitor [2017/10/06 18:29] – [Solution] tallpaul
+++ 184_notes:examples:week7_charging_capacitor [2017/10/09 15:37] – dmcpadden
@@ Line 1: / Line 1: @@
 =====Looking at a Capacitor as it's Charging=====
-Suppose you have a parallel plate capacitor that is disconnected from any power source and is discharged. At time $t=0$, the capacitor is connected to a battery. Make qualitative graphs of current ($I$) in the wire, charge ($Q$) on the positive plate, and voltage ($\Delta V$) across the capacitor over time.
+Suppose you have a parallel plate capacitor that is disconnected from any power source and is discharged. At time $t=0$, the capacitor is connected to a battery. Sketch the graphs of current ($I$) in the wire, charge ($Q$) on the positive plate, and voltage ($\Delta V$) across the capacitor over time.
 ===Facts===
@@ Line 11: / Line 11: @@
 ===Approximations & Assumptions===
-  * The power source is connected correctly with respect to the capacitor and there are no other circuit elements (except for the wire).
+  * There are no other circuit elements (except for the wire).
   * The wire itself has a small resistance, just so we do not have infinite current at $t=0$.
   * Practically speaking, the capacitor becomes "fully charged" (with respect to the potential of the battery) at some finite time.
@@ Line 21: / Line 21: @@
 ====Solution====
-Before the capacitor is connected, we know that it is discharged, so there is a net neutral charge both on the wire and on the parallel plates. At time $t=0$, when the power source is connected, it brings with it an electric field, which the charges on the wire and capacitor do not immediately oppose, since they have not had time to accumulate. At time $t=0$ (or maybe slightly after 0, once the electric field has propagated at the speed of light), the electric field in the wire may look like this:
+Before the capacitor is connected, we know that it is discharged, so there is a net neutral charge both on the wire and on the parallel plates. At time $t=0$, when the power source is connected, it sets up an electric field in the wire (quasi-steady state), which the charges on the wire and capacitor do not immediately oppose, since there is initially no charge on the capacitor. At time $t=0$ (or maybe slightly after 0, once the electric field has propagated at the speed of light), the electric field in the wire may look like this:
 {{ 184_notes:7_cap_time_1.png?300 |Capacitor Circuit at time t=0}}
-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:
+Slightly after time $t=0$, charges start to move (setting up a current) 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. The charges on the capacitor create an electric field that opposes the electric field in the wire and diminishes 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. 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:
+Lastly, we wish to produce a graph of the potential difference between the plates. We can pull from a [[184_projects:project_5|previous in-class project]] in which we found the electric field from a large plate of charge. If the plates are large and close together, the electric field is constant in the middle, and depends on the charge of the plate like so:
 $$E_{plate} = \frac{Q_{plate}}{2\epsilon_0 A_{plate}}$$
-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:
+Notice that we do not necessarily need to remember this equation -- the electric field of everything we have looked at so far has been directly proportional to the charge (not $Q^2$ or $Q^5$ or anything strange).
+We also remember that potential difference across a constant electric field is just the electric field times the distance:
+$$\Delta V = E\cdot \Delta d$$
+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}}