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184_notes:examples:week7_charging_capacitor [2017/10/04 18:58] – [Example: Application of Node Rule] tallpaul | 184_notes:examples:week7_charging_capacitor [2017/10/06 18:29] – [Solution] tallpaul | ||
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=====Looking at a Capacitor as it's Charging===== | =====Looking at a Capacitor as it's Charging===== | ||
- | I, V, Q over time. | + | Suppose you have a parallel plate capacitor that is disconnected from any power source and is discharged. At time $t=0$, |
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- | Suppose you have the circuit below. Nodes are labeled for simplicity | + | |
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- | {{ 184_notes: | + | |
===Facts=== | ===Facts=== | ||
- | * $I_1=8 \text{ A}$, $I_2=3 \text{ A}$, and $I_3=4 \text{ A}$. | + | * The capacitor is discharged. |
- | * $I_1$, $I_2$, and $I_3$ are directed as pictured. | + | * The capacitor is made up of parallel plates. |
+ | * The capacitor is connected to a power source at $t=0$. | ||
===Lacking=== | ===Lacking=== | ||
- | * All other currents (including their directions). | + | * Graphs of $I$, $Q$, $\Delta V$. |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The current | + | * The power source |
- | * All current | + | * The wire itself has a small resistance, just so we do not have infinite |
- | * No resistance in the battery | + | * Practically speaking, |
===Representations=== | ===Representations=== | ||
- | * We represent the situation with diagram given. | + | * We represent the setup below. The capacitor is pictured both disconnected and hooked up to the power source. |
- | * We represent | + | |
+ | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | Let's start with node $A$. Incoming current is $I_1$, and outgoing | + | 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 |
- | $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$ | + | |
+ | {{ 184_notes: | ||
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+ | 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, | ||
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+ | {{ 184_notes: | ||
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+ | During this time, the current | ||
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+ | {{ 184_notes: | ||
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+ | Based on what we have argued so far, the graph of the current in the circuit should look something like this: | ||
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+ | {{ 184_notes: | ||
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+ | 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 (" | ||
- | We do a similar analysis for node $B$. Incoming current is $I_{A\rightarrow B}$, and outgoing current is $I_3$. Since $I_{A\rightarrow B}=5 \text{ A}$ and $I_3=4 \text{ A}$, we need $I_{B\rightarrow D}$ to be outgoing to balance. To satisfy the Node Rule, we set | + | {{ 184_notes: |
- | $$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$ | + | |
- | For node $C$, incoming current is $I_2$ and $I_3$. There is no outgoing current defined yet! $I_{C\rightarrow D}$ must be outgoing | + | 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: |
- | $$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$ | + | $$E_{plate} = \frac{Q_{plate}}{2\epsilon_0 A_{plate}}$$ |
+ | 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: | ||
- | Lastly, we look at node $D$. Incoming current is $I_{B\rightarrow D}$ and $I_{C\rightarrow D}$. Since there is no outgoing current defined yet, $I_{D\rightarrow battery}$ must be outgoing to balance. To satisfy the Node Rule, we set | + | {{ 184_notes: |
- | $$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$ | + | |
- | Notice | + | That's it! We can even do one more check to be sure we are happy with our result. |
- | {{ 184_notes: | + | We know that capacitance is a constant, since it is an intrinsic property of the capacitor' |