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184_notes:examples:week7_charging_capacitor [2017/10/06 18:34] – [Solution] tallpaul | 184_notes:examples:week7_charging_capacitor [2017/10/09 15:37] – dmcpadden | ||
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=====Looking at a Capacitor as it's Charging===== | =====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. | + | 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. |
===Facts=== | ===Facts=== | ||
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===Approximations & Assumptions=== | ===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 $t=0$. | * 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" | * Practically speaking, the capacitor becomes "fully charged" | ||
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====Solution==== | ====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: |
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- | Slightly after time $t=0$, | + | Slightly after time $t=0$, |
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- | 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. | + | Lastly, we wish to produce a graph of the potential difference between the plates. We can pull from a [[184_projects: |
$$E_{plate} = \frac{Q_{plate}}{2\epsilon_0 A_{plate}}$$ | $$E_{plate} = \frac{Q_{plate}}{2\epsilon_0 A_{plate}}$$ | ||
- | 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. | + | 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 |
- | 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: | + | 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: | {{ 184_notes: |