A battery, two capacitors, and a switch

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SUMMARY

The discussion revolves around the behavior of two capacitors, C1 and C2, when a switch is flipped from position A to position B in a circuit powered by a 200V battery. Initially, capacitor C1 charges to 200V, storing a charge calculated by the formula Q=CV. Upon switching to position B, the charge is conserved, and C1 and C2 share the total charge, resulting in a voltage equalization across both capacitors. The participants clarify that C1 and C2 are in parallel after the switch is flipped, despite initial confusion regarding their configuration.

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  • Familiarity with the concept of charge conservation in electrical circuits
  • Knowledge of circuit configurations, specifically parallel and series arrangements
  • Proficiency in using the formula Q=CV for capacitors
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Homework Statement



If the switch is flipped from position A to position B, what's the charge on capacitor C1 and C2?

Please see the attachment below

clip_image002.jpg


Homework Equations



Q=CV


The Attempt at a Solution



I know that the capacitor C1 gains potential difference of 200V (= potential difference of a battery) and has charge Q = C1*200, but I'm not sure what happens once the flip is switched. My assumption is that it must discharge because it is no longer connected to the battery, but does it discharges until Q = 0 or until the charge on C2 = charge on C1. Could somebody please explain to me what exactly happens once the switch is in position B
Thank you!
 
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Charge moves down the potential hill. (conventional)
 
from conservation of charge: amount of charge stored when in position A must be same as in position B, the only different is that now you have two capacitors to share the storage.. now, assuming that the second capacitor is initially uncharged, then the total charge in the two-capacitor system must be the same as before... due to how the circuit is set up the voltage across the two capacitor must also equal...
 
mjsd said:
from conservation of charge: amount of charge stored when in position A must be same as in position B, the only different is that now you have two capacitors to share the storage.. now, assuming that the second capacitor is initially uncharged, then the total charge in the two-capacitor system must be the same as before... due to how the circuit is set up the voltage across the two capacitor must also equal...

I got it. Thank you very much. I knew about the voltage, but I forgot to consider charge conservation. Thanks!
 
I'm trying to solve a similar problem and I understand everything except the following part:

Why are C1 & C2 in parallel after flipping the switch to B?

Doesn't the circuit just look like this after the switch goes to B (ignoring the part of the circuit where they connect to the battery because that is not functional now):

Untitled-1-4.jpg


So aren't they really in serial?

Thanks in advance. :)
 
Last edited:

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