A battery, two capacitors, and a switch

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Homework Help Overview

The discussion revolves around a circuit involving a battery, two capacitors (C1 and C2), and a switch. The original poster is trying to understand the behavior of the capacitors when the switch is flipped from position A (connected to the battery) to position B (disconnected from the battery).

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to determine the charge on the capacitors after the switch is flipped, questioning whether C1 discharges completely or until the charge on C2 equals that of C1. Some participants discuss conservation of charge and how it applies to the system when the switch is in position B. Others raise questions about the configuration of the capacitors, specifically whether they are in parallel or series after the switch is flipped.

Discussion Status

Participants are exploring the implications of charge conservation and the configuration of the capacitors. Some have provided insights into the behavior of the system, while others are seeking clarification on specific aspects of the circuit arrangement.

Contextual Notes

The original poster mentions an attachment that presumably contains a diagram of the circuit, which may be relevant for understanding the configuration of the capacitors. There is also a reference to the initial conditions of the capacitors, particularly the assumption that C2 is initially uncharged.

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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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