Calculating Charge with Capacitors in Series

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Homework Help Overview

The discussion revolves around calculating the charge on a capacitor in a series circuit, specifically focusing on the charge on the right plate of capacitor C2 after a switch has been closed. Participants are exploring the implications of the equivalent capacitance in the context of the given options for charge values.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants attempt to calculate the charge using the formula Q = CeqV and discuss the interpretation of the resulting expression. Questions arise regarding the correct representation of the charge formula and the role of equivalent capacitance in the calculations.

Discussion Status

Some participants have provided insights into the calculation of equivalent capacitance and have pointed out misunderstandings in the application of formulas. There is an ongoing exploration of the correct approach to determining the charge, with multiple interpretations being discussed.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for their calculations. There is a focus on understanding the relationships between capacitance, charge, and voltage in series configurations.

PeachBanana
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Homework Statement



What is the charge on the right
plate of C2 long after the switch is
closed?

1. CV
2. CV/2
3. CV/3
4. 2CV/3
5. 0

Homework Equations



Q = CeqV

The Attempt at a Solution



Q = 1/3c * 2V
so I thought
Q = 2V/3C
 

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PeachBanana said:

Homework Statement



What is the charge on the right
plate of C2 long after the switch is
closed?

1. CV
2. CV/2
3. CV/3
4. 2CV/3
5. 0

Homework Equations



Q = CeqV

The Attempt at a Solution



Q = 1/3c * 2V
so I thought
Q = 2V/3C
Is that (2V)/(3C) or (2V/3)C ?
 
I wanted (2V/3)C but written that way it looks like

2CV/3 which is one of the answers but I'm trying to understand why "C" is in the numerator. Is it because Ceq = 1 / C?
 
PeachBanana said:
I wanted (2V/3)C but written that way it looks like

2CV/3 which is one of the answers but I'm trying to understand why "C" is in the numerator. Is it because Ceq = 1 / C?

What's the net (equivalent) capacitance of the two given capacitors?
 
Ceq = 1/C + 1/2C = 3/2C
 
PeachBanana said:
Ceq = 1/C + 1/2C = 3/2C

Not quite. Two capacitors Ca and Cb in series yield:

$$C_{eq} = \frac{1}{\frac{1}{Ca} + \frac{1}{Cb}}$$

You've forgotten to take the overall reciprocal.
 
I see what I did incorrectly now. My final answer should have been 2C/3.
 

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