Capacitor problem, easy but just can't seem to get it

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Homework Help Overview

The problem involves two capacitors, with capacitances of 0.40E-6F and 0.70E-6F, connected in series to a 12 V battery. Participants are tasked with calculating the potential difference and charge on each capacitor.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between total voltage, total capacitance, and charge in series configurations. There is mention of using the equation V = Q/C to find the potential difference across each capacitor. Some participants express confusion about the initial steps needed to approach the problem.

Discussion Status

There is ongoing dialogue with some participants providing guidance on how to set up the equations for the problem. However, confusion remains for some, indicating that further clarification may be needed. Multiple interpretations of the problem setup are being explored.

Contextual Notes

Some participants question the assumptions regarding the equations for capacitors in series and whether they are applying them correctly. There is a sense of uncertainty about the initial steps to take in solving the problem.

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A 0.40E-6F and 0.70E-6F capacitor are connected in series to a 12 V battery. Calculate the potential difference and charge on each capacitor.

I know this problem is really easy, but I just can't seem to get it started. How do you find the potential difference. Can someone help me out?
 
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supermenscher said:
A 0.40E-6F and 0.70E-6F capacitor are connected in series to a 12 V battery. Calculate the potential difference and charge on each capacitor.

I know this problem is really easy, but I just can't seem to get it started. How do you find the potential difference. Can someone help me out?
V = Q/C. You know the total voltage and total capacitance (I believe capacitors in series add up like resistors in parallel), and so you can calculate the total charge. Since they are in series, I would think that the charge on both capacitors would be the same, so you've got the charge values you need. Now, apply the V = Q/C equation again on each capacitor individually to get the voltage across each one, and you're done.
 
Thank you very much for your explanation, but it is still confusing me...do you think you could show me the first couple steps of what you mean to get me started...that would really help me out.
 
supermenscher said:
A 0.40E-6F and 0.70E-6F capacitor are connected in series to a 12 V battery. Calculate the potential difference and charge on each capacitor.

I know this problem is really easy, but I just can't seem to get it started. How do you find the potential difference. Can someone help me out?
Let [itex]C_1[/itex] and [itex]C_2[/itex] represent the capacitances of the capacitors. Let [itex]C_t[/itex] represent the capacitance of both capacitors in series.

[tex]C_t = (\frac{1}{C_1} + \frac{1}{C_2})^{-1}[/tex]

[tex]V_t = \frac{Q_t}{C_t}[/tex]

[tex]Q_t = V_tC_t[/tex]

[tex]Q_1 = Q_2 = Q_t[/tex]

[tex]V_1 = \frac{Q_1}{C_1}[/tex]

[tex]V_2 = \frac{Q_2}{C_2}[/tex]

The above is exactly what I said before, but in equation form. As long as I'm not mistaken, lines 1 and 4 are correct, and so you'll get the right answers. If I am mistaken, and one of them are wrong, you'll probably want to find the right equations somewhere for capacitance and charge in series.
 

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