Battery provides a potential difference problem

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Homework Help Overview

The problem involves a capacitor with a capacitance of 27 µF connected to a battery providing a potential difference of 118 V. The original poster seeks to determine the amount of charge that will pass through the capacitor after a switch is closed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between capacitance, charge, and voltage, referencing the equation C = Q/V. There is an attempt to manipulate this equation to find the charge Q. One participant also raises the concept of charge conservation in the context of the capacitor and battery.

Discussion Status

The discussion is active, with participants engaging in mathematical reasoning and clarifying concepts related to capacitors. Some guidance has been provided regarding the relationship between charge and voltage, and there is an acknowledgment of the need to consider additional factors such as current flow and resistance, although not all participants are familiar with these concepts.

Contextual Notes

One participant notes a lack of access to the image referenced in the original post, which may limit the discussion. Additionally, there is mention of the time decay of capacitors, indicating that some participants may not have covered this topic yet.

secret5437
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Homework Statement



The capacitor in Fig. 25-25 has a capacitance of 27 µF and is initially uncharged. The battery provides a potential difference of 118 V. After switch S is closed, how much charge will pass through it? (picture is in the attachment)


Homework Equations



C = Q/V
U = Q2/2e = QV/2 = CV2/2


The Attempt at a Solution



I think this is really easy, but for some reason, i just can't get it.
 

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Can't see the image.
But let's start with how much charge IS in the capacitor?

Write out the equation, and tell me. I will be online for another 10 minutes.
We can solve this in 5 minutes.
 


i'm trying to figure out how much charge runs through the capacitor.
it has a capacitance of 27 microfarads. and the battery provides a potential difference of 118 V.

I'm pretty sure the equation is C = Q/V
which would be simplified to C x V = Q
so that would be 27 x 118...wouldnt it?
 


good job. Don't forget your units.

Now remember that the charge entering the one side of the capacitor has to equal that returning to the battery from the other side.

Have you learned about time decay of capacitors yet?
 


nope.
 


Normally what you would do is integrate the current flowing from the battery to the capacitor. This means taking into account the resistance of the wires and anything in between the capacitor and battery.

But if you haven't had this, LET CHEAT! We really don't need to know all the minutia.

The cap will hold a certain level of charge. Now THINK. If you have two plates of a capacitor, one having negative charges pumped into it. The other plate will have negative charges pulled away from it.

Potential difference is just the voltage difference between two things. It does not matter what the gauge offset is.

SO if the switch IS one leg. That means 1/2 the charges flow through it.

See ya.
 


got it! thanks.
 

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