Battery provides a potential difference problem

[secret5437]

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.

 

[Mr.Amin]

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.

 

[secret5437]

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?

 

[Mr.Amin]

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?

 

[secret5437]

nope.

 

[Mr.Amin]

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.

 

[secret5437]

got it! thanks.