GS CBT 9 PS section #40

[yjj8817]

40) A parallel plate capacitor is made of two thin plates of area A separated by a distance d has capacitance C1. If a third identical plate is inserted midway between the two others (see Figure 1), what is the new capacitance C2?

1. C2 = C1
2. C2 > C1
3. C2 < C1
4. C2 = 0

INCORRECT:

Correct Answer: A

So they are saying that since inserting another plate in the middle reduces the distance to d/2 and since the plates are connected in series, the equation comes out to 1/C2 = 1/(2*C1) + 1/(2*C1).

This might be a silly question, but shouldn't capacitor plates be connected to the wires to act as capacitor plates and store charges? Can someone clarify this? Thank you!

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[sazerac]

The third plate is not infinitely thin. It can have a positive charge on the top and a corresponding negative charge on the bottom of the same plate.

Or, think of the top third of the new plate as Plate1, the bottom third of the plate as Plate2, and the middle third of the plate as a really really fat wire connecting them, if that helps.

 

[yjj8817]

The third plate is not infinitely thin. It can have a positive charge on the top and a corresponding negative charge on the bottom of the same plate.

Or, think of the top third of the new plate as Plate1, the bottom third of the plate as Plate2, and the middle third of the plate as a really really fat wire connecting them, if that helps.

Thanks! But if the positive charges are on top and negative charges are on bottom, then doesn't that makes 2 positive sides facing each other and same for the 2 negative sides?

 

[sazerac]

Thanks! But if the positive charges are on top and negative charges are on bottom, then doesn't that makes 2 positive sides facing each other and same for the 2 negative sides?

It's just an example. The positive could be in the bottom, or wherever positive charges feel compelled to go.

 

[yjj8817]

It's just an example. The positive could be in the bottom, or wherever positive charges feel compelled to go.

Ok thanks so in conclusion since charges can be on the middle plate, the middle plate can act as a capacitor plate and store charge? So you treat it just like every other capacitor plates.

Is this correct way of thinking?

 

[sazerac]

Yup

 

[yjj8817]

For this question, shouldn't there be a potential difference between the middle plate and the charged top
plate for the charge to move? Like + on the charged plate and - on middle plate.

Since electric field doesn't necessarily need two opposite charges to exist, I guess the middle plate can still store charge?

 

[sazerac]

There is a minus charge on the top portion of the middle plate. There is a corresponding plus charge on the bottom of the middle plate. The charge of the middle plate as a whole will be zero.

 

[yjj8817]

There is a minus charge on the top portion of the middle plate. There is a corresponding plus charge on the bottom of the middle plate. The charge of the middle plate as a whole will be zero.

How do you know this?

 

[sazerac]

How do you know this?

How do I know what?

How do I know the inserted plate has a net charge of zero? I think it's a logical assumption since nobody said otherwise.

How to I know the charges will migrate within the plate? Because of the EMF supplied by the power source, and therefore the charges on the original plates. All the positive charge will migrate to the bottom, attracted to the negative charge on the plate below.

I think you may be over-thinking the whole problem. Think of the device inserted as a plate, connected to a wire, connected to another plate. Kind of like a letter H turned sideways. Now make the wire shorter. Now make the wire infinitesimally short. Solder those two plates together. That is essentially the object that has been inserted into the existing capacitor.

How does this compare to the original system? Well, according to the equations, it splits the original capacitor into two mini-capacitors of double strength, and when they are connected in series they re-create the original capacitor. Neato. Done and done.