Exploring the Effects of Dielectric Insertion on Capacitor Behavior

[Korisnik]

The capacitor and dielectric

Homework Statement

A small dielectric plate is pushed between parallel plates of an (1) isolated capacitor to the half of its length. If the dielectric is let to move freely, what's going to happen to it? (2) What would happen with the dielectric if the capacitor was connected to the voltage source?

Homework Equations

$E=\frac{1}{2} CU^2=\frac{Q^2}{2C}~~(1)$
where $U$ is voltage, $C$ capacity, $Q$ charge on plates, $E$ energy of the electric field of the capacitor.

$C=ε_0ε_r\frac{S}{d}~~(2)$
where $ε$ is permitivity, $S$ area of one plate, $d$ distance between the plates, $C$ capacity of the capacitor.

The Attempt at a Solution

When the dielectric enters the capacitor, the capacity, by the (2) equation increases, therefore by the (1) equation the energy of the part of the capacitor where the dielectric is - decreases, because $Q$ is constant (isolated capacitor), and $C$ is reverse-proportional to the energy $E$. Because the energy of the first part (where the dielectric is) decreased and is lower than on the other part of the capacitor, the dielectric tends to go to the lower energy and is therefore pushed out of the capacitor.
For the second question the same rules are applied but this time the reverse thing happens because the energy increases ... the dielectric get's pushed in.

Is any of this correct?

 

[rude man]

'Fraid not.

For the isolated capacitor, you correctly state that the stored energy after dielectric insertion is lower. So what happened to the "lost" energy?

For the capacitor with constant potential V, the energy is 1/2 C V^2 so what happened to C after insertion (see your equation 2), what is the new stored energy and what had to be done to get it?

 

[Korisnik]

'Fraid not.

For the isolated capacitor, you correctly state that the stored energy after dielectric insertion is lower. So what happened to the "lost" energy?

For the capacitor with constant potential V, the energy is 1/2 C V^2 so what happened to C after insertion (see your equation 2), what is the new stored energy and what had to be done to get it?

Lost energy was spent in the voltage decrease, so maybe... the distance between the plates [STRIKE]increased[/STRIKE] decreased, but that's impossible because they probably cannot be [STRIKE]away[/STRIKE] closer. So the charges probably neutralized, or something... I don't know.

Where there's constant voltage, and capacity of the plates increased, the energy increased. That was done through bringing of the new charge from the voltage source...

EDIT: Or maybe I was on the wrong track; maybe THE SYSTEM wants to decrease ITS energy, therefore it pulls in the dielectric in the first case, and pushes out in the second?

 

[rude man]

Part (a): The total energy of the system remains unchanged. Conservation of energy. The energy stored in the field is reduced, so some other form of energy must be produced. Think about what form that new energy consists of. Hint: think of letting go of the dielectric when it's just beginning to enter the plate area. What happens? What's the situation when the dielectric is centered inbetween the plates?

Part (b): Your statement is correct, but you haven't described what happens as the dielectric is inserted.This is a more difficult problem since there are two sources of external energy: the force on the dielectric times the distance traveled, plus the energy supplied by the battery. You need to isolate the two since you're interested in the former energy source only.

 

[Korisnik]

Part (a): The total energy of the system remains unchanged. Conservation of energy. The energy stored in the field is reduced, so some other form of energy must be produced. Think about what form that new energy consists of. Hint: think of letting go of the dielectric when it's just beginning to enter the plate area. What happens? What's the situation when the dielectric is centered inbetween the plates?

Part (b): Your statement is correct, but you haven't described what happens as the dielectric is inserted.This is a more difficult problem since there are two sources of external energy: the force on the dielectric times the distance traveled, plus the energy supplied by the battery. You need to isolate the two since you're interested in the former energy source only.

I think that when I'm putting the dielectric into the condenser it pulls it in, because the molecules in the dielectric polarise, in both 1st and 2nd problem. It's definitely easier to push it in in the case where the voltage is constant as more charge comes to the plates (faster) but it gets harder after because it's polarised already. I've never held a condenser or have put in a dielectric... I don't know. Nowhere in my book is that process described, the question is as I've written it.

The only thing I can think of to decrease the voltage in the first problem is that an electron jumps over to the other plate. (Because if for example the energy of the first part of the plate is decreased, the energy on the other half is increased.)

 

[rude man]

You need to use Gaussian surfaces and get a quantitative feel for what's going on rather than engage in qualitative speculation.

Electrons are not jumping anywhere.

Contemplate the possibility that it takes positive force to insert the dielectric in one case and negative in the case of another.