# Capacitance W/ Dielectric Problem

1. Mar 7, 2006

### minhngo

A parallel-plate capacitor has rectangular plates of length L = 19 cm and width W = 3 cm . The region between the plates is filled with a dielectric slab of dielectric constant K= 4 which can slide along the length of the capacitor. Initially, the slab completely fills the rectangular region, and the capacitor holds a charge of 0.1 µC. How far should the dielectric slab be pulled so that the stored energy is double its initial value?

http://www.webassign.net/tipler4/25-33.gif

Ok. I know there are two capacitance that are parallel with each other. One with a dielectric (1) with length L-X and the other without (2) with length X.

Q1+Q2=Qo
V1=V2=Vo

The doubling of stored energy confuses me, and I really dont know where to begin.

Last edited: Mar 7, 2006
2. Mar 7, 2006

### Staff: Mentor

Stored energy in a capacitor is given by

1/2 CV2, where V = voltage, and C = capacitance. So doubling the capacitance, C, one could double the stored energy at the same voltage.

3. Mar 7, 2006

### lightgrav

Energy stored in a capacitor is (1/2) Q_final V_final .

V is MORE with no dielectric between the plates,
because the di-electric molecules polarize and orient themselbves
so that the E-field with a dielectric is E_dielectric = E_original / K .
The thing that stays the same is the plysical distance between the plates.
(Recall dV = E dot ds)

4. Mar 7, 2006

### minhngo

How can one capacitance have a higher voltage than the other one? Shouldnt potential across a parallel circuit be the same? Sorry for any confusion.

5. Mar 7, 2006

### lightgrav

You're right, the two pieces have the same voltage (they're one conductor)
But that Voltage is NOT the same as it used to be ... the puller does Work!
The charges on the plates are denser where the dielectric is,
and sparser near the open space. The Voltage increases as the slab is pulled.

The key point is that the Voltage source has been disconnected.
The Areas add to the total Area, the distance is the same, Q's add.

Last edited: Mar 7, 2006
6. Mar 7, 2006

### minhngo

Ah I see. so if the dielectric were to be completely removed, the potential would be much greater b/c there is no dielectric to lessen the electric field between the plates. This would explain why there is more potential energy afterwards. Ok thanks.

EDIT: Ui=(Q^2)/(2KCo)

Cf=C1+C2
Cf=KCo(1-R)+Co(R)

R=X/L (ratio for capacitance)

Uf=Q^2/Cf

Uf=2Ui

I believe this is right. Can someone verify for me?

Last edited: Mar 7, 2006
7. Mar 7, 2006

### lightgrav

Removing the dielectric is essentially the same as pulling the plates 4x farther apart, "unzipping" it.