Action of a dielectric in a capacitor

[mahela007]

Homework Statement

A prallel plate capacitor placed in air has a plate area A and a plate separation d. It is cahred to charge Q by connecting a constant foltage source across the plates. The voltage source is then disconnected and a slab of dielectric constant K and thickness h is inserted between the plates as shown in the figure.

write down expressions for the electric field intensity
(a) in the gap between the upper plate and the dielectric slag.
(b) in the dielectric slab
(c)in the gap between the dielectric and the lower plate.

Homework Equations

The capacitance of a parallel plate capacitor is C = AE₀/D
Equations of Gauss's theorem.

The Attempt at a Solution

I assumed that the charge on one surface of the dielectric is equal and opposite to the charge on the plate closest to that surface of the dielectric. Then Gauss's theorem can be applied to the charges on metal plate and those on the surface of the dielectric. The field between the two can be found by vectorial addition of the electric fields produced by the charges on each object. (the individual fields can be calculated using Gauss's law)

My first assumption may be incorrect (about charges on the dielectric and the plates being equal and opposite).. but that's the only way I can see to solve this problem.

Unfortunately, I don't get the right answer...
Thanks.

 

[tiny-tim]

hi mahela007!

My first assumption may be incorrect (about charges on the dielectric and the plates being equal and opposite)..

you're right! …

it is incorrect …​

that only works for conductors ​

hint: the electric displacement field D depends only on the charge on the capacitor …

inserting a dielectric (which, unlike a conductor, has no free charge ) doesn't change D …

work out the electric intensity E from that ​

 

[eczeno]

in a dielectric, the charges are not as free to move as in a conductor, so your assumption does not hold. however, the charges can move a little bit and so there is a small charge build up on the surface of the dielectric (mostly from either 'turning' of polar molecules or from slight 'rearrangements' of the electron configurations of non-polar molecules). this means the electric field inside the dielectric will be lowered (by a factor K), but not to zero as in a conductor. the field outside the dielectric will not be affected.hope this helps

 

[mahela007]

hi mahela007!


you're right! …

it is incorrect …​

that only works for conductors ​

hint: the electric displacement field D depends only on the charge on the capacitor …

inserting a dielectric (which, unlike a conductor, has no free charge ) doesn't change D …

work out the electric intensity E from that ​

Sorry.. is there is typo in the 4th line? Are you referring to the electric field as D or the separation of the plates as D?

 

[tiny-tim]

i'm referring to the electric displacement field …

D = εE = ε₀E + P​

D depends only on free charge (which doesn't exist in a dielectric)

P (the polarisation field or dipole moment field) depends on the dipole moments (bound charge) in a dielectric

so D is the same whether there's a dielectric present or not: if there is, then the P field acts in opposition to the D field, and makes the total electric field E smaller