What is the dielectric constant of the slab on a parallel plate capacitor ?

[mba444]

Homework Statement

When a certain air-filled parallel-plate ca-
pacitor is connected across a battery, it ac-
quires a charge (on each plate) of magnitude
214 μC. While the battery connection is
maintained, a dielectric slab is inserted into
the space between the capacitor plates and
completely fills this region. This results in
the accumulation of an additional charge of
272 μC on each plate.

http://images.upload2world.com/get-6-2009-upload2world_com_ohteyhd.jpg

What is the dielectric constant of the slab?

Homework Equations

i used
k= Uf/Ui
where
Uf=Q2^2/2C
Ui=Q1^2/2C

The Attempt at a Solution

what i did is that i k=[Q2^2/2C]/[Q1^2/2C] .. in this case the 2C will cancel each other .. we will be left out with k=[Q2^2/Q1^2] .. which as a result gave me 1.61551

i really don't know were I am going wroing .. so please help mee !

thanx in advance

 

[cepheid]

I would take a step by step approach.

1. By how much did the capacitance increase in the second case from the first case?

2. How does the capcitance depend upon the dielectric constant in the case of a parallel plate capacitor?

3. Therefore, what must the increase in dielectric constant of the gap have been?

 

[mba444]

i didnt undcerstand

 

[drizzle]

shouldn't the dielectric constant be like; k=C/C_o?
where C is the capacitance with the dielectric material in between the plats and C_o the same but in vaccum

 

[mba444]

Ok

so it will be k= Q/Qo.. were the volt cancel each other

 

[drizzle]

no, I don't think V is the same, there is V_o and V, where V is less than V_o [due to the inverse electric field induced by putting the dielectric material in the capacitor, which reduces the main E, thus the sum of E is less than before and as a result the applied voltage decreases as well]