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

1. Jun 19, 2009

### mba444

1. The problem statement, all variables and given/known data
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.

What is the dielectric constant of the slab?

2. Relevant equations

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

3. 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

Last edited by a moderator: May 4, 2017
2. Jun 19, 2009

### cepheid

Staff Emeritus
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?

3. Jun 20, 2009

### mba444

i didnt undcerstand

4. Jun 20, 2009

### drizzle

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

5. Jun 20, 2009

### mba444

Ok

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

6. Jun 20, 2009

### drizzle

no, I don't think V is the same, there is Vo and V, where V is less than Vo [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]