# Homework Help: What is the capacitance

1. Jul 21, 2017

1. The problem statement, all variables and given/known data
The plates of an air filled parallel plate capacitor with a plate area of .0016 m^2 and a separation of .009 m are charged to a 145 V potential difference. After the plates are disconnected from the source, a porcelain dialectric with k = 6.5 is inserted between the plates of the capacitor

a) what is the charge on the capacitor before and after the dialectic is inserted?
b) what is the capacitance of the capacitor after the dialectic has been inserted?
c) what is the potential difference between the plates of the capacitor after the dialectric is inserted?
d) what is the magnitude of the change in the energy stored in the capacitor after the dialectric is inserted?

2. Relevant equations

q = vc
Ue = (1/2) CV^2

c = epsilon 0 * A / d

3. The attempt at a solution
so for A)

I have q = vc ---> q = (145) ((epsilon 0 * A )/ d )

plugging in, q = (145) (8.55x10^-12) (.0016) / .009 = 2.204 x 10^-10

my books answer was 2.28 x 10^-10

They didn't even answer the second part of question A, which was what is that charge after. Since they only listed "2.28 x 10^-10 " as the answer to A, I'm going to guess that is the before AND after charge? Is this because if you disconnect from source, inserting the dialectric isnt going to change anything? Why was my answer off from the books, did I even do it right? I mean it was just plugging in the equation but still, it seems wrong to me..

Anyways I thought my formula would now be q = kvc with k being 6.5 but I guess not?

2. Jul 21, 2017

### TSny

Looks like a slip in entering the value of ε0.

It changes some things. But, you are right. Once the capacitor is disconnected from the source, the charge on each plate is "trapped".

That formula q = kvc is correct if c is the capacitance without the dielectric. But after the dielectric is inserted, is the potential difference still 145 V?

3. Jul 21, 2017

### scottdave

If the dielectric has neutral charge, before inserting, then it will not change the amount of charge in the capacitor. The dielectric does change the capacitance, so what happens to the voltage?