How Does Inserting a Dielectric Change Capacitor Voltage?

[arod2812]

Homework Statement

An empty capacitor is connected to a 11.7-V battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material (κ = 3.1) is inserted between the plates.
Find the magnitude of the amount by which the potential difference across the plates changes.

Homework Equations

I did the following steps but did not get the right answer. Is there another step to the problem?

The Attempt at a Solution

q=CV, therefore C=q/V
C is affected by the k... therefore C(dielectric)= k*C
then the equation reads: q/V = k (q/11.7)... and the Q's cancel out.

The final equation I worked with is (1/V)=3.1 (1/11.7). The answer I got however is not correct. help.

 

[berkeman]

Looks like you are on the right track. Since Q is constant (no place for the charge to go), CV must be constant. Since C goes up by 3.1x, V must do the inverse. What answer exactly did you try?

 

[arod2812]

the answer i tried was 3.77... originally i had tried 36.27 which was wrong. But it turns out that both are wrong. What do you think?

 

[berkeman]

Well, that's pretty much what I get for 11.7/3.1. And unless I'm missing something, that seems like the right way to get it -- even talks about using the dielectric insertion method for measuring k here:

http://en.wikipedia.org/wiki/High-k

How many sig figs does this homework program expect? Maybe they just want 3.8V?