How Does Inserting a Dielectric Affect Capacitor Voltage?

[rcrx]

Homework Statement

An empty capacitor is connected to a 12.0-V battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material (K = 2.8) is inserted between the plates. Find the amount by which the potential difference across the plates changes.

Homework Equations

C2 = KC1
q = CV

The Attempt at a Solution

I know that the charge will stay the same and voltage will drop.

I have no idea how to set this up mathematically. This is where I have hit a road block.

 

[Delphi51]

You are not give C1, so you must think of it as a "known" that may appear in the answer. From that and one of your formulas, you can calculate the charge.

Great observation to see that the charge remains the same with the dialectric is inserted!

Use the formula with the V in it to find the potential on the cap with dielectric.

 

[rcrx]

Thanks for the reply. But how do I set it up? I am so confused. I know it is simple, but some problems I just can't wrap my head around mathematically.

Any and all help is very much appreciated!

 

[Delphi51]

Well, to calculate the charge, I just meant to use your formula q = CV.
To avoid confusion, I would replace C with C1. Put in your potential and you've got an expression for q with no unknowns (except C1, which we are treating as a known for now).

The first formula gives the new C2.
Use the 2nd formula again to find the new V.